CPLEX 我如何编写此类索引的方程式

Question

我正在尝试在 cplex ide 中编写以下约束。

x_ij +x_ji == 1（汽车i前面的汽车j或汽车j前面的汽车i）对于所有i和j 这样 i<j

他们都被分配了起始位置，bently是1honda是2bmw是3audi是4

我试过的..

{string} cars = ["bently" "honda" "bmw" "audi"];

dvar int x[cars][cars] in 0..1;

    forall(i in cars)
      forall(j in cars:i<j)
        x[i][j] + x[j][i] == 1;

提前致谢..

Answer 1

可以

{string} cars = {"bently" ,"honda" ,"bmw" ,"audi"};

dvar int x[cars][cars] in 0..1;

subject to
{
    forall(ordered  i,j in cars)

        x[i][j] + x[j][i] == 1;

}

帮助？

或

using CP;

{string} cars = {"bently" ,"honda" ,"bmw" ,"audi"};

dvar int position[cars] in 0..3;

subject to
{
    allDifferent(position);

}

?

Answer 2

如果你标记你的约束，你会得到放松，这将帮助你调试：

forall(i in cars)
     ctA:-aps <= ((numcars - sum(i in cars, j in cars:i < j) x[i][j]) - inicarposition[i]);
   forall(i in cars)
     ctB:((numcars - sum(i in cars, j in cars:i < j) x[i][j]) - inicarposition[i]) <= aps;

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