Microsoft Solver Foundation Services声明性语法

时间:2009-09-09 17:14:27

标签: c# ms-solver-foundation

我有以下简单的问题,我想用它来试验[MS Solver Foundation] [1]:

我有10个插槽需要填充1到5范围内的整数。我想强制执行两个约束:

  • slot [n]!= slot [n + 1]
  • 所有广告位的总和应超过20

我可以简单地创建以下决定:

Decision s1 = new Decision(Domain.IntegerRange(1, 5), "slot1");
Decision s2 = new Decision(Domain.IntegerRange(1, 5), "slot2");
Decision s3 = new Decision(Domain.IntegerRange(1, 5), "slot3");
Decision s4 = new Decision(Domain.IntegerRange(1, 5), "slot4");
Decision s5 = new Decision(Domain.IntegerRange(1, 5), "slot5");
Decision s6 = new Decision(Domain.IntegerRange(1, 5), "slot6");
Decision s7 = new Decision(Domain.IntegerRange(1, 5), "slot7");
Decision s8 = new Decision(Domain.IntegerRange(1, 5), "slot8");
Decision s9 = new Decision(Domain.IntegerRange(1, 5), "slot9");
Decision s10 = new Decision(Domain.IntegerRange(1, 5), "slot10");

然后手动设置约束,如

model.AddConstraints("neighbors not equal",
               s1 != s2, s2 != s3, s3 != s4, s4 != s5,
               s5 != s6, s6 != s7, s7!= s8, s8 != s9, s9 != s10
               );

model.AddConstraint("sum",
              s1 + s2 + s3 + s4 + s5 + s6 + s7 + s8 + s9 + s10 > 20 );

但是,我必须想象有一种更好的方法可以做到这一点 - 希望有更类似于声明性语法的东西。

2 个答案:

答案 0 :(得分:2)

代码。

SolverContext context = SolverContext.GetContext();
Model model = context.CreateModel();

Decision[] slot = new Decision[10];

for (int i = 0; i < slot.Length; i++)
{
    slot[i]  = new Decision(Domain.IntegerRange(1, 5), "slot" + i.ToString());
    model.AddDecision(slot[i]);
    if (i > 0) model.AddConstraint("neighbors not equal", slot[i-1] != slot[i]);
}

model.AddConstraint("sum", Model.Sum(slot) > 20);

Solution solution = context.Solve();

答案 1 :(得分:-1)

讨论已移至我们的官方论坛:

http://code.msdn.microsoft.com/solverfoundation/Thread/View.aspx?ThreadId=2256

冷凝管

相关问题