旅行商动态规划lingo代码
时间: 2023-08-04 10:04:15 浏览: 70
下面是一个使用Lingo编写的旅行商动态规划的示例代码:
```
model:
sets:
N /1..n/; // 城市集合
S /1..2**n/; // 子集合集合
parameters:
c(N,N); // 城市之间的距离矩阵
M = 9999; // 一个大的数值,表示无穷大
variables:
x(N,N,S) binary; // 是否从城市i到城市j经过子集合s
u(N,S) integer; // 子集合s中以城市i结尾的最短路径长度
z integer; // 最短路径长度
equations:
obj: z =e= sum((i,j), c(i,j)*x(i,j,2**n));
c1(i): sum((j,s), x(i,j,s)) =e= 1, i =e= 1..n;
c2(s): sum((j), x(1,j,s)) =e= sum((i), x(i,1,s)), s =e= 2..2**(n-1);
c3(i,s): u(i,s) =l= sum((j|s), x(j,i,s)), (i,s) =e= 2..n, 2..2**n;
c4(i): u(i,2**n) =e= 1, i =e= 2..n;
c5(i,s): u(i,s) =g= u(j,s-{i})+x(j,i,s), (i,j,s) =e= 2..n, 2..n, 2..2**n;
c6(i,j,s): x(i,j,s) =l= u(i,s- {j}), (i,j,s) =e= 2..n, 2..n, 2..2**n;
c7(i,j): x(i,j,2**n) =l= x(j,i,2**n) =l= 1, (i,j) =e= 2..n;
c8(i,j,s): x(i,j,s) =e= 0, (i,j,s) =e= 1..n, 1..n, 1..2**n;
c9(i,j): x(i,j,2**n) =e= 0, (i,j) =e= 1..n;
c10(i,j,s): x(i,j,s) + x(j,i,s) =l= 1, (i,j,s) =e= 2..n, 2..n, 1..2**n;
c11(i): u(i,1) =e= 0, i =e= 2..n;
c12(s): sum((i), x(i,i,s)) =e= 0, s =e= 2..2**n;
c13: sum((j), x(1,j,2**n)) =e= 1;
c14: sum((i), x(i,1,2**n)) =e= 1;
binary variables:
x;
integer variables:
u,z;
data:
c = [M, 3, 6, 7;
5, M, 2, 3;
6, 4, M, 2;
3, 7, 5, M]; // 城市之间的距离矩阵
end
```
这是一个简化的旅行商问题,其中城市之间的距离由矩阵`c`表示。动态规划的思想是通过递归地计算子问题的最优解来构建整个问题的最优解。这个代码使用了一系列约束条件来确保每个城市都被访问一次,并且最终返回最短路径的长度。你可以根据自己的需求进行修改和扩展。请注意,这里使用了一个大数M来表示无穷大,你可以根据实际情况进行调整。