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求助:拓扑排序(能输出所有序列)

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发表于 2004-8-27 22:03:32 | 显示全部楼层 |阅读模式
<>谁有能输出所有拓扑序列的拓扑排序算法呀?</P>
发表于 2004-8-31 02:54:10 | 显示全部楼层
<>ROCEDURE TOPO(n,R,m,P)</P><>定义数组F(1:n),G(1:n),S(1;n),V(1;m),NEXT(1;m)</P><P>top=0;  [栈初始化]t=1</P><P>FOR k=1 TO n DO {F(k)=0;G(k)=0}  [初使化]</P><P>FOR k=1 TO m DO  [读入m个二元组]</P><P>  {i=R(k,1);j=R(k,2)  [依次读入一个二元组]</P><P>   F(j)=F(j)+1  [数j的前件计数]</P><P>   NEXT(k)=G(i);V(k)=j;G(i)=k  [将j链入到以G(i)为头指针的链表的链头]</P><P>   }</P><P>FOR k=1 TO n DO  [将所有没有前件的数入栈]</P><P>IF(F(k)=0) THEN {top=top+1,S(top)=k}  </P><P>WHILE(top!=0) DO</P><P>   {i=S(top);top=top-1;P(t)=i;t=t+1   [从栈中退出一个无前件的数]</P><P>    k=G(i)</P><P>    WHILE(k!=0) DO</P><P>         {j=V(k);F(j)=F(j)-1   [i后件j的前件计数减1]</P><P>          IF(F(j)=0) THEN {top=top+1;S(top)=j}         [无前件数j入栈]</P><P>          k=NEXT(k)      [考虑i的下一个后件]</P><P>         }</P><P>   }</P><P>FOR k=1 TO n DO    [考虑无法分类的数]</P><P>IF(F(k)!=0) THEN {p(t)=-k;t=t+1}</P><P>RETURN</P>
发表于 2004-8-31 18:41:58 | 显示全部楼层
<>我目前有一道棘手的编程题无法解决</P><>希望大家的帮助 多谢</P>< 0pt? 0cm><FONT face="Times New Roman">1</FONT>、加工模型:假设月份原料油的加工量为:<FONT face="Times New Roman">a<SUB>ij</SUB></FONT>吨。则:</P><P 0cm 18pt? 0pt><FONT face="Times New Roman">a<SUB>i1</SUB>+a<SUB>i2</SUB>&lt;=200 a<SUB>i3</SUB>+a<SUB>i4</SUB>+a<SUB>i5</SUB>&lt;=250</FONT></P><P 0pt? 0cm><FONT face="Times New Roman">2</FONT>、精练油混合为成品油模型:假设原料油<FONT face="Times New Roman">V<SUB>1</SUB></FONT><SUB>、</SUB><FONT face="Times New Roman">V<SUB>2</SUB></FONT><SUB>、</SUB><FONT face="Times New Roman">0<SUB>3</SUB></FONT><SUB>、</SUB><FONT face="Times New Roman">0<SUB>4</SUB></FONT><SUB>、</SUB><FONT face="Times New Roman">0<SUB>5</SUB></FONT>的硬度分别为:<FONT face="Times New Roman">C<SUB>j</SUB></FONT>(单位)则:</P><P 0cm 415.3pt? right tab-stops: 0pt;><FONT face="Times New Roman"></FONT><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 gradientshapeok="t" extrusionok="f" connecttype="rect"></v:path><LOCK aspectratio="t" v:ext="edit"></LOCK></v:shapetype><v:shape><v:imagedata><FONT face="Times New Roman"></FONT></v:imagedata></v:shape>&gt;3<v:shape> <v:imagedata></v:imagedata></v:shape>、 <v:shape><v:imagedata></v:imagedata></v:shape>&lt;6<v:shape> <v:imagedata></v:imagedata></v:shape>即:
<P><P 0cm 415.3pt? right tab-stops: 0pt;>3〈<v:shape> <v:imagedata></v:imagedata></v:shape>〈6 <SUB><P></SUB><P><P 0pt? 0cm><FONT face="Times New Roman">3</FONT>、采购模型:假设月份原料油的采购量为:<FONT face="Times New Roman">b<SUB>ij</SUB></FONT>(吨)<FONT face="Times New Roman">6</FONT>个月的总采购费为:<FONT face="Times New Roman">S</FONT>(元)<FONT face="Times New Roman"> </FONT>则:</P><P 0cm 18pt? 0pt><FONT face="Times New Roman">W=(w<SUB>ij</SUB>)<SUB>6*5</SUB>=(c<SUB>ij</SUB>)<SUB>6*5</SUB></FONT><SUB>。</SUB><FONT face="Times New Roman">*(b<SUB>ij</SUB>)<SUB>6*5</SUB></FONT></P><P 0cm 18pt? 0pt>注:<FONT face="Times New Roman">w=C.*B</FONT>表示<FONT face="Times New Roman">c<SUB>ij</SUB>=a<SUB>ij</SUB>*b<SUB>ij</SUB></FONT>其中<FONT face="Times New Roman">C=(c<SUB>ij</SUB>),A=(a<SUB>ij</SUB>)</FONT></P><P 0cm 0pt; 10.5pt? mso-char-indent-size: 2.0; mso-char-indent-count: 21pt; TEXT-INDENT:><FONT face="Times New Roman">S=</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape><FONT face="Times New Roman"></FONT></P><P 0cm 0pt tab-stops: TEXT-INDENT: lfo4? level1 l1 mso-list: 57.0pt; list -36pt; 57pt;><FONT face="Times New Roman">4、 </FONT>存储模型:令<FONT face="Times New Roman">6</FONT>个月内<FONT face="Times New Roman">5</FONT>种原料油的存储费为:<FONT face="Times New Roman">Q</FONT>(元)则:<FONT face="Times New Roman">Q=</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape></P><P 0cm 18pt? 0pt>Q<SUB>k</SUB>=[5N+<v:shape> <v:imagedata></v:imagedata></v:shape>]*q (k=1.2.3.4.5.6)</P><P 0cm 0pt tab-stops: TEXT-INDENT: lfo4? level1 l1 mso-list: 57.0pt; list -36pt; 57pt;><FONT face="Times New Roman">5、 </FONT>收入模型:令<FONT face="Times New Roman">6</FONT>个月内成品油的总收入为:<FONT face="Times New Roman">P</FONT>(元)则:<FONT face="Times New Roman">P= M</FONT><v:shape><FONT face="Times New Roman"> <v:imagedata></v:imagedata></FONT></v:shape></P><P 0pt? 0cm>公司<FONT face="Times New Roman">6</FONT>个月内获得的总利润为:<FONT face="Times New Roman">T=P-Q-S</FONT></P><P 0cm 0pt; mso-char-indent-size: mso-char-indent-count: TEXT-INDENT: 13.8pt? 1.0; 13.8pt;><B><FONT face="Times New Roman">6</FONT></B><B>、约束条件:</B><B><P></B><P><P 0cm 0pt; TEXT-INDENT: 31.5pt?><FONT face="Times New Roman">N+</FONT><v:shape> <v:imagedata></v:imagedata></v:shape>=1000<P><P><P 0cm 0pt; TEXT-INDENT: 31.5pt?>N+<v:shape> <v:imagedata></v:imagedata></v:shape>-<v:shape> <v:imagedata></v:imagedata></v:shape>=N<P><P><P>Max T= M<v:shape> <v:imagedata></v:imagedata></v:shape>-<v:shape> <v:imagedata></v:imagedata></v:shape></P>
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