两个指定顶点之间最短路问题

见下图，用点表示城市，现有 A,B1,B2 ,C1,C2 ,C3, D共 7 个城市。点与点之间的连线表示城市间有道路相连。连线旁的数字表示道路的长度。现计划从城市A到城市D 铺设一条天然气管道，请设计出最小价格管道铺设方案。

lingo程序：

model:
sets:
cities/A,B1,B2,C1,C2,C3,D/;
roads(cities,cities)/A B1,A B2,B1 C1,B1 C2,B1 C3,B2 C1,
B2 C2,B2 C3,C1 D,C2 D,C3 D/:w,x;!路径长度

endsets
data:
w=2 4 3 3 1 2 3 1 1 3 4;
enddata
n=@size(cities); !城市的个数;
min=@sum(roads:w*x);
@for(cities(i)|i #ne#1 #and# i #ne#n:
@sum(roads(i,j):x(i,j))=@sum(roads(j,i):x(j,i)));
@sum(roads(i,j)|i #eq#1:x(i,j))=1;
@sum(roads(i,j)|j #eq#n:x(i,j))=1;
end

运行结果：

Global optimal solution found.
Objective value:                              6.000000
Infeasibilities:                              0.000000
Total solver iterations:                             0

Model Class:                                        LP

Total variables:                     11
Nonlinear variables:                  0
Integer variables:                    0

Total constraints:                    8
Nonlinear constraints:                0

Total nonzeros:                      33
Nonlinear nonzeros:                   0

Variable           Value        Reduced Cost
N        7.000000            0.000000
W( A, B1)        2.000000            0.000000
W( A, B2)        4.000000            0.000000
W( B1, C1)        3.000000            0.000000
W( B1, C2)        3.000000            0.000000
W( B1, C3)        1.000000            0.000000
W( B2, C1)        2.000000            0.000000
W( B2, C2)        3.000000            0.000000
W( B2, C3)        1.000000            0.000000
W( C1, D)        1.000000            0.000000
W( C2, D)        3.000000            0.000000
W( C3, D)        4.000000            0.000000
X( A, B1)        1.000000            0.000000
X( A, B2)        0.000000            0.000000
X( B1, C1)        1.000000            0.000000
X( B1, C2)        0.000000            2.000000
X( B1, C3)        0.000000            1.000000
X( B2, C1)        0.000000            1.000000
X( B2, C2)        0.000000            4.000000
X( B2, C3)        0.000000            3.000000
X( C1, D)        1.000000            0.000000
X( C2, D)        0.000000            0.000000
X( C3, D)        0.000000            0.000000

Row    Slack or Surplus      Dual Price
1        0.000000            0.000000
2        6.000000           -1.000000
3        0.000000           -4.000000
4        0.000000           -2.000000
5        0.000000            0.000000
6        0.000000            1.000000
7        0.000000           -1.000000
8        0.000000           -2.000000
9        0.000000           -2.000000

由此，我们可以很容易得到以下结论：
A和D之间的最短路径为：A→B1→C1→D.