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[英]How to solve the Cumulative Traveling Salesman Problem using or-tools in python?
[英]or-tools: Trivial problem is infeasible in cpp, but works in python
我遇到了一個優化問題,在該問題中,瑣碎的可行約束導致我的cpp程序在嘗試解決時返回“不可行”。
為了演示,我創建了一個由3名護士和5個插槽組成的護士時間表優化程序。
我有兩個瑣碎的約束條件:1)第一位護士占用第一個空位,以及2)每個插槽中最多允許一位護士。
一次使用一個約束時,這些約束導致or-tools返回一個可行的解決方案,但是當我同時使用兩個約束時,則得到了一個不可行的解決方案。 即使同時使用了兩個約束,完全相同的問題在python API中也能正常工作。
我懷疑我在設置第一個約束( cp_model.AddEquality(LinearExpr(slots[0][0]), 1);
)時以某種方式濫用了AddEquality
,但是我無法弄清楚問題出在哪里。
請幫忙。
#include <iostream>
#include <vector>
#include "ortools/sat/cp_model.h"
#include "ortools/sat/sat_parameters.pb.h"
namespace operations_research {
namespace sat {
void slots(bool add_sum, bool add_const) {
CpModelBuilder cp_model;
const int num_nurses = 3;
const int num_slots = 5;
std::vector<std::vector<IntVar>> slots(num_nurses);
for (int n = 0; n < num_nurses; n++) {
for (int d = 0; d < num_slots; d++) {
const IntVar var = cp_model.NewIntVar({0, 1});
slots[n].push_back(var);
}
}
if (add_const) {
// trival constraint
cp_model.AddEquality(LinearExpr(slots[0][0]), 1);
}
if (add_sum) {
// make the first row sum to one; should be trivial too
std::vector<IntVar> this_nurse_vals(num_nurses);
for (int n = 0; n < num_nurses; n++) {
const IntVar var = slots[n][0];
this_nurse_vals.push_back(var);
}
cp_model.AddEquality(LinearExpr::Sum(this_nurse_vals), 1);
}
// solve
const CpSolverResponse response = Solve(cp_model.Build());
LOG(INFO) << CpSolverResponseStats(response);
for (int d = 0; d < num_slots; d++) {
for (int n = 0; n < num_nurses; n++) {
std::cout << SolutionIntegerValue(response, slots[n][d]);
}
std::cout << std::endl;
}
std::cout << std::endl;
// [END solve]
}
} // namespace sat
} // namespace operations_research
// ----- MAIN -----
int main(int argc, char **argv) {
operations_research::sat::slots(false, true); // works
operations_research::sat::slots(true, false); // works
operations_research::sat::slots(true, true); // infeasible
return EXIT_SUCCESS;
}
// [END program]
在python中可以正常工作的相同程序:
from ortools.sat.python import cp_model
num_nurses = 3
num_slots = 5
model = cp_model.CpModel()
# make vars
slots = {}
for n in range(num_nurses):
for d in range(num_slots):
slots[(n, d)] = model.NewIntVar(0, 1, "slot")
model.Add(slots[(0, 0)] == 1)
model.Add(sum(slots[(n, 0)] for n in range(num_nurses)) == 1)
solver = cp_model.CpSolver()
solver.Solve(model)
solution = []
for d in range(num_slots):
solution.append([])
for n in range(num_nurses):
solution[d].append(solver.Value(slots[(n, d)]))
print(solution)
你的護士太多了。
這個:
std::vector<IntVar> this_nurse_vals(num_nurses);
用num_nurses
元素創建一個向量。
然后,您將另一個num_nurses
元素push_back
num_nurses
,從而獲得所需的兩倍。
要么以一個空向量開始,然后將push_back
插入其中:
std::vector<IntVar> this_nurse_vals;
for (int n = 0; n < num_nurses; n++) {
this_nurse_vals.push_back(IntVar(slots[n][0]));
}
或以“完整”向量開始並分配給它:
std::vector<IntVar> this_nurse_vals(num_nurses);
for (int n = 0; n < num_nurses; n++) {
this_nurse_vals[n] = IntVar(slots[n][0]);
}
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