最短路问题的 LINGO 求解

设图共有个节点，其赋权图的邻接矩阵为 . 表示节点到的权值为 .当为有向图时，；当为无向图时，和分别由图得到，通常不一样。当，表示节点与节点不连通。令。假设图的所有权值现求节点到节点的最短路，其线性规划模型为：最短路问题的LINGO求解

模型一 决策变量：设目标函数为寻找一条节点到节点的通路，使其上权值和最小，故目标函数为： 1.对节点恰有一条路出去，却不能有路回来，故有：且

2.对节点恰有一条路到达，却不能有路出去，故有：2.对节点恰有一条路到达，却不能有路出去，故有：且 3.对除起始点和目标点之外，其它点进入和出去的路是一样多（可都为0），则： 4.对不通的路不取，约束为：

总的线性规划模型为：

示例演示 例1 现有11个点的无向图见图14.1，求点1到点11的最短路。图1 无向图最短路示意图

其Lingo实现程序为： model: sets: point/1..11/:u; road(point,point):W,X; endsets data: W=0,2,8,1,0,0,0,0,0,0,0, 2,0,6,0,1,0,0,0,0,0,0, 8,6,0,7,5,1,2,0,0,0,0, 1,0,7,0,0,0,9,0,0,0,0, 0,1,5,0,0,3,0,2,9,0,0, 0,0,1,0,3,0,4,0,6,0,0, 0,0,2,9,0,4,0,0,3,1,0, 0,0,0,0,2,0,0,0,7,0,9, 0,0,0,0,9,6,3,7,0,1,2, 0,0,0,0,0,0,1,0,1,0,4, 0,0,0,0,0,0,0,9,2,4,0; enddata

min=@sum(road(i,j):w(i,j)*x(i,j)); !最短路; @for(point(i)|i#ne#1#and#i#ne#11:@sum(point(k):x(k,i))=@sum(point(j):x(i,j))); @sum(point(j)|j#ne#1:x(1,j))=1; !起始点要出去; @sum(point(k)|k#ne#1:x(k,1))=0; !不能回到起始点; @sum(point(k)|k#ne#11:x(k,11))=1; !要到达目标点; @sum(point(j)|j#ne#11:x(11,j))=0; !目标点不能出去; @for(road(i,j):x(i,j)<=W(i,j)); !不能到达的路不考虑; @for(road(i,j):@bin(x(i,j))); end 结果为minZ=13 x(1,2)=1 x(2,5)=1; x(5,6)=1 x(6,3)=1 x(3,7)=1 x(7,10)=1 x(10,9)=1 x(9,11)=1 故路径为12563710911

模型二 不必考虑起始点不回去，结束点不出去，统一考虑所有中间点不出现圈，添加约束为：总模型为：

其Lingo实现程序为： model: sets: point/1..11/; road(point,point):W,X; endsets data: W=0,2,8,1,0,0,0,0,0,0,0, 2,0,6,0,1,0,0,0,0,0,0, 8,6,0,7,5,1,2,0,0,0,0, 1,0,7,0,0,0,9,0,0,0,0, 0,1,5,0,0,3,0,2,9,0,0, 0,0,1,0,3,0,4,0,6,0,0, 0,0,2,9,0,4,0,0,3,1,0, 0,0,0,0,2,0,0,0,7,0,9, 0,0,0,0,9,6,3,7,0,1,2, 0,0,0,0,0,0,1,0,1,0,4, 0,0,0,0,0,0,0,9,2,4,0; enddata

min=@sum(road(i,j):w(i,j)*x(i,j)); !最短路; n=@size(point); @for(point(i)|i#ne#1#and#i#ne#11:@sum(point(k):x(k,i))=@sum(point(j):x(i,j))); @sum(point(j)|j#ne#1:x(1,j))=1; !起始点要出去; @sum(point(k)|k#ne#11:x(k,11))=1; !要到达目标点; @for(road(i,j):u(i)-u(j)+n*x(i,j)<=n-1); !不出现圈; @for(road(i,j):x(i,j)<=W(i,j)); !不能到达的路不考虑; @for(road(i,j):@bin(x(i,j))); end

最短路问题的 LINGO 求解