本文整理汇总了C#中Solver.Fail方法的典型用法代码示例。如果您正苦于以下问题：C# Solver.Fail方法的具体用法？C# Solver.Fail怎么用？C# Solver.Fail使用的例子？那么恭喜您, 这里精选的方法代码示例或许可以为您提供帮助。您也可以进一步了解该方法所在类Solver的用法示例。
在下文中一共展示了Solver.Fail方法的1个代码示例，这些例子默认根据受欢迎程度排序。您可以为喜欢或者感觉有用的代码点赞，您的评价将有助于系统推荐出更棒的C#代码示例。
示例1: Next
public override Decision Next(Solver solver)
{
int large = 100000;
int number_of_variables = vars_.Length;
long last_slot = last_slot_var_.Max();
// Lets build a bipartite graph with equal number of nodes left and right.
// Variables will be on the left, slots on the right.
// We will add dummy variables when needed.
// Arcs will have a cost x is slot x is possible for a variable, a large
// number otherwise. For dummy variables, the cost will be 0 always.
LinearSumAssignment matching = new LinearSumAssignment();
for (int speaker = 0; speaker < number_of_variables; ++speaker)
{
IntVar var = vars_[speaker];
for (int value = first_slot_; value <= last_slot; ++value)
{
if (var.Contains(value))
{
matching.AddArcWithCost(speaker, value - first_slot_, value);
}
else
{
matching.AddArcWithCost(speaker, value - first_slot_, large);
}
}
}
// The Matching algorithms expect the same number of left and right nodes.
// So we fill the rest with dense zero-cost arcs.
for (int dummy = number_of_variables;
dummy <= last_slot - first_slot_; ++dummy) {
for (int value = first_slot_; value <= last_slot; ++value)
{
matching.AddArcWithCost(dummy, value - first_slot_, 0);
}
}
if (matching.Solve() == LinearSumAssignment.OPTIMAL &&
matching.OptimalCost() < large) // No violated arcs.
{
for (int speaker = 0; speaker < number_of_variables; ++speaker)
{
vars_[speaker].SetValue(matching.RightMate(speaker) + first_slot_);
}
} else {
solver.Fail();
}
return null;
}