我正在使用 choco 求解器库来生成一组谜题。我需要运行求解器，检查有多少个解决方案，如果有多个，添加一个额外的约束。重复这将给我一组具有独特解决方案的约束（线索）。
但是，一旦我运行 model.getSolver(findAllSolutions()) 任何其他检查都会返回零解决方案。
我猜我需要以某种方式重置模型求解器，但找不到实现这一点的方法 - 如果必须，我宁愿不生成新模型并重新创建现有约束。
原始代码有 110 个 IntVar 和大量约束，但我创建了一个小得多的示例。
注意：在实际应用程序中，我使用 model.getSolver().findAllSolutions(new SolutionCounter(model,2)) 来加快速度，但我在这里省略了该步骤。
Model model = new Model();
// setup two doors A and B, one has the value 0 the other 1
IntVar doorA = model.intVar("Door A", 0, 1);
IntVar doorB = model.intVar("Door B", 0, 1);
model.allDifferent(new IntVar[]{doorA, doorB}).post();
// setup two windows A and B, one has the value 0 the other 1
IntVar windowA = model.intVar("Window A", 0, 1);
IntVar windowB = model.intVar("Window B", 0, 1);
model.allDifferent(new IntVar[]{windowA, windowB}).post();
// assign the first constraint and count the solutions
model.arithm(doorA,"=",0).post();
// this should force door B to be 1 - there are two remaining solutions
List<Solution> solutions = model.getSolver().findAllSolutions();
System.out.println("results after first clue");
for (Solution s : solutions) {
System.out.println(">"+s.toString());
}
assertEquals("First clue leaves two solutions",2,solutions.size());
// add second clue
model.arithm(windowA,"=",1).post();
// this should force window B to by 0 - only one valid solution
List<Solution> solutions2 = model.getSolver().findAllSolutions();
System.out.println("results after second clue");
for (Solution s : solutions2) {
System.out.println(">"+s.toString());
}
assertEquals("Second clue leaves one solution",1,solutions2.size());