历年全国数学建模题
<P><a href="http://bt.5qzone.net/download.php?id=231014&file" target="_blank" >http://bt.5qzone.net/download.php?id=231014&file</A>=数学建模竞赛题目与解答.torrent&id2=1125993213&action=1</P><BR> <P>谢谢</P> <P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char">求解最小生成树的方法虽然很多，但是利用LINGO建立相应的整数规划模型是一种新的尝试。这对于处理非标准的MST问题非常方便。我们主要参考了文。<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-INDENT: 21pt">在图论中，称无圈的连通图为树。在一个连通图G中，称包含图G全部顶点的树为图G的生成树。生成树上各边的权之和称为该生成树的权。连通图G的权最小的生成树称为图G的最小生成树。<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-INDENT: 21pt">许多实际问题都可以归结为最小生成树。例如,如何修筑一些公路把若干个城镇连接起来；如何架设通讯网络将若干个地区连接起来；如何修筑水渠将水源和若干块待灌溉的土地连接起来等等。为了说明问题,以下面的问题作为范例。<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-INDENT: 21pt">范例：假设某电话公司计划在六个村庄架设电话线，各村庄之间的距离如图所示。试求出使电话线总长度最小的架线方案。<p></p></P>
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<P 0cm 0cm 0pt; TEXT-ALIGN: left; mso-pagination: widow-orphan" align=left> <p></p></P></DIV></TD></TR></TABLE></v:textbox></v:shape><v:group><v:oval></v:oval><v:oval></v:oval><v:oval></v:oval><v:oval></v:oval><v:oval></v:oval><v:oval></v:oval><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line><v:line></v:line></v:group><w:wrap type="square" anchorx="page"></w:wrap></v:group>为了便于计算机求解，特作如下规定：（1）节点V1表示树根；（2）当两个节点之间没有线路时，规定两个节点之间的距离为M（较大的值）。<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char"> MST的整数规划模型如下：<p></p></P>
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<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char"><B> 例7.7 分配问题（指派问题，Assignment Problem）<p></p></B></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-INDENT: 21pt; LINE-HEIGHT: 16pt; mso-line-height-rule: exactly">这是个给n个人分配n项工作以获得某个最高总效果的问题。第i个人完成第j项工作需要平均时间<SUB><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></v:imagedata></v:shape></SUB>。要求给每个人分配一项工作，并要求分配完这些工作，以使完成全部任务的总时间为最小。该问题可表示如下：<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: center" align=center><SUB><v:shape><v:imagedata></v:imagedata></v:shape></SUB><p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; LINE-HEIGHT: 16pt; mso-line-height-rule: exactly"> 显然，此问题可看作是运输问题的特殊情况。可将此问题看作具有n个源和n个汇的问题，每个源有1单位的可获量，而每个汇有1单位的需要量。从表面看，这问题要求用整数规划以保证<SUB><v:shape> <v:imagedata></v:imagedata></v:shape></SUB>能取0或1。然而，幸运的是，此问题是运输问题的特例，因此即使不限制<SUB><v:shape> <v:imagedata></v:imagedata></v:shape></SUB>取0或1，最优解也将取0或1。如果把婚姻看作分配问题，丹茨证明，整数性质证明一夫一妻会带来最美满幸福的生活！显然，分配问题可以作为线性规划问题来求解，尽管模型可能很大。例如，给100人分配100项工作将使所得的模型具有10000个变量。这时，如采用专门算法效果会更好。时间复杂度为<SUB><v:shape> <v:imagedata></v:imagedata></v:shape></SUB>的匈牙利算法便是好选择，这是由Kuhu（1955）提出的。<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left>model:<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> !7个工人，7个工作的分配问题;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left>sets:<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> workers/w1..w7/;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> jobs/j1..j7/;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> links(workers,jobs): cost,volume;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left>endsets<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> !目标函数;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> min=@sum(links: cost*volume);<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> !每个工人只能有一份工作;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> @for(workers(I):<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> @sum(jobs(J): volume(I,J))=1;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> );<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> !每份工作只能有一个工人;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> @for(jobs(J):<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> @sum(workers(I): volume(I,J))=1;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> );<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left>data:<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> cost= 6 2 6 7 4 2 5<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> 4 9 5 3 8 5 8<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> 5 2 1 9 7 4 3<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> 7 6 7 3 9 2 7<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> 2 3 9 5 7 2 6<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> 5 5 2 2 8 11 4<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left> 9 2 3 12 4 5 10;<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left>enddata<p></p></P>
<P 0cm 0cm 0pt; LAYOUT-GRID-MODE: char; TEXT-ALIGN: left; mso-layout-grid-align: none" align=left>end<p></p></P> <P>不是很爽</P> <p>打不开啊 能不能发到我油箱去啊 <a href="mailto:ljg-578@163.com">ljg-578@163.com</a> </p><p>谢谢了啊~~~</p> 真的打不开啊~~~~~~~~~~~~~~~~~~~~ ~~~~~~~~~~~~~~~~~~~~~~~~~~~···· 没有打开啊 怎么回事 G的权最小的生成树称为图G的最小生成树。
许多实际问题都可以归结为最小生成树。例如,如何修筑一些公路把若干个城镇连接起来；如何架设通讯网络将若干个地区连接起来；如何修筑水渠将水源和若干块待灌溉的土地连接起来等等。为了说明问题,以下面的问题作为范例。
范例：假设某电话公司计划在六 能不能把“商业中的 订货问”论文发我邮箱里792608375@qq.com 确实打不开呀
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