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Giải bài tập

Phần này đưa ra một ví dụ cho thấy cách giải một bài tập bằng cách dùng cả trình giải MIP và trình giải CP-SAT.

Ví dụ:

Trong ví dụ này có 5 worker (được đánh số từ 0 đến 4) và bốn công việc (được đánh số) 0 – 3). Lưu ý rằng có một worker nhiều hơn ví dụ trong ví dụ Tổng quan.

Chi phí phân công nhân viên vào các nhiệm vụ được thể hiện trong bảng sau.

Worker	Nhiệm vụ 0	Nhiệm vụ 1	Nhiệm vụ 2	Nhiệm vụ 3
0	90	80	75	70
1	35	85	55	65
2	125	95	90	95
3	45	110	95	115
4	50	100	90	100

Vấn đề là chỉ định mỗi nhân viên cho tối đa một công việc, không có hai nhân viên nào đang hoạt động cùng một công việc, vừa giảm thiểu tổng chi phí. Vì có nhiều nhân viên hơn công việc, thì một worker sẽ không được giao một công việc.

Giải pháp MIP

Các phần sau đây mô tả cách giải bài tập bằng cách sử dụng Trình bao bọc MPSolver.

Nhập thư viện

Mã sau đây nhập các thư viện bắt buộc.

Python

from ortools.linear_solver import pywraplp

C++

#include <memory>
#include <vector>

#include "ortools/base/logging.h"
#include "ortools/linear_solver/linear_solver.h"

Java

import com.google.ortools.Loader;
import com.google.ortools.linearsolver.MPConstraint;
import com.google.ortools.linearsolver.MPObjective;
import com.google.ortools.linearsolver.MPSolver;
import com.google.ortools.linearsolver.MPVariable;

C#

using System;
using Google.OrTools.LinearSolver;

Tạo dữ liệu

Mã sau đây sẽ tạo dữ liệu cho bài toán này.

Python

costs = [
    [90, 80, 75, 70],
    [35, 85, 55, 65],
    [125, 95, 90, 95],
    [45, 110, 95, 115],
    [50, 100, 90, 100],
]
num_workers = len(costs)
num_tasks = len(costs[0])

C++

const std::vector<std::vector<double>> costs{
    {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
    {45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = costs.size();
const int num_tasks = costs[0].size();

Java

double[][] costs = {
    {90, 80, 75, 70},
    {35, 85, 55, 65},
    {125, 95, 90, 95},
    {45, 110, 95, 115},
    {50, 100, 90, 100},
};
int numWorkers = costs.length;
int numTasks = costs[0].length;

C#

int[,] costs = {
    { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);

Mảng costs tương ứng với bảng chi phí để phân công nhân viên vào nhiệm vụ, như đã trình bày ở trên.

Khai báo trình giải mã MIP

Mã sau đây khai báo trình giải MIP.

Python

# Create the mip solver with the SCIP backend.
solver = pywraplp.Solver.CreateSolver("SCIP")

if not solver:
    return

C++

// Create the mip solver with the SCIP backend.
std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP"));
if (!solver) {
  LOG(WARNING) << "SCIP solver unavailable.";
  return;
}

Java

// Create the linear solver with the SCIP backend.
MPSolver solver = MPSolver.createSolver("SCIP");
if (solver == null) {
  System.out.println("Could not create solver SCIP");
  return;
}

C#

Solver solver = Solver.CreateSolver("SCIP");
if (solver is null)
{
    return;
}

Tạo biến

Mã sau đây tạo các biến số nguyên nhị phân cho bài toán này.

Python

# x[i, j] is an array of 0-1 variables, which will be 1
# if worker i is assigned to task j.
x = {}
for i in range(num_workers):
    for j in range(num_tasks):
        x[i, j] = solver.IntVar(0, 1, "")

C++

// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
std::vector<std::vector<const MPVariable*>> x(
    num_workers, std::vector<const MPVariable*>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    x[i][j] = solver->MakeIntVar(0, 1, "");
  }
}

Java

// x[i][j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
MPVariable[][] x = new MPVariable[numWorkers][numTasks];
for (int i = 0; i < numWorkers; ++i) {
  for (int j = 0; j < numTasks; ++j) {
    x[i][j] = solver.makeIntVar(0, 1, "");
  }
}

C#

// x[i, j] is an array of 0-1 variables, which will be 1
// if worker i is assigned to task j.
Variable[,] x = new Variable[numWorkers, numTasks];
for (int i = 0; i < numWorkers; ++i)
{
    for (int j = 0; j < numTasks; ++j)
    {
        x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}");
    }
}

Tạo các quy tắc ràng buộc

Mã sau đây tạo ra các điều kiện ràng buộc cho bài toán này.

Python

# Each worker is assigned to at most 1 task.
for i in range(num_workers):
    solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1)

# Each task is assigned to exactly one worker.
for j in range(num_tasks):
    solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1)

C++

// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
  LinearExpr worker_sum;
  for (int j = 0; j < num_tasks; ++j) {
    worker_sum += x[i][j];
  }
  solver->MakeRowConstraint(worker_sum <= 1.0);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
  LinearExpr task_sum;
  for (int i = 0; i < num_workers; ++i) {
    task_sum += x[i][j];
  }
  solver->MakeRowConstraint(task_sum == 1.0);
}

Java

// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i) {
  MPConstraint constraint = solver.makeConstraint(0, 1, "");
  for (int j = 0; j < numTasks; ++j) {
    constraint.setCoefficient(x[i][j], 1);
  }
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j) {
  MPConstraint constraint = solver.makeConstraint(1, 1, "");
  for (int i = 0; i < numWorkers; ++i) {
    constraint.setCoefficient(x[i][j], 1);
  }
}

C#

// Each worker is assigned to at most one task.
for (int i = 0; i < numWorkers; ++i)
{
    Constraint constraint = solver.MakeConstraint(0, 1, "");
    for (int j = 0; j < numTasks; ++j)
    {
        constraint.SetCoefficient(x[i, j], 1);
    }
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < numTasks; ++j)
{
    Constraint constraint = solver.MakeConstraint(1, 1, "");
    for (int i = 0; i < numWorkers; ++i)
    {
        constraint.SetCoefficient(x[i, j], 1);
    }
}

Tạo hàm mục tiêu

Mã sau đây sẽ tạo hàm mục tiêu cho bài toán này.

Python

objective_terms = []
for i in range(num_workers):
    for j in range(num_tasks):
        objective_terms.append(costs[i][j] * x[i, j])
solver.Minimize(solver.Sum(objective_terms))

C++

MPObjective* const objective = solver->MutableObjective();
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    objective->SetCoefficient(x[i][j], costs[i][j]);
  }
}
objective->SetMinimization();

Java

MPObjective objective = solver.objective();
for (int i = 0; i < numWorkers; ++i) {
  for (int j = 0; j < numTasks; ++j) {
    objective.setCoefficient(x[i][j], costs[i][j]);
  }
}
objective.setMinimization();

C#

Objective objective = solver.Objective();
for (int i = 0; i < numWorkers; ++i)
{
    for (int j = 0; j < numTasks; ++j)
    {
        objective.SetCoefficient(x[i, j], costs[i, j]);
    }
}
objective.SetMinimization();

Giá trị của hàm mục tiêu là tổng chi phí trên tất cả các biến được chỉ định giá trị 1 theo trình giải.

Gọi trình giải

Mã sau đây gọi trình giải.

Python

print(f"Solving with {solver.SolverVersion()}")
status = solver.Solve()

C++

const MPSolver::ResultStatus result_status = solver->Solve();

Java

MPSolver.ResultStatus resultStatus = solver.solve();

C#

Solver.ResultStatus resultStatus = solver.Solve();

In giải pháp

Mã sau đây in giải pháp cho sự cố.

Python

if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE:
    print(f"Total cost = {solver.Objective().Value()}\n")
    for i in range(num_workers):
        for j in range(num_tasks):
            # Test if x[i,j] is 1 (with tolerance for floating point arithmetic).
            if x[i, j].solution_value() > 0.5:
                print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}")
else:
    print("No solution found.")

C++

// Check that the problem has a feasible solution.
if (result_status != MPSolver::OPTIMAL &&
    result_status != MPSolver::FEASIBLE) {
  LOG(FATAL) << "No solution found.";
}

LOG(INFO) << "Total cost = " << objective->Value() << "\n\n";

for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    // Test if x[i][j] is 0 or 1 (with tolerance for floating point
    // arithmetic).
    if (x[i][j]->solution_value() > 0.5) {
      LOG(INFO) << "Worker " << i << " assigned to task " << j
                << ".  Cost = " << costs[i][j];
    }
  }
}

Java

// Check that the problem has a feasible solution.
if (resultStatus == MPSolver.ResultStatus.OPTIMAL
    || resultStatus == MPSolver.ResultStatus.FEASIBLE) {
  System.out.println("Total cost: " + objective.value() + "\n");
  for (int i = 0; i < numWorkers; ++i) {
    for (int j = 0; j < numTasks; ++j) {
      // Test if x[i][j] is 0 or 1 (with tolerance for floating point
      // arithmetic).
      if (x[i][j].solutionValue() > 0.5) {
        System.out.println(
            "Worker " + i + " assigned to task " + j + ".  Cost = " + costs[i][j]);
      }
    }
  }
} else {
  System.err.println("No solution found.");
}

C#

// Check that the problem has a feasible solution.
if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE)
{
    Console.WriteLine($"Total cost: {solver.Objective().Value()}\n");
    for (int i = 0; i < numWorkers; ++i)
    {
        for (int j = 0; j < numTasks; ++j)
        {
            // Test if x[i, j] is 0 or 1 (with tolerance for floating point
            // arithmetic).
            if (x[i, j].SolutionValue() > 0.5)
            {
                Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
            }
        }
    }
}
else
{
    Console.WriteLine("No solution found.");
}

Đây là kết quả của chương trình.

Total cost =  265.0

Worker 0 assigned to task 3.  Cost = 70
Worker 1 assigned to task 2.  Cost = 55
Worker 2 assigned to task 1.  Cost = 95
Worker 3 assigned to task 0.  Cost = 45

Hoàn tất chương trình

Sau đây là các chương trình hoàn chỉnh cho giải pháp MIP.

Python

from ortools.linear_solver import pywraplp


def main():
    # Data
    costs = [
        [90, 80, 75, 70],
        [35, 85, 55, 65],
        [125, 95, 90, 95],
        [45, 110, 95, 115],
        [50, 100, 90, 100],
    ]
    num_workers = len(costs)
    num_tasks = len(costs[0])

    # Solver
    # Create the mip solver with the SCIP backend.
    solver = pywraplp.Solver.CreateSolver("SCIP")

    if not solver:
        return

    # Variables
    # x[i, j] is an array of 0-1 variables, which will be 1
    # if worker i is assigned to task j.
    x = {}
    for i in range(num_workers):
        for j in range(num_tasks):
            x[i, j] = solver.IntVar(0, 1, "")

    # Constraints
    # Each worker is assigned to at most 1 task.
    for i in range(num_workers):
        solver.Add(solver.Sum([x[i, j] for j in range(num_tasks)]) <= 1)

    # Each task is assigned to exactly one worker.
    for j in range(num_tasks):
        solver.Add(solver.Sum([x[i, j] for i in range(num_workers)]) == 1)

    # Objective
    objective_terms = []
    for i in range(num_workers):
        for j in range(num_tasks):
            objective_terms.append(costs[i][j] * x[i, j])
    solver.Minimize(solver.Sum(objective_terms))

    # Solve
    print(f"Solving with {solver.SolverVersion()}")
    status = solver.Solve()

    # Print solution.
    if status == pywraplp.Solver.OPTIMAL or status == pywraplp.Solver.FEASIBLE:
        print(f"Total cost = {solver.Objective().Value()}\n")
        for i in range(num_workers):
            for j in range(num_tasks):
                # Test if x[i,j] is 1 (with tolerance for floating point arithmetic).
                if x[i, j].solution_value() > 0.5:
                    print(f"Worker {i} assigned to task {j}." + f" Cost: {costs[i][j]}")
    else:
        print("No solution found.")


if __name__ == "__main__":
    main()

C++

#include <memory>
#include <vector>

#include "ortools/base/logging.h"
#include "ortools/linear_solver/linear_solver.h"

namespace operations_research {
void AssignmentMip() {
  // Data
  const std::vector<std::vector<double>> costs{
      {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
      {45, 110, 95, 115}, {50, 100, 90, 100},
  };
  const int num_workers = costs.size();
  const int num_tasks = costs[0].size();

  // Solver
  // Create the mip solver with the SCIP backend.
  std::unique_ptr<MPSolver> solver(MPSolver::CreateSolver("SCIP"));
  if (!solver) {
    LOG(WARNING) << "SCIP solver unavailable.";
    return;
  }

  // Variables
  // x[i][j] is an array of 0-1 variables, which will be 1
  // if worker i is assigned to task j.
  std::vector<std::vector<const MPVariable*>> x(
      num_workers, std::vector<const MPVariable*>(num_tasks));
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      x[i][j] = solver->MakeIntVar(0, 1, "");
    }
  }

  // Constraints
  // Each worker is assigned to at most one task.
  for (int i = 0; i < num_workers; ++i) {
    LinearExpr worker_sum;
    for (int j = 0; j < num_tasks; ++j) {
      worker_sum += x[i][j];
    }
    solver->MakeRowConstraint(worker_sum <= 1.0);
  }
  // Each task is assigned to exactly one worker.
  for (int j = 0; j < num_tasks; ++j) {
    LinearExpr task_sum;
    for (int i = 0; i < num_workers; ++i) {
      task_sum += x[i][j];
    }
    solver->MakeRowConstraint(task_sum == 1.0);
  }

  // Objective.
  MPObjective* const objective = solver->MutableObjective();
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      objective->SetCoefficient(x[i][j], costs[i][j]);
    }
  }
  objective->SetMinimization();

  // Solve
  const MPSolver::ResultStatus result_status = solver->Solve();

  // Print solution.
  // Check that the problem has a feasible solution.
  if (result_status != MPSolver::OPTIMAL &&
      result_status != MPSolver::FEASIBLE) {
    LOG(FATAL) << "No solution found.";
  }

  LOG(INFO) << "Total cost = " << objective->Value() << "\n\n";

  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      // Test if x[i][j] is 0 or 1 (with tolerance for floating point
      // arithmetic).
      if (x[i][j]->solution_value() > 0.5) {
        LOG(INFO) << "Worker " << i << " assigned to task " << j
                  << ".  Cost = " << costs[i][j];
      }
    }
  }
}
}  // namespace operations_research

int main(int argc, char** argv) {
  operations_research::AssignmentMip();
  return EXIT_SUCCESS;
}

Java

package com.google.ortools.linearsolver.samples;
import com.google.ortools.Loader;
import com.google.ortools.linearsolver.MPConstraint;
import com.google.ortools.linearsolver.MPObjective;
import com.google.ortools.linearsolver.MPSolver;
import com.google.ortools.linearsolver.MPVariable;

/** MIP example that solves an assignment problem. */
public class AssignmentMip {
  public static void main(String[] args) {
    Loader.loadNativeLibraries();
    // Data
    double[][] costs = {
        {90, 80, 75, 70},
        {35, 85, 55, 65},
        {125, 95, 90, 95},
        {45, 110, 95, 115},
        {50, 100, 90, 100},
    };
    int numWorkers = costs.length;
    int numTasks = costs[0].length;

    // Solver
    // Create the linear solver with the SCIP backend.
    MPSolver solver = MPSolver.createSolver("SCIP");
    if (solver == null) {
      System.out.println("Could not create solver SCIP");
      return;
    }

    // Variables
    // x[i][j] is an array of 0-1 variables, which will be 1
    // if worker i is assigned to task j.
    MPVariable[][] x = new MPVariable[numWorkers][numTasks];
    for (int i = 0; i < numWorkers; ++i) {
      for (int j = 0; j < numTasks; ++j) {
        x[i][j] = solver.makeIntVar(0, 1, "");
      }
    }

    // Constraints
    // Each worker is assigned to at most one task.
    for (int i = 0; i < numWorkers; ++i) {
      MPConstraint constraint = solver.makeConstraint(0, 1, "");
      for (int j = 0; j < numTasks; ++j) {
        constraint.setCoefficient(x[i][j], 1);
      }
    }
    // Each task is assigned to exactly one worker.
    for (int j = 0; j < numTasks; ++j) {
      MPConstraint constraint = solver.makeConstraint(1, 1, "");
      for (int i = 0; i < numWorkers; ++i) {
        constraint.setCoefficient(x[i][j], 1);
      }
    }

    // Objective
    MPObjective objective = solver.objective();
    for (int i = 0; i < numWorkers; ++i) {
      for (int j = 0; j < numTasks; ++j) {
        objective.setCoefficient(x[i][j], costs[i][j]);
      }
    }
    objective.setMinimization();

    // Solve
    MPSolver.ResultStatus resultStatus = solver.solve();

    // Print solution.
    // Check that the problem has a feasible solution.
    if (resultStatus == MPSolver.ResultStatus.OPTIMAL
        || resultStatus == MPSolver.ResultStatus.FEASIBLE) {
      System.out.println("Total cost: " + objective.value() + "\n");
      for (int i = 0; i < numWorkers; ++i) {
        for (int j = 0; j < numTasks; ++j) {
          // Test if x[i][j] is 0 or 1 (with tolerance for floating point
          // arithmetic).
          if (x[i][j].solutionValue() > 0.5) {
            System.out.println(
                "Worker " + i + " assigned to task " + j + ".  Cost = " + costs[i][j]);
          }
        }
      }
    } else {
      System.err.println("No solution found.");
    }
  }

  private AssignmentMip() {}
}

C#

using System;
using Google.OrTools.LinearSolver;

public class AssignmentMip
{
    static void Main()
    {
        // Data.
        int[,] costs = {
            { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
        };
        int numWorkers = costs.GetLength(0);
        int numTasks = costs.GetLength(1);

        // Solver.
        Solver solver = Solver.CreateSolver("SCIP");
        if (solver is null)
        {
            return;
        }

        // Variables.
        // x[i, j] is an array of 0-1 variables, which will be 1
        // if worker i is assigned to task j.
        Variable[,] x = new Variable[numWorkers, numTasks];
        for (int i = 0; i < numWorkers; ++i)
        {
            for (int j = 0; j < numTasks; ++j)
            {
                x[i, j] = solver.MakeIntVar(0, 1, $"worker_{i}_task_{j}");
            }
        }

        // Constraints
        // Each worker is assigned to at most one task.
        for (int i = 0; i < numWorkers; ++i)
        {
            Constraint constraint = solver.MakeConstraint(0, 1, "");
            for (int j = 0; j < numTasks; ++j)
            {
                constraint.SetCoefficient(x[i, j], 1);
            }
        }
        // Each task is assigned to exactly one worker.
        for (int j = 0; j < numTasks; ++j)
        {
            Constraint constraint = solver.MakeConstraint(1, 1, "");
            for (int i = 0; i < numWorkers; ++i)
            {
                constraint.SetCoefficient(x[i, j], 1);
            }
        }

        // Objective
        Objective objective = solver.Objective();
        for (int i = 0; i < numWorkers; ++i)
        {
            for (int j = 0; j < numTasks; ++j)
            {
                objective.SetCoefficient(x[i, j], costs[i, j]);
            }
        }
        objective.SetMinimization();

        // Solve
        Solver.ResultStatus resultStatus = solver.Solve();

        // Print solution.
        // Check that the problem has a feasible solution.
        if (resultStatus == Solver.ResultStatus.OPTIMAL || resultStatus == Solver.ResultStatus.FEASIBLE)
        {
            Console.WriteLine($"Total cost: {solver.Objective().Value()}\n");
            for (int i = 0; i < numWorkers; ++i)
            {
                for (int j = 0; j < numTasks; ++j)
                {
                    // Test if x[i, j] is 0 or 1 (with tolerance for floating point
                    // arithmetic).
                    if (x[i, j].SolutionValue() > 0.5)
                    {
                        Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
                    }
                }
            }
        }
        else
        {
            Console.WriteLine("No solution found.");
        }
    }
}

Giải pháp CP SAT

Các phần sau đây mô tả cách dùng trình giải quyết CP-SAT để giải bài tập này.

Nhập thư viện

Mã sau đây nhập các thư viện bắt buộc.

Python

import io

import pandas as pd

from ortools.sat.python import cp_model

C++

#include <stdlib.h>

#include <vector>

#include "ortools/base/logging.h"
#include "ortools/sat/cp_model.h"
#include "ortools/sat/cp_model.pb.h"
#include "ortools/sat/cp_model_solver.h"

Java

import com.google.ortools.Loader;
import com.google.ortools.sat.CpModel;
import com.google.ortools.sat.CpSolver;
import com.google.ortools.sat.CpSolverStatus;
import com.google.ortools.sat.LinearExpr;
import com.google.ortools.sat.LinearExprBuilder;
import com.google.ortools.sat.Literal;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;

C#

using System;
using System.Collections.Generic;
using Google.OrTools.Sat;

Khai báo mô hình

Mã sau đây khai báo mô hình CP-SAT.

Python

model = cp_model.CpModel()

C++

CpModelBuilder cp_model;

Java

CpModel model = new CpModel();

C#

CpModel model = new CpModel();

Tạo dữ liệu

Mã sau đây thiết lập dữ liệu cho sự cố này.

Python

  data_str = """
worker  task  cost
    w1    t1    90
    w1    t2    80
    w1    t3    75
    w1    t4    70
    w2    t1    35
    w2    t2    85
    w2    t3    55
    w2    t4    65
    w3    t1   125
    w3    t2    95
    w3    t3    90
    w3    t4    95
    w4    t1    45
    w4    t2   110
    w4    t3    95
    w4    t4   115
    w5    t1    50
    w5    t2   110
    w5    t3    90
    w5    t4   100
"""

  data = pd.read_table(io.StringIO(data_str), sep=r"\s+")

C++

const std::vector<std::vector<int>> costs{
    {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
    {45, 110, 95, 115}, {50, 100, 90, 100},
};
const int num_workers = static_cast<int>(costs.size());
const int num_tasks = static_cast<int>(costs[0].size());

Java

int[][] costs = {
    {90, 80, 75, 70},
    {35, 85, 55, 65},
    {125, 95, 90, 95},
    {45, 110, 95, 115},
    {50, 100, 90, 100},
};
final int numWorkers = costs.length;
final int numTasks = costs[0].length;

final int[] allWorkers = IntStream.range(0, numWorkers).toArray();
final int[] allTasks = IntStream.range(0, numTasks).toArray();

C#

int[,] costs = {
    { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
};
int numWorkers = costs.GetLength(0);
int numTasks = costs.GetLength(1);

Mảng costs tương ứng với bảng chi phí để phân công nhân viên vào nhiệm vụ, như đã trình bày ở trên.

Tạo biến

Mã sau đây tạo các biến số nguyên nhị phân cho bài toán này.

Python

x = model.new_bool_var_series(name="x", index=data.index)

C++

// x[i][j] is an array of Boolean variables. x[i][j] is true
// if worker i is assigned to task j.
std::vector<std::vector<BoolVar>> x(num_workers,
                                    std::vector<BoolVar>(num_tasks));
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    x[i][j] = cp_model.NewBoolVar();
  }
}

Java

Literal[][] x = new Literal[numWorkers][numTasks];
for (int worker : allWorkers) {
  for (int task : allTasks) {
    x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]");
  }
}

C#

BoolVar[,] x = new BoolVar[numWorkers, numTasks];
// Variables in a 1-dim array.
for (int worker = 0; worker < numWorkers; ++worker)
{
    for (int task = 0; task < numTasks; ++task)
    {
        x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}");
    }
}

Tạo các quy tắc ràng buộc

Mã sau đây tạo ra các điều kiện ràng buộc cho bài toán này.

Python

# Each worker is assigned to at most one task.
for unused_name, tasks in data.groupby("worker"):
    model.add_at_most_one(x[tasks.index])

# Each task is assigned to exactly one worker.
for unused_name, workers in data.groupby("task"):
    model.add_exactly_one(x[workers.index])

C++

// Each worker is assigned to at most one task.
for (int i = 0; i < num_workers; ++i) {
  cp_model.AddAtMostOne(x[i]);
}
// Each task is assigned to exactly one worker.
for (int j = 0; j < num_tasks; ++j) {
  std::vector<BoolVar> tasks;
  for (int i = 0; i < num_workers; ++i) {
    tasks.push_back(x[i][j]);
  }
  cp_model.AddExactlyOne(tasks);
}

Java

// Each worker is assigned to at most one task.
for (int worker : allWorkers) {
  List<Literal> tasks = new ArrayList<>();
  for (int task : allTasks) {
    tasks.add(x[worker][task]);
  }
  model.addAtMostOne(tasks);
}

// Each task is assigned to exactly one worker.
for (int task : allTasks) {
  List<Literal> workers = new ArrayList<>();
  for (int worker : allWorkers) {
    workers.add(x[worker][task]);
  }
  model.addExactlyOne(workers);
}

C#

// Each worker is assigned to at most one task.
for (int worker = 0; worker < numWorkers; ++worker)
{
    List<ILiteral> tasks = new List<ILiteral>();
    for (int task = 0; task < numTasks; ++task)
    {
        tasks.Add(x[worker, task]);
    }
    model.AddAtMostOne(tasks);
}

// Each task is assigned to exactly one worker.
for (int task = 0; task < numTasks; ++task)
{
    List<ILiteral> workers = new List<ILiteral>();
    for (int worker = 0; worker < numWorkers; ++worker)
    {
        workers.Add(x[worker, task]);
    }
    model.AddExactlyOne(workers);
}

Tạo hàm mục tiêu

Mã sau đây sẽ tạo hàm mục tiêu cho bài toán này.

Python

model.minimize(data.cost.dot(x))

C++

LinearExpr total_cost;
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    total_cost += x[i][j] * costs[i][j];
  }
}
cp_model.Minimize(total_cost);

Java

LinearExprBuilder obj = LinearExpr.newBuilder();
for (int worker : allWorkers) {
  for (int task : allTasks) {
    obj.addTerm(x[worker][task], costs[worker][task]);
  }
}
model.minimize(obj);

C#

LinearExprBuilder obj = LinearExpr.NewBuilder();
for (int worker = 0; worker < numWorkers; ++worker)
{
    for (int task = 0; task < numTasks; ++task)
    {
        obj.AddTerm((IntVar)x[worker, task], costs[worker, task]);
    }
}
model.Minimize(obj);

Giá trị của hàm mục tiêu là tổng chi phí trên tất cả các biến được chỉ định giá trị 1 theo trình giải.

Gọi trình giải

Mã sau đây gọi trình giải.

Python

solver = cp_model.CpSolver()
status = solver.solve(model)

C++

const CpSolverResponse response = Solve(cp_model.Build());

Java

CpSolver solver = new CpSolver();
CpSolverStatus status = solver.solve(model);

C#

CpSolver solver = new CpSolver();
CpSolverStatus status = solver.Solve(model);
Console.WriteLine($"Solve status: {status}");

In giải pháp

Mã sau đây in giải pháp cho sự cố.

Python

if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
    print(f"Total cost = {solver.objective_value}\n")
    selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
    for unused_index, row in selected.iterrows():
        print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
elif status == cp_model.INFEASIBLE:
    print("No solution found")
else:
    print("Something is wrong, check the status and the log of the solve")

C++

if (response.status() == CpSolverStatus::INFEASIBLE) {
  LOG(FATAL) << "No solution found.";
}

LOG(INFO) << "Total cost: " << response.objective_value();
LOG(INFO);
for (int i = 0; i < num_workers; ++i) {
  for (int j = 0; j < num_tasks; ++j) {
    if (SolutionBooleanValue(response, x[i][j])) {
      LOG(INFO) << "Task " << i << " assigned to worker " << j
                << ".  Cost: " << costs[i][j];
    }
  }
}

Java

// Check that the problem has a feasible solution.
if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
  System.out.println("Total cost: " + solver.objectiveValue() + "\n");
  for (int i = 0; i < numWorkers; ++i) {
    for (int j = 0; j < numTasks; ++j) {
      if (solver.booleanValue(x[i][j])) {
        System.out.println(
            "Worker " + i + " assigned to task " + j + ".  Cost: " + costs[i][j]);
      }
    }
  }
} else {
  System.err.println("No solution found.");
}

C#

// Check that the problem has a feasible solution.
if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
{
    Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n");
    for (int i = 0; i < numWorkers; ++i)
    {
        for (int j = 0; j < numTasks; ++j)
        {
            if (solver.Value(x[i, j]) > 0.5)
            {
                Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
            }
        }
    }
}
else
{
    Console.WriteLine("No solution found.");
}

Đây là kết quả của chương trình.

Total cost = 265

Worker  0  assigned to task  3   Cost =  70
Worker  1  assigned to task  2   Cost =  55
Worker  2  assigned to task  1   Cost =  95
Worker  3  assigned to task  0   Cost =  45

Hoàn tất chương trình

Dưới đây là các chương trình hoàn chỉnh cho giải pháp CP-SAT.

Python

import io

import pandas as pd

from ortools.sat.python import cp_model


def main() -> None:
    # Data
    data_str = """
  worker  task  cost
      w1    t1    90
      w1    t2    80
      w1    t3    75
      w1    t4    70
      w2    t1    35
      w2    t2    85
      w2    t3    55
      w2    t4    65
      w3    t1   125
      w3    t2    95
      w3    t3    90
      w3    t4    95
      w4    t1    45
      w4    t2   110
      w4    t3    95
      w4    t4   115
      w5    t1    50
      w5    t2   110
      w5    t3    90
      w5    t4   100
  """

    data = pd.read_table(io.StringIO(data_str), sep=r"\s+")

    # Model
    model = cp_model.CpModel()

    # Variables
    x = model.new_bool_var_series(name="x", index=data.index)

    # Constraints
    # Each worker is assigned to at most one task.
    for unused_name, tasks in data.groupby("worker"):
        model.add_at_most_one(x[tasks.index])

    # Each task is assigned to exactly one worker.
    for unused_name, workers in data.groupby("task"):
        model.add_exactly_one(x[workers.index])

    # Objective
    model.minimize(data.cost.dot(x))

    # Solve
    solver = cp_model.CpSolver()
    status = solver.solve(model)

    # Print solution.
    if status == cp_model.OPTIMAL or status == cp_model.FEASIBLE:
        print(f"Total cost = {solver.objective_value}\n")
        selected = data.loc[solver.boolean_values(x).loc[lambda x: x].index]
        for unused_index, row in selected.iterrows():
            print(f"{row.task} assigned to {row.worker} with a cost of {row.cost}")
    elif status == cp_model.INFEASIBLE:
        print("No solution found")
    else:
        print("Something is wrong, check the status and the log of the solve")


if __name__ == "__main__":
    main()

C++

#include <stdlib.h>

#include <vector>

#include "ortools/base/logging.h"
#include "ortools/sat/cp_model.h"
#include "ortools/sat/cp_model.pb.h"
#include "ortools/sat/cp_model_solver.h"

namespace operations_research {
namespace sat {

void IntegerProgrammingExample() {
  // Data
  const std::vector<std::vector<int>> costs{
      {90, 80, 75, 70},   {35, 85, 55, 65},   {125, 95, 90, 95},
      {45, 110, 95, 115}, {50, 100, 90, 100},
  };
  const int num_workers = static_cast<int>(costs.size());
  const int num_tasks = static_cast<int>(costs[0].size());

  // Model
  CpModelBuilder cp_model;

  // Variables
  // x[i][j] is an array of Boolean variables. x[i][j] is true
  // if worker i is assigned to task j.
  std::vector<std::vector<BoolVar>> x(num_workers,
                                      std::vector<BoolVar>(num_tasks));
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      x[i][j] = cp_model.NewBoolVar();
    }
  }

  // Constraints
  // Each worker is assigned to at most one task.
  for (int i = 0; i < num_workers; ++i) {
    cp_model.AddAtMostOne(x[i]);
  }
  // Each task is assigned to exactly one worker.
  for (int j = 0; j < num_tasks; ++j) {
    std::vector<BoolVar> tasks;
    for (int i = 0; i < num_workers; ++i) {
      tasks.push_back(x[i][j]);
    }
    cp_model.AddExactlyOne(tasks);
  }

  // Objective
  LinearExpr total_cost;
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      total_cost += x[i][j] * costs[i][j];
    }
  }
  cp_model.Minimize(total_cost);

  // Solve
  const CpSolverResponse response = Solve(cp_model.Build());

  // Print solution.
  if (response.status() == CpSolverStatus::INFEASIBLE) {
    LOG(FATAL) << "No solution found.";
  }

  LOG(INFO) << "Total cost: " << response.objective_value();
  LOG(INFO);
  for (int i = 0; i < num_workers; ++i) {
    for (int j = 0; j < num_tasks; ++j) {
      if (SolutionBooleanValue(response, x[i][j])) {
        LOG(INFO) << "Task " << i << " assigned to worker " << j
                  << ".  Cost: " << costs[i][j];
      }
    }
  }
}
}  // namespace sat
}  // namespace operations_research

int main(int argc, char** argv) {
  operations_research::sat::IntegerProgrammingExample();
  return EXIT_SUCCESS;
}

Java

package com.google.ortools.sat.samples;
import com.google.ortools.Loader;
import com.google.ortools.sat.CpModel;
import com.google.ortools.sat.CpSolver;
import com.google.ortools.sat.CpSolverStatus;
import com.google.ortools.sat.LinearExpr;
import com.google.ortools.sat.LinearExprBuilder;
import com.google.ortools.sat.Literal;
import java.util.ArrayList;
import java.util.List;
import java.util.stream.IntStream;

/** Assignment problem. */
public class AssignmentSat {
  public static void main(String[] args) {
    Loader.loadNativeLibraries();
    // Data
    int[][] costs = {
        {90, 80, 75, 70},
        {35, 85, 55, 65},
        {125, 95, 90, 95},
        {45, 110, 95, 115},
        {50, 100, 90, 100},
    };
    final int numWorkers = costs.length;
    final int numTasks = costs[0].length;

    final int[] allWorkers = IntStream.range(0, numWorkers).toArray();
    final int[] allTasks = IntStream.range(0, numTasks).toArray();

    // Model
    CpModel model = new CpModel();

    // Variables
    Literal[][] x = new Literal[numWorkers][numTasks];
    for (int worker : allWorkers) {
      for (int task : allTasks) {
        x[worker][task] = model.newBoolVar("x[" + worker + "," + task + "]");
      }
    }

    // Constraints
    // Each worker is assigned to at most one task.
    for (int worker : allWorkers) {
      List<Literal> tasks = new ArrayList<>();
      for (int task : allTasks) {
        tasks.add(x[worker][task]);
      }
      model.addAtMostOne(tasks);
    }

    // Each task is assigned to exactly one worker.
    for (int task : allTasks) {
      List<Literal> workers = new ArrayList<>();
      for (int worker : allWorkers) {
        workers.add(x[worker][task]);
      }
      model.addExactlyOne(workers);
    }

    // Objective
    LinearExprBuilder obj = LinearExpr.newBuilder();
    for (int worker : allWorkers) {
      for (int task : allTasks) {
        obj.addTerm(x[worker][task], costs[worker][task]);
      }
    }
    model.minimize(obj);

    // Solve
    CpSolver solver = new CpSolver();
    CpSolverStatus status = solver.solve(model);

    // Print solution.
    // Check that the problem has a feasible solution.
    if (status == CpSolverStatus.OPTIMAL || status == CpSolverStatus.FEASIBLE) {
      System.out.println("Total cost: " + solver.objectiveValue() + "\n");
      for (int i = 0; i < numWorkers; ++i) {
        for (int j = 0; j < numTasks; ++j) {
          if (solver.booleanValue(x[i][j])) {
            System.out.println(
                "Worker " + i + " assigned to task " + j + ".  Cost: " + costs[i][j]);
          }
        }
      }
    } else {
      System.err.println("No solution found.");
    }
  }

  private AssignmentSat() {}
}

C#

using System;
using System.Collections.Generic;
using Google.OrTools.Sat;

public class AssignmentSat
{
    public static void Main(String[] args)
    {
        // Data.
        int[,] costs = {
            { 90, 80, 75, 70 }, { 35, 85, 55, 65 }, { 125, 95, 90, 95 }, { 45, 110, 95, 115 }, { 50, 100, 90, 100 },
        };
        int numWorkers = costs.GetLength(0);
        int numTasks = costs.GetLength(1);

        // Model.
        CpModel model = new CpModel();

        // Variables.
        BoolVar[,] x = new BoolVar[numWorkers, numTasks];
        // Variables in a 1-dim array.
        for (int worker = 0; worker < numWorkers; ++worker)
        {
            for (int task = 0; task < numTasks; ++task)
            {
                x[worker, task] = model.NewBoolVar($"worker_{worker}_task_{task}");
            }
        }

        // Constraints
        // Each worker is assigned to at most one task.
        for (int worker = 0; worker < numWorkers; ++worker)
        {
            List<ILiteral> tasks = new List<ILiteral>();
            for (int task = 0; task < numTasks; ++task)
            {
                tasks.Add(x[worker, task]);
            }
            model.AddAtMostOne(tasks);
        }

        // Each task is assigned to exactly one worker.
        for (int task = 0; task < numTasks; ++task)
        {
            List<ILiteral> workers = new List<ILiteral>();
            for (int worker = 0; worker < numWorkers; ++worker)
            {
                workers.Add(x[worker, task]);
            }
            model.AddExactlyOne(workers);
        }

        // Objective
        LinearExprBuilder obj = LinearExpr.NewBuilder();
        for (int worker = 0; worker < numWorkers; ++worker)
        {
            for (int task = 0; task < numTasks; ++task)
            {
                obj.AddTerm((IntVar)x[worker, task], costs[worker, task]);
            }
        }
        model.Minimize(obj);

        // Solve
        CpSolver solver = new CpSolver();
        CpSolverStatus status = solver.Solve(model);
        Console.WriteLine($"Solve status: {status}");

        // Print solution.
        // Check that the problem has a feasible solution.
        if (status == CpSolverStatus.Optimal || status == CpSolverStatus.Feasible)
        {
            Console.WriteLine($"Total cost: {solver.ObjectiveValue}\n");
            for (int i = 0; i < numWorkers; ++i)
            {
                for (int j = 0; j < numTasks; ++j)
                {
                    if (solver.Value(x[i, j]) > 0.5)
                    {
                        Console.WriteLine($"Worker {i} assigned to task {j}. Cost: {costs[i, j]}");
                    }
                }
            }
        }
        else
        {
            Console.WriteLine("No solution found.");
        }

        Console.WriteLine("Statistics");
        Console.WriteLine($"  - conflicts : {solver.NumConflicts()}");
        Console.WriteLine($"  - branches  : {solver.NumBranches()}");
        Console.WriteLine($"  - wall time : {solver.WallTime()}s");
    }
}