单循环赛程编排的数学模型

hanbi · 发表于 2004-7-24 07:47:43

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>摘 要: 用邻接矩阵的方式建立了单循环赛程编排的数学模型，给出了尽量公平原则的最优上限和下限，并以图论方法详细描述了<v:shapetype> <v:stroke joinstyle="miter"></v:stroke><v:formulas><v:f eqn="if lineDrawn pixelLineWidth 0 "></v:f><v:f eqn="sum @0 1 0 "></v:f><v:f eqn="sum 0 0 @1 "></v:f><v:f eqn="prod @2 1 2 "></v:f><v:f eqn="prod @3 21600 pixelWidth "></v:f><v:f eqn="prod @3 21600 pixelHeight "></v:f><v:f eqn="sum @0 0 1 "></v:f><v:f eqn="prod @6 1 2 "></v:f><v:f eqn="prod @7 21600 pixelWidth "></v:f><v:f eqn="sum @8 21600 0 "></v:f><v:f eqn="prod @7 21600 pixelHeight "></v:f><v:f eqn="sum @10 21600 0 "></v:f></v:formulas><v:path connecttype="rect" gradientshapeok="t" extrusionok="f"></v:path><lock aspectratio="t" v:ext="edit"></lock></v:shapetype><v:shape><v:imagedata src="./mathmodel1.files/image001.wmz" title=""></v:imagedata></v:shape>为偶数和<v:shape> <v:imagedata src="./mathmodel1.files/image001.wmz" title=""></v:imagedata></v:shape>为奇数情形均衡解编排的全部过程．作为特例，得到了<v:shape> <v:imagedata src="./mathmodel1.files/image004.wmz" title=""></v:imagedata></v:shape>情形的均衡解．
关键词: 赛程；邻接矩阵；均衡解
中图分类号: O157.6 文献标识码: A 文章编号: 1672-0318(2003)01- 
 
　　单循环赛程编排的数学建模是2002年9月全国大学生数学建模竞赛所提出的问题之一，无论是参赛论文还是竞赛委员会所提供的参考答案，都没有给出令人满意的解答和建设性的一般结果．作者以邻接矩阵的方式建立起这一问题的数学模型，得到了“尽量公平原则“的最优上限和下限，并借助图论方法给出了赛程均衡解直观、完整的编排过程．
 
1 问题的提出
 
在某些情况下，需要在同一场地上进行单循环比赛，尽量公平的赛程安排是充分调动参赛各队的积极性，提高比赛水平的前提．
例如，某年级有5个班，每班一支足球队在同一块场地上进行单循环比赛，共要进行10场比赛，下面是随便安排的一个赛程：
记5支球队为A，B，C，D，E，在表1左半栏的右上三角的10个空格中，随手填上1，2，…，10，得到一个赛程，即第1场A对B，第2场B对C，…，第10场C对E． 
 
表1 5支球队的一个赛程安排
<DIV align=center>
<TABLE cellSpacing=0 cellPadding=0 border=0>

<TR>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=25>
A</TD>
<TD vAlign=top width=25>
B</TD>
<TD vAlign=top width=28>
C</TD>
<TD vAlign=top width=25>
D</TD>
<TD vAlign=top width=28>
E</TD>
<TD vAlign=top width=168>
每两场比赛间隔的场次数</TD></TR>
<TR>
<TD vAlign=top width=25>
<H3>A</H3></TD>
<TD vAlign=top width=25>
X</TD>
<TD vAlign=top width=25>
1</TD>
<TD vAlign=top width=28>
9</TD>
<TD vAlign=top width=25>
3</TD>
<TD vAlign=top width=28>
6</TD>
<TD vAlign=top width=168>
1, 2, 2</TD></TR>
<TR>
<TD vAlign=top width=25>
B</TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=25>
X</TD>
<TD vAlign=top width=28>
2</TD>
<TD vAlign=top width=25>
5</TD>
<TD vAlign=top width=28>
8</TD>
<TD vAlign=top width=168>
0, 2, 2</TD></TR>
<TR>
<TD vAlign=top width=25>
C</TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=28>
X</TD>
<TD vAlign=top width=25>
7</TD>
<TD vAlign=top width=28>
10</TD>
<TD vAlign=top width=168>
4, 1, 0</TD></TR>
<TR>
<TD vAlign=top width=25>
D</TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=28>
 </TD>
<TD vAlign=top width=25>
X</TD>
<TD vAlign=top width=28>
4</TD>
<TD vAlign=top width=168>
0, 0, 1</TD></TR>
<TR>
<TD vAlign=top width=25>
E</TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=28>
 </TD>
<TD vAlign=top width=25>
 </TD>
<TD vAlign=top width=28>
X</TD>
<TD vAlign=top width=168>
1, 1, 1</TD></TR></TABLE></DIV>
 
这个赛程的公平性如何呢？从表1右半栏各队每2场比赛之间相隔的场次数不难看出，这个赛程对A，E有利，对D则不公平．若<v:shape> <v:imagedata src="./mathmodel1.files/image006.wmz" title=""></v:imagedata></v:shape>个队实力相当，且在同一场地上进行单循环比赛，建立尽量公平赛程编排的数学模型就尤为必要．
 
2 模型的建立
设<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>个队为<v:shape> <v:imagedata src="./mathmodel1.files/image010.wmz" title=""></v:imagedata></v:shape>，…，<v:shape> <v:imagedata src="./mathmodel1.files/image012.wmz" title=""></v:imagedata></v:shape>，在同一场地上进行单循环比赛的场数共有：
<v:line></v:line><v:shape><v:imagedata src="./mathmodel1.files/image015.wmz" title=""></v:imagedata></v:shape>． （1）
我们将赛程转化成邻接矩阵：
1）以参赛队表示矩阵的行，比赛场次表示矩阵的列；
2）在某场所对应的列中，参加比赛的队取1值，没参加比赛的队取0值．
例如，表1中所给赛程的邻接矩阵为：
 <v:shape><v:imagedata src="./mathmodel1.files/image017.wmz" title=""></v:imagedata></v:shape>
<v:shape><v:imagedata src="./mathmodel1.files/image019.wmz" title=""></v:imagedata></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image021.wmz" title=""></v:imagedata></v:shape> （2）
其中每行中的“0”的个数为该队休息的场次数．一般赛程可转化如下：
设<v:shape> <v:imagedata src="./mathmodel1.files/image023.wmz" title=""></v:imagedata></v:shape>为第<v:shape> <v:imagedata src="./mathmodel1.files/image025.wmz" title=""></v:imagedata></v:shape>行第<v:shape> <v:imagedata src="./mathmodel1.files/image027.wmz" title=""></v:imagedata></v:shape>列的元素为1、其他元素均为0的<v:shape> <v:imagedata src="./mathmodel1.files/image029.wmz" title=""></v:imagedata></v:shape>矩阵，那么参赛队<v:shape> <v:imagedata src="./mathmodel1.files/image031.wmz" title=""></v:imagedata></v:shape>的赛程安排可以表示为：
<v:shape><v:imagedata src="./mathmodel1.files/image033.wmz" title=""></v:imagedata></v:shape> （3）
整个赛程可表示为邻接矩阵：<v:shape> <v:imagedata src="./mathmodel1.files/image035.wmz" title=""></v:imagedata></v:shape>． 　 　　（4）
其中<v:shape> <v:imagedata src="./mathmodel1.files/image037.wmz" title=""></v:imagedata></v:shape>满足：
1）<v:shape> <v:imagedata src="./mathmodel1.files/image039.wmz" title=""></v:imagedata></v:shape>（<v:shape> <v:imagedata src="./mathmodel1.files/image041.wmz" title=""></v:imagedata></v:shape>,<v:shape> <v:imagedata src="./mathmodel1.files/image043.wmz" title=""></v:imagedata></v:shape>）；
2）<v:shape> <v:imagedata src="./mathmodel1.files/image045.wmz" title=""></v:imagedata></v:shape>= <v:shape><v:imagedata src="./mathmodel1.files/image047.wmz" title=""></v:imagedata></v:shape>（即<v:shape> <v:imagedata src="./mathmodel1.files/image049.wmz" title=""></v:imagedata></v:shape>的行和为每个队参加比赛的场次数<v:shape> <v:imagedata src="./mathmodel1.files/image051.wmz" title=""></v:imagedata></v:shape>）；
3）<v:shape> <v:imagedata src="./mathmodel1.files/image053.wmz" title=""></v:imagedata></v:shape>（即<v:shape> <v:imagedata src="./mathmodel1.files/image049.wmz" title=""></v:imagedata></v:shape>的列和为每场参加比赛的队的个数）．
 
 
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hanbi · 发表于 2004-7-24 07:48:20

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> <

>3 尽量公平原则的定量表示<

center" align=center> 为了实现尽量公平原则，首先每个队任2场比赛间间隔的场次数应大于等于1；其次对每个参赛队来说，他们的最优期望是每2场比赛之间的间隔场次数要尽可能地大和尽可能地平均分布．显然，在同一场地上进行单循环比赛的<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>个队中，每个队均要比赛<v:shape> <v:imagedata src="./mathmodel1.files/image057.wmz" title=""></v:imagedata></v:shape>场，共有<v:shape> <v:imagedata src="./mathmodel1.files/image059.wmz" title=""></v:imagedata></v:shape>次间隔．设参赛队<v:shape> <v:imagedata src="./mathmodel1.files/image031.wmz" title=""></v:imagedata></v:shape>每两场比赛之间间隔的场次数分别为<v:shape> <v:imagedata src="./mathmodel1.files/image062.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image064.wmz" title=""></v:imagedata></v:shape>，…，<v:shape> <v:imagedata src="./mathmodel1.files/image066.wmz" title=""></v:imagedata></v:shape>，则根据式(1)可知：<v:shape><v:imagedata src="./mathmodel1.files/image068.wmz" title=""></v:imagedata></v:shape>． （5）令<v:shape> <v:imagedata src="./mathmodel1.files/image070.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image072.wmz" title=""></v:imagedata></v:shape>，则有<v:shape><v:imagedata src="./mathmodel1.files/image074.wmz" title=""></v:imagedata></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image076.wmz" title=""></v:imagedata></v:shape>即 <v:shape><v:imagedata src="./mathmodel1.files/image078.wmz" title=""></v:imagedata></v:shape>　　　　 　　　　　 　　　（6）令<v:shape> <v:imagedata src="./mathmodel1.files/image080.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image082.wmz" title=""></v:imagedata></v:shape>，即<v:shape> <v:imagedata src="./mathmodel1.files/image084.wmz" title=""></v:imagedata></v:shape>和<v:shape> <v:imagedata src="./mathmodel1.files/image086.wmz" title=""></v:imagedata></v:shape>分别为整个赛程中所有队间隔场次数的最大值和最小值，有：定理1 　<v:shape> <v:imagedata src="./mathmodel1.files/image088.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image090.wmz" title=""></v:imagedata></v:shape>．证明 先证<v:shape> <v:imagedata src="./mathmodel1.files/image088.wmz" title=""></v:imagedata></v:shape>．由（6）式可知：<v:shape> <v:imagedata src="./mathmodel1.files/image093.wmz" title=""></v:imagedata></v:shape>，我们只需证明<v:shape> <v:imagedata src="./mathmodel1.files/image095.wmz" title=""></v:imagedata></v:shape>．采用反证法，若<v:shape> <v:imagedata src="./mathmodel1.files/image097.wmz" title=""></v:imagedata></v:shape>，则1）当<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>为奇数时，有<v:shape> <v:imagedata src="./mathmodel1.files/image100.wmz" title=""></v:imagedata></v:shape>，注意到<v:shape><v:imagedata src="./mathmodel1.files/image102.wmz" title=""></v:imagedata></v:shape>，<v:shape><v:imagedata src="./mathmodel1.files/image104.wmz" title=""></v:imagedata></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image076.wmz" title=""></v:imagedata></v:shape>，且<v:shape> <v:imagedata src="./mathmodel1.files/image107.wmz" title=""></v:imagedata></v:shape>（<v:shape> <v:imagedata src="./mathmodel1.files/image109.wmz" title=""></v:imagedata></v:shape>），于是我们有<v:shape><v:imagedata src="./mathmodel1.files/image111.wmz" title=""></v:imagedata></v:shape>（<v:shape> <v:imagedata src="./mathmodel1.files/image113.wmz" title=""></v:imagedata></v:shape>），由此可知，所有参赛队都必须在同一场次同时参赛，这与比赛的单循环规则相矛盾．2）当<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>为偶数时，有<v:shape> <v:imagedata src="./mathmodel1.files/image116.wmz" title=""></v:imagedata></v:shape>，考察前<v:shape> <v:imagedata src="./mathmodel1.files/image118.wmz" title=""></v:imagedata></v:shape>场没参赛队的赛程，由于<v:shape><v:imagedata src="./mathmodel1.files/image120.wmz" title=""></v:imagedata></v:shape>，同理可得前<v:shape> <v:imagedata src="./mathmodel1.files/image118.wmz" title=""></v:imagedata></v:shape>场没有参赛的队所进行的比赛也与比赛的单循环规则相矛盾．综合1）和2）可得，<v:shape><v:imagedata src="./mathmodel1.files/image123.wmz" title=""></v:imagedata></v:shape>．　　为了证明<v:shape> <v:imagedata src="./mathmodel1.files/image090.wmz" title=""></v:imagedata></v:shape>，我们考察最极端的情形，即有一个队<v:shape> <v:imagedata src="./mathmodel1.files/image126.wmz" title=""></v:imagedata></v:shape>在第一场和最后一场都有参赛，这样间隔的场次数总和也才可能为最多，注意到<v:shape><v:imagedata src="./mathmodel1.files/image128.wmz" title=""></v:imagedata></v:shape>于是我们有<v:shape><v:imagedata src="./mathmodel1.files/image090.wmz" title=""></v:imagedata></v:shape>．　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　　我们将每个参赛队的赛程安排视为一次博弈，对每个参赛队来说，他们的最优战略是希望自己的间隔场次数在<v:shape> <v:imagedata src="./mathmodel1.files/image131.wmz" title=""></v:imagedata></v:shape>和<v:shape> <v:imagedata src="./mathmodel1.files/image133.wmz" title=""></v:imagedata></v:shape>之间，其赛程安排的均衡解便是能达到这样上下限确界的赛程． 4 赛程均衡解的编排 利用前面所建立的模型，我们将赛程的安排转化为对邻接矩阵的编排，具体方法如下：1）先假定前<v:shape> <v:imagedata src="./mathmodel1.files/image135.wmz" title=""></v:imagedata></v:shape>场比赛的对阵，即确定出赛程邻接矩阵的前<v:shape> <v:imagedata src="./mathmodel1.files/image135.wmz" title=""></v:imagedata></v:shape>列；2）下一场参赛的队必须是一个已连续间隔了<v:shape> <v:imagedata src="./mathmodel1.files/image138.wmz" title=""></v:imagedata></v:shape>（<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>为奇数）或<v:shape> <v:imagedata src="./mathmodel1.files/image141.wmz" title=""></v:imagedata></v:shape>（<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>为偶数）场比赛的队和另一个已连续间隔了<v:shape> <v:imagedata src="./mathmodel1.files/image144.wmz" title=""></v:imagedata></v:shape>场比赛的队．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image146.wmz" title=""></v:imagedata></v:shape>时，由定理1可知，<v:shape><v:imagedata src="./mathmodel1.files/image148.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image150.wmz" title=""></v:imagedata></v:shape>，我们希望能编排出一个间隔场次数最大值为2、最小值为1的赛程．不妨假设第1场比赛的是1和5 ，由于每2场比赛之间有间隔，因此第二场比赛只能是2，3，4中的任意2个队，不妨设为2和3，这样第3场比赛就只能是4和1或者4和5，不妨设为4和5，于是我们得到该赛程邻接矩阵<v:shape> <v:imagedata src="./mathmodel1.files/image152.wmz" title=""></v:imagedata></v:shape>的前3列：<v:shape><v:imagedata src="./mathmodel1.files/image154.wmz" title=""></v:imagedata></v:shape>由此类推，第<v:shape> <v:imagedata src="./mathmodel1.files/image156.wmz" title=""></v:imagedata></v:shape>列中为1的元素在第<v:shape> <v:imagedata src="./mathmodel1.files/image158.wmz" title=""></v:imagedata></v:shape>列中的对应行元素必须为0，第<v:shape> <v:imagedata src="./mathmodel1.files/image160.wmz" title=""></v:imagedata></v:shape>列中为0的元素在第<v:shape> <v:imagedata src="./mathmodel1.files/image162.wmz" title=""></v:imagedata></v:shape>列中的对应行元素根据间隔场次数至少为1且不超过2来选取为1或0，于是我们得到邻接矩阵<v:shape> <v:imagedata src="./mathmodel1.files/image152.wmz" title=""></v:imagedata></v:shape>：<v:shape><v:imagedata src="./mathmodel1.files/image165.wmz" title=""></v:imagedata></v:shape>即赛程<v:shape> <v:imagedata src="./mathmodel1.files/image167.wmz" title=""></v:imagedata></v:shape>，分布情况如表2所示： 表2 <v:shape><v:imagedata src="./mathmodel1.files/image146.wmz" title=""></v:imagedata></v:shape>情形的一个均衡解<DIV align=center><TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 106.65pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-border-alt: solid windowtext .5pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 24.4pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid" vAlign=top width=33>队</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 108pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=144>1 2 3 4 5</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 133.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=178>每2场比赛间隔的场次数</TD></TR><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 24.4pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=33 rowSpan=5>12345</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 108pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=144 rowSpan=5>X 4 7 10 1X 2 8 6 X 5 9X 3X</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 133.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=178>2，2，2</TD></TR><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 133.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=178>1，1，1</TD></TR><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 133.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=178>2，1，1</TD></TR><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 133.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=178>1，2，1</TD></TR><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 133.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=178>1，2，2</TD></TR></TABLE></DIV> 上述方法编制邻接矩阵的关键在于，对2个已连续间隔了<v:shape> <v:imagedata src="./mathmodel1.files/image144.wmz" title=""></v:imagedata></v:shape>场比赛的队取0或1的判别，当<v:shape> <v:imagedata src="./mathmodel1.files/image008.wmz" title=""></v:imagedata></v:shape>值较大时，操作不当很难得到那样的均衡解，利用计算机编程是解决这一问题的

hanbi · 发表于 2004-7-24 07:48:58

最有效方法之一，下面我们利用图论方法[1]、借助几何图形来完善我们的编排．<

21pt; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt"><v:shape><v:imagedata src="./mathmodel1.files/image172.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image174.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．<

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val><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2323"></v:textbox></v:shape><v:shapetype><v:path connecttype="custom" gradientshapeok="t" textboxrect="3163,3163,18437,18437" connectlocs="10800,0;3163,3163;0,10800;3163,18437;10800,21600;18437,18437;21600,10800;18437,3163"></v:path></v:shapetype><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2327"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.437pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2328"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.437pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2329"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 7.437pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2330"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2331"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image184.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 6.843pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2340"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image185.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2341"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image186.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2342"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image187.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.437pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2343"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2344"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2345"></v:textbox></v:shape><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image189.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.468pt; LEFT: auto; MARGIN-LEFT: 7pt; WIDTH: 7.062pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2758"></v:textbox></v:shape></v:group><v:line></v:line></v:group><v:group><v:group><v:group><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2085"></v:textbox></v:shape><v

val></v:oval><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2218"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2219"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2220"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2221"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2222"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2233"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image192.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2234"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image193.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 7.625pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2236"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2237"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image194.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2238"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2250"></v:textbox></v:shape></v:group><v:line></v:line></v:group><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image195.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.468pt; LEFT: auto; MARGIN-LEFT: 6.843pt; WIDTH: 5.562pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2210"></v:textbox></v:shape></v:group> 1）如图1所示，将队1视为圆的中心，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的2个队，以竖直线为开始，横线按从上至下的次序编排，得第一轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image199.wmz" title=""></v:imagedata></v:shape>．<

21pt"> 2）将1不动，3换至2的位置，其它点随3逆时针换位使图1变为图2，从而得第二轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image201.wmz" title=""></v:imagedata></v:shape>． 注：也可从4开始按顺时针方向换位．3）按上述方法依次在已得图的基础上换位即可得到整个赛程．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image203.wmz" title=""></v:imagedata></v:shape>时按上述方法可得一个赛程：<v:shape><v:imagedata src="./mathmodel1.files/image205.wmz" title=""></v:imagedata></v:shape> 分布情况如表3所示： 表3 <v:shape><v:imagedata src="./mathmodel1.files/image207.wmz" title=""></v:imagedata></v:shape>情形均衡解的场次分布<DIV align=center><TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 14.4pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-border-alt: solid windowtext .5pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid" vAlign=top width=63>队</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=209>间隔的场次数</TD></TR><TR 129.5pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=63>12345678</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=209>3,3,3,3,3,34,4,4,3,2,22,4,4,4,3,24,4,3,2,2,22,2,4,4,4,34,3,2,2,2,42,2,2,4,4,43,2,2,2,4,4</TD></TR></TABLE></DIV> <v:shape><v:imagedata src="./mathmodel1.files/image209.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image211.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．1）如图3所示，让圆内的点1表示队1，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的两个队，以1与上顶点(即2的位置)的连线为开始，然后是横线按从上至下的次序，最后以竖直线（即2、<v:shape> <v:imagedata src="./mathmodel1.files/image215.wmz" title=""></v:imagedata></v:shape>位置的连线）结束，得第一轮的赛程

hanbi · 发表于 2004-7-24 07:48:59

最有效方法之一，下面我们利用图论方法[1]、借助几何图形来完善我们的编排．<

21pt; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt"><v:shape><v:imagedata src="./mathmodel1.files/image172.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image174.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．<

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val></v:oval><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2218"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2219"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2220"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2221"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2222"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2233"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image192.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2234"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image193.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 7.625pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2236"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2237"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image194.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2238"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2250"></v:textbox></v:shape></v:group><v:line></v:line></v:group><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image195.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.468pt; LEFT: auto; MARGIN-LEFT: 6.843pt; WIDTH: 5.562pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2210"></v:textbox></v:shape></v:group> 1）如图1所示，将队1视为圆的中心，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的2个队，以竖直线为开始，横线按从上至下的次序编排，得第一轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image199.wmz" title=""></v:imagedata></v:shape>．<

21pt"> 2）将1不动，3换至2的位置，其它点随3逆时针换位使图1变为图2，从而得第二轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image201.wmz" title=""></v:imagedata></v:shape>． 注：也可从4开始按顺时针方向换位．3）按上述方法依次在已得图的基础上换位即可得到整个赛程．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image203.wmz" title=""></v:imagedata></v:shape>时按上述方法可得一个赛程：<v:shape><v:imagedata src="./mathmodel1.files/image205.wmz" title=""></v:imagedata></v:shape> 分布情况如表3所示： 表3 <v:shape><v:imagedata src="./mathmodel1.files/image207.wmz" title=""></v:imagedata></v:shape>情形均衡解的场次分布<DIV align=center><TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 14.4pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-border-alt: solid windowtext .5pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid" vAlign=top width=63>队</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=209>间隔的场次数</TD></TR><TR 129.5pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=63>12345678</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=209>3,3,3,3,3,34,4,4,3,2,22,4,4,4,3,24,4,3,2,2,22,2,4,4,4,34,3,2,2,2,42,2,2,4,4,43,2,2,2,4,4</TD></TR></TABLE></DIV> <v:shape><v:imagedata src="./mathmodel1.files/image209.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image211.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．1）如图3所示，让圆内的点1表示队1，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的两个队，以1与上顶点(即2的位置)的连线为开始，然后是横线按从上至下的次序，最后以竖直线（即2、<v:shape> <v:imagedata src="./mathmodel1.files/image215.wmz" title=""></v:imagedata></v:shape>位置的连线）结束，得第一轮的赛程

hanbi · 发表于 2004-7-24 07:49:01

最有效方法之一，下面我们利用图论方法[1]、借助几何图形来完善我们的编排．<

21pt; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt"><v:shape><v:imagedata src="./mathmodel1.files/image172.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image174.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．<

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val></v:oval><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2218"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2219"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2220"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2221"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2222"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2233"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image192.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2234"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image193.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 7.625pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2236"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2237"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image194.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2238"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2250"></v:textbox></v:shape></v:group><v:line></v:line></v:group><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image195.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.468pt; LEFT: auto; MARGIN-LEFT: 6.843pt; WIDTH: 5.562pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2210"></v:textbox></v:shape></v:group> 1）如图1所示，将队1视为圆的中心，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的2个队，以竖直线为开始，横线按从上至下的次序编排，得第一轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image199.wmz" title=""></v:imagedata></v:shape>．<

21pt"> 2）将1不动，3换至2的位置，其它点随3逆时针换位使图1变为图2，从而得第二轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image201.wmz" title=""></v:imagedata></v:shape>． 注：也可从4开始按顺时针方向换位．3）按上述方法依次在已得图的基础上换位即可得到整个赛程．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image203.wmz" title=""></v:imagedata></v:shape>时按上述方法可得一个赛程：<v:shape><v:imagedata src="./mathmodel1.files/image205.wmz" title=""></v:imagedata></v:shape> 分布情况如表3所示： 表3 <v:shape><v:imagedata src="./mathmodel1.files/image207.wmz" title=""></v:imagedata></v:shape>情形均衡解的场次分布<DIV align=center><TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 14.4pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-border-alt: solid windowtext .5pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid" vAlign=top width=63>队</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=209>间隔的场次数</TD></TR><TR 129.5pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=63>12345678</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=209>3,3,3,3,3,34,4,4,3,2,22,4,4,4,3,24,4,3,2,2,22,2,4,4,4,34,3,2,2,2,42,2,2,4,4,43,2,2,2,4,4</TD></TR></TABLE></DIV> <v:shape><v:imagedata src="./mathmodel1.files/image209.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image211.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．1）如图3所示，让圆内的点1表示队1，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的两个队，以1与上顶点(即2的位置)的连线为开始，然后是横线按从上至下的次序，最后以竖直线（即2、<v:shape> <v:imagedata src="./mathmodel1.files/image215.wmz" title=""></v:imagedata></v:shape>位置的连线）结束，得第一轮的赛程

hanbi · 发表于 2004-7-24 07:49:01

最有效方法之一，下面我们利用图论方法[1]、借助几何图形来完善我们的编排．<

21pt; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt"><v:shape><v:imagedata src="./mathmodel1.files/image172.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image174.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．<

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val></v

val><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2323"></v:textbox></v:shape><v:shapetype><v:path connecttype="custom" gradientshapeok="t" textboxrect="3163,3163,18437,18437" connectlocs="10800,0;3163,3163;0,10800;3163,18437;10800,21600;18437,18437;21600,10800;18437,3163"></v:path></v:shapetype><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2327"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.437pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2328"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.437pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2329"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 7.437pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2330"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2331"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image184.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 6.843pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2340"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image185.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2341"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image186.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2342"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image187.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.437pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2343"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2344"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2345"></v:textbox></v:shape><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image189.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.468pt; LEFT: auto; MARGIN-LEFT: 7pt; WIDTH: 7.062pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2758"></v:textbox></v:shape></v:group><v:line></v:line></v:group><v:group><v:group><v:group><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2085"></v:textbox></v:shape><v

val></v:oval><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2218"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2219"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2220"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2221"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2222"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2233"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image192.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2234"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image193.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 7.625pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2236"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2237"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image194.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2238"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2250"></v:textbox></v:shape></v:group><v:line></v:line></v:group><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image195.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.468pt; LEFT: auto; MARGIN-LEFT: 6.843pt; WIDTH: 5.562pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2210"></v:textbox></v:shape></v:group> 1）如图1所示，将队1视为圆的中心，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的2个队，以竖直线为开始，横线按从上至下的次序编排，得第一轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image199.wmz" title=""></v:imagedata></v:shape>．<

21pt"> 2）将1不动，3换至2的位置，其它点随3逆时针换位使图1变为图2，从而得第二轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image201.wmz" title=""></v:imagedata></v:shape>． 注：也可从4开始按顺时针方向换位．3）按上述方法依次在已得图的基础上换位即可得到整个赛程．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image203.wmz" title=""></v:imagedata></v:shape>时按上述方法可得一个赛程：<v:shape><v:imagedata src="./mathmodel1.files/image205.wmz" title=""></v:imagedata></v:shape> 分布情况如表3所示： 表3 <v:shape><v:imagedata src="./mathmodel1.files/image207.wmz" title=""></v:imagedata></v:shape>情形均衡解的场次分布<DIV align=center><TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 14.4pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-border-alt: solid windowtext .5pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid" vAlign=top width=63>队</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=209>间隔的场次数</TD></TR><TR 129.5pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=63>12345678</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=209>3,3,3,3,3,34,4,4,3,2,22,4,4,4,3,24,4,3,2,2,22,2,4,4,4,34,3,2,2,2,42,2,2,4,4,43,2,2,2,4,4</TD></TR></TABLE></DIV> <v:shape><v:imagedata src="./mathmodel1.files/image209.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image211.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．1）如图3所示，让圆内的点1表示队1，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的两个队，以1与上顶点(即2的位置)的连线为开始，然后是横线按从上至下的次序，最后以竖直线（即2、<v:shape> <v:imagedata src="./mathmodel1.files/image215.wmz" title=""></v:imagedata></v:shape>位置的连线）结束，得第一轮的赛程

hanbi · 发表于 2004-7-24 07:50:57

<v:shape> <v:imagedata src="./mathmodel1.files/image199.wmz" title=""></v:imagedata></v:shape>．<

21pt"> <

21pt"> 2）将1不动，3换至2的位置，其它点随3逆时针换位使图1变为图2，从而得第二轮的赛程<v:shape> <v:imagedata src="./mathmodel1.files/image201.wmz" title=""></v:imagedata></v:shape>． 注：也可从4开始按顺时针方向换位．3）按上述方法依次在已得图的基础上换位即可得到整个赛程．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image203.wmz" title=""></v:imagedata></v:shape>时按上述方法可得一个赛程：<v:shape><v:imagedata src="./mathmodel1.files/image205.wmz" title=""></v:imagedata></v:shape> 分布情况如表3所示： 表3 <v:shape><v:imagedata src="./mathmodel1.files/image207.wmz" title=""></v:imagedata></v:shape>情形均衡解的场次分布<DIV align=center><TABLE medium none; BORDER-TOP: medium none; MARGIN-LEFT: 14.4pt; BORDER-LEFT: medium none; BORDER-BOTTOM: medium none; BORDER-COLLAPSE: collapse; mso-padding-alt: 0cm 5.4pt 0cm 5.4pt; mso-border-alt: solid windowtext .5pt" cellSpacing=0 cellPadding=0 border=1><TR><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid" vAlign=top width=63>队</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: windowtext 0.5pt solid; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; mso-border-left-alt: solid windowtext .5pt" vAlign=top width=209>间隔的场次数</TD></TR><TR 129.5pt"><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: windowtext 0.5pt solid; WIDTH: 47.5pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=63>12345678</TD><TD windowtext 0.5pt solid; PADDING-RIGHT: 5.4pt; BORDER-TOP: medium none; PADDING-LEFT: 5.4pt; PADDING-BOTTOM: 0cm; BORDER-LEFT: medium none; WIDTH: 156.75pt; PADDING-TOP: 0cm; BORDER-BOTTOM: windowtext 0.5pt solid; HEIGHT: 129.5pt; mso-border-left-alt: solid windowtext .5pt; mso-border-top-alt: solid windowtext .5pt" vAlign=top width=209>3,3,3,3,3,34,4,4,3,2,22,4,4,4,3,24,4,3,2,2,22,2,4,4,4,34,3,2,2,2,42,2,2,4,4,43,2,2,2,4,4</TD></TR></TABLE></DIV>

hanbi · 发表于 2004-7-24 07:51:32

<

21pt; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt"><v:shape><v:imagedata src="./mathmodel1.files/image209.wmz" title=""></v:imagedata></v:shape>．<v:shape> <v:imagedata src="./mathmodel1.files/image211.wmz" title=""></v:imagedata></v:shape>(<v:shape> <v:imagedata src="./mathmodel1.files/image176.wmz" title=""></v:imagedata></v:shape>为正整数)的情形．<

21pt; mso-char-indent-count: 2.0; mso-char-indent-size: 10.5pt">1）如图3所示，让圆内的点1表示队1，其余<v:shape> <v:imagedata src="./mathmodel1.files/image197.wmz" title=""></v:imagedata></v:shape>个队对称地分布于圆周，标有连线的两点表示该轮对阵的两个队，以1与上顶点(即2的位置)的连线为开始，然后是横线按从上至下的次序，最后以竖直线（即2、<v:shape> <v:imagedata src="./mathmodel1.files/image215.wmz" title=""></v:imagedata></v:shape>位置的连线）结束，得第一轮的赛程<

center" align=center><v:shape><v:imagedata src="./mathmodel1.files/image217.wmz" title=""></v:imagedata></v:shape>．<v:group><v:group><v:group><v

val></v

val><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.406pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2529"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2530"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2531"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.5pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2532"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2540"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.562pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2541"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image219.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.5pt; LEFT: auto; MARGIN-LEFT: 7pt; WIDTH: 7.562pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2542"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image220.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2543"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image221.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2544"></v:textbox></v:shape><v:line></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image222.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 6.843pt; WIDTH: 4.812pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2551"></v:textbox></v:shape><v:line></v:line><v:shape><v:imagedata src="./mathmodel1.files/image223.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.5pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2554"></v:textbox></v:shape><v:group><v:line><v:stroke endarrowlength="long" endarrowwidth="narrow" endarrow="block"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.406pt; LEFT: auto; MARGIN-LEFT: 6.937pt; TOP: auto; HEIGHT: 8.406pt; mso-next-textbox: #_x0000_s2557"></v:textbox></v:shape></v:group></v:group><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2525"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2528"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2545"></v:textbox></v:shape></v:group><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape></v:group><v:group><v:group><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2601"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2621"></v:textbox></v:shape><v:group><v:group><v

val></v:oval><v:line></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line></v:line><v:group><v:line><v:stroke endarrowlength="long" endarrowwidth="narrow" endarrow="block"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 6.937pt; TOP: auto; HEIGHT: 8.406pt; mso-next-textbox: #_x0000_s2633"></v:textbox></v:shape></v:group><v:shape><v:imagedata src="./mathmodel1.files/image225.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.437pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 5.562pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2764"></v:textbox></v:shape></v:group><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.406pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2605"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2606"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2607"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.5pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2608"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image226.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.656pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2616"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image227.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.562pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.437pt; mso-next-textbox: #_x0000_s2617"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image219.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.5pt; LEFT: auto; MARGIN-LEFT: 7pt; WIDTH: 7.562pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2618"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2619"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image221.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2620"></v:textbox></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.5pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2630"></v:textbox></v:shape></v:group></v:group><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.343pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2604"></v:textbox></v:shape></v:group> 2）在图3中，让1不动，上半圆周左边弧段上的每一个点（如果存在的话也包括<v:shape> <v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata></v:shape>轴与圆周的交点）依次按左半圆回路<v:shape><v:imagedata src="./mathmodel1.files/image231.wmz" title=""></v:imagedata></v:shape>．顺时针换位至2的位置，左半圆回路上的其它点也随着顺时针换位，但右半圆弧上除上下顶点外的其它点均不动．然后按（1）中方式对应点相连，得第二部份赛程：<v:shape><v:imagedata src="./mathmodel1.files/image233.wmz" title=""></v:imagedata></v:shape>（如图4）．<v:group><v:group><v:oval></v:oval><v:group><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image181.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2483"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image183.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.406pt; mso-next-textbox: #_x0000_s2484"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image182.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2485"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image184.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2486"></v:textbox></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image191.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2494"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image188.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.75pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2495"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image235.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2496"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image180.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.281pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2497"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image236.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 6.937pt; WIDTH: 6.875pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2498"></v:textbox></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image237.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.781pt; LEFT: auto; MARGIN-LEFT: 6.843pt; WIDTH: 4.031pt; TOP: auto; HEIGHT: 7.125pt; mso-next-textbox: #_x0000_s2505"></v:textbox></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image223.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.687pt; LEFT: auto; MARGIN-LEFT: 7.187pt; WIDTH: 8.312pt; TOP: auto; HEIGHT: 4.468pt; mso-next-textbox: #_x0000_s2508"></v:textbox></v:shape><v:group><v:line></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line><v:stroke dashstyle="shortDot" endcap="round"></v:stroke></v:line><v:line></v:line><v:group><v:line><v:stroke endarrowlength="long" endarrowwidth="narrow" endarrow="block"></v:stroke></v:line><v:shape><v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.593pt; LEFT: auto; MARGIN-LEFT: 6.937pt; TOP: auto; HEIGHT: 8.406pt; mso-next-textbox: #_x0000_s2511"></v:textbox></v:shape></v:group></v:group></v:group><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.843pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.25pt; mso-next-textbox: #_x0000_s2479"></v:textbox></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image179.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.187pt; TOP: auto; HEIGHT: 3.562pt; mso-next-textbox: #_x0000_s2482"></v:textbox></v:shape><v:shape></v:shape><v:shape><v:imagedata src="./mathmodel1.files/image178.wmz" title=""></v:imagedata><v:textbox style="MARGIN-TOP: 3.531pt; LEFT: auto; MARGIN-LEFT: 7.531pt; TOP: auto; HEIGHT: 5.281pt; mso-next-textbox: #_x0000_s2499"></v:textbox></v:shape></v:group><v:shape></v:shape><w:wrap type="square"></w:wrap></v:group>3）在图3中，让1不动，上半圆周右边弧段上的每一个点（如果存在的话也包括<v:shape> <v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata></v:shape>轴与圆周的交点）依次按右半圆回路<v:shape><v:imagedata src="./mathmodel1.files/image240.wmz" title=""></v:imagedata></v:shape>．逆时针换位至2的位置，右半圆回路上的其它点也随着换位，但左半圆弧上除上下顶点外的其它点均不动．然后按（1）中方式对应点相连，且从后向前编排，得第四部份赛程：

hanbi · 发表于 2004-7-24 07:52:00

<

center" align=center><v:shape><v:imagedata src="./mathmodel1.files/image242.wmz" title=""></v:imagedata></v:shape>（如图5）．<

21pt; TEXT-ALIGN: center" align=center>4）在图3中，以1和下顶点(即<v:shape> <v:imagedata src="./mathmodel1.files/image244.wmz" title=""></v:imagedata></v:shape>的位置)的连线开始向左、右两边扩展分别得赛程<v:shape> <v:imagedata src="./mathmodel1.files/image246.wmz" title=""></v:imagedata></v:shape>和<v:shape> <v:imagedata src="./mathmodel1.files/image248.wmz" title=""></v:imagedata></v:shape>．每个队<v:shape> <v:imagedata src="./mathmodel1.files/image156.wmz" title=""></v:imagedata></v:shape>在该轮的编排方法是：让<v:shape> <v:imagedata src="./mathmodel1.files/image251.wmz" title=""></v:imagedata></v:shape>不动，下半圆周左（右）边弧段上的其它点（如果存在的话也包括<v:shape> <v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata></v:shape>轴与圆周的交点）沿下半圆回路逆（顺）时针换位，上半圆周上的其它点不动，先以1，<v:shape> <v:imagedata src="./mathmodel1.files/image251.wmz" title=""></v:imagedata></v:shape>相连，然后按下半圆回路换位方向和换位前距点<v:shape> <v:imagedata src="./mathmodel1.files/image251.wmz" title=""></v:imagedata></v:shape>由近至远的点的次序依次横竖相连，<v:shape> <v:imagedata src="./mathmodel1.files/image246.wmz" title=""></v:imagedata></v:shape>按原序编排，<v:shape> <v:imagedata src="./mathmodel1.files/image248.wmz" title=""></v:imagedata></v:shape>由右向左编排，于是可得第三部分赛程<v:shape> <v:imagedata src="./mathmodel1.files/image258.wmz" title=""></v:imagedata></v:shape>，<

>从而可求得整个赛程<v:shape> <v:imagedata src="./mathmodel1.files/image260.wmz" title=""></v:imagedata></v:shape>．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image262.wmz" title=""></v:imagedata></v:shape>时，由图6按上述方法编排，得1）<v:shape> <v:imagedata src="./mathmodel1.files/image264.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image266.wmz" title=""></v:imagedata></v:shape>，<v:shape><v:imagedata src="./mathmodel1.files/image268.wmz" title=""></v:imagedata></v:shape>；

hanbi · 发表于 2004-7-24 07:52:23

<

center" align=center><v:shape><v:imagedata src="./mathmodel1.files/image242.wmz" title=""></v:imagedata></v:shape>（如图5）．<

21pt; TEXT-ALIGN: center" align=center>4）在图3中，以1和下顶点(即<v:shape> <v:imagedata src="./mathmodel1.files/image244.wmz" title=""></v:imagedata></v:shape>的位置)的连线开始向左、右两边扩展分别得赛程<v:shape> <v:imagedata src="./mathmodel1.files/image246.wmz" title=""></v:imagedata></v:shape>和<v:shape> <v:imagedata src="./mathmodel1.files/image248.wmz" title=""></v:imagedata></v:shape>．每个队<v:shape> <v:imagedata src="./mathmodel1.files/image156.wmz" title=""></v:imagedata></v:shape>在该轮的编排方法是：让<v:shape> <v:imagedata src="./mathmodel1.files/image251.wmz" title=""></v:imagedata></v:shape>不动，下半圆周左（右）边弧段上的其它点（如果存在的话也包括<v:shape> <v:imagedata src="./mathmodel1.files/image224.wmz" title=""></v:imagedata></v:shape>轴与圆周的交点）沿下半圆回路逆（顺）时针换位，上半圆周上的其它点不动，先以1，<v:shape> <v:imagedata src="./mathmodel1.files/image251.wmz" title=""></v:imagedata></v:shape>相连，然后按下半圆回路换位方向和换位前距点<v:shape> <v:imagedata src="./mathmodel1.files/image251.wmz" title=""></v:imagedata></v:shape>由近至远的点的次序依次横竖相连，<v:shape> <v:imagedata src="./mathmodel1.files/image246.wmz" title=""></v:imagedata></v:shape>按原序编排，<v:shape> <v:imagedata src="./mathmodel1.files/image248.wmz" title=""></v:imagedata></v:shape>由右向左编排，于是可得第三部分赛程<v:shape> <v:imagedata src="./mathmodel1.files/image258.wmz" title=""></v:imagedata></v:shape>，<

>从而可求得整个赛程<v:shape> <v:imagedata src="./mathmodel1.files/image260.wmz" title=""></v:imagedata></v:shape>．例如，<v:shape> <v:imagedata src="./mathmodel1.files/image262.wmz" title=""></v:imagedata></v:shape>时，由图6按上述方法编排，得1）<v:shape> <v:imagedata src="./mathmodel1.files/image264.wmz" title=""></v:imagedata></v:shape>，<v:shape> <v:imagedata src="./mathmodel1.files/image266.wmz" title=""></v:imagedata></v:shape>，<v:shape><v:imagedata src="./mathmodel1.files/image268.wmz" title=""></v:imagedata></v:shape>；

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