التعيين كمشكلة الحد الأدنى من التكلفة

يمكنك استخدام أداة حلّ الحدّ الأدنى للتكلفة لحلّ الحالات الخاصة من مشكلة تعيين.

في الواقع، يمكن أن يعرض تدفق الحد الأدنى للتكلفة حلاً أسرع من أداة حل MIP أو CP-SAT. ومع ذلك، يمكن أن يحل MIP وCP-SAT فئة أكبر من المشكلات من الحد الأدنى لتدفق التكلفة، لذلك في معظم الحالات، فإن MIP أو CP-SAT هي أفضل الخيارات.

تعرض الأقسام التالية برامج بايثون تحل مشكلات التعيين التالية باستخدام أداة حل الحد الأدنى لتدفق التكلفة:

• مثال على الحد الأدنى للمهمة الخطية.
• مشكلة في المهام الدراسية مع فِرق العاملين.

مثال على التعيين الخطي

يوضح هذا القسم كيفية حل المثال الموضح في القسم أداة حل المهام الخطية، باعتبارها مشكلة تدفّق أدنى تكلفة.

استيراد المكتبات

يستورد الرمز التالي المكتبة المطلوبة.

Python

from ortools.graph.python import min_cost_flow

C++

#include <cstdint>
#include <vector>

#include "ortools/graph/min_cost_flow.h"

Java

import com.google.ortools.Loader;
import com.google.ortools.graph.MinCostFlow;
import com.google.ortools.graph.MinCostFlowBase;

C#

using System;
using Google.OrTools.Graph;

تعريف أداة الحلّ

تنشئ التعليمة البرمجية التالية أداة حل الحد الأدنى لتدفق التكلفة.

Python

# Instantiate a SimpleMinCostFlow solver.
smcf = min_cost_flow.SimpleMinCostFlow()

C++

// Instantiate a SimpleMinCostFlow solver.
SimpleMinCostFlow min_cost_flow;

Java

// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();

C#

// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();

إنشاء البيانات

يتكون مخطط التدفق للمشكلة من رسم بياني ثنائي الأجزاء لمصفوفة التكلفة (راجع نظرة عامة على المهمة للاطّلاع على مثال مختلف قليلاً)، مع إضافة المصدر والمستودع.

ملاحظة: يختلف ترقيم العاملين والمهام قليلاً مقارنةً بالقسم أداة حل المهام الخطية، لأن أداة حل الحد الأدنى للتكاليف تتطلب ترقيم جميع العُقد في الرسم البياني بأرقام مختلفة.

تحتوي البيانات على الصفائف الأربع التالية، المقابلة لعُقد البداية وعُقد النهاية والسعة وتكاليف المشكلة. طول كل صفيف هو عدد الأقواس في الرسم البياني.

Python

# Define the directed graph for the flow.
start_nodes = (
    [0, 0, 0, 0] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4] + [5, 6, 7, 8]
)
end_nodes = (
    [1, 2, 3, 4] + [5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8] + [9, 9, 9, 9]
)
capacities = (
    [1, 1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1]
)
costs = (
    [0, 0, 0, 0]
    + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115]
    + [0, 0, 0, 0]
)

source = 0
sink = 9
tasks = 4
supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]

C++

// Define four parallel arrays: sources, destinations, capacities,
// and unit costs between each pair. For instance, the arc from node 0
// to node 1 has a capacity of 15.
const std::vector<int64_t> start_nodes = {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2,
                                          3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
const std::vector<int64_t> end_nodes = {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8,
                                        5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
const std::vector<int64_t> capacities = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                                         1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
const std::vector<int64_t> unit_costs = {0,  0,   0,  0,   90,  76, 75, 70,
                                         35, 85,  55, 65,  125, 95, 90, 105,
                                         45, 110, 95, 115, 0,   0,  0,  0};

const int64_t source = 0;
const int64_t sink = 9;
const int64_t tasks = 4;
// Define an array of supplies at each node.
const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};

Java

// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes =
    new int[] {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
int[] endNodes =
    new int[] {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
int[] capacities =
    new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int[] unitCosts = new int[] {
    0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0};

int source = 0;
int sink = 9;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};

C#

// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes = { 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8 };
int[] endNodes = { 1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9 };
int[] capacities = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
int[] unitCosts = { 0,   0,  0,  0,   90, 76,  75, 70,  35, 85, 55, 65,
                    125, 95, 90, 105, 45, 110, 95, 115, 0,  0,  0,  0 };

int source = 0;
int sink = 9;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };

لتوضيح كيفية إعداد البيانات، يتم تقسيم كل صفيف إلى ثلاث مصفوفات فرعية:

• يتجاوب الصفيف الأول مع الأقواس التي تخرج من المصدر.
• تتجاوب الصفيفة الثانية مع الأقواس بين العاملين والمهام. بالنسبة إلى costs، تمثّل هذه القيمة مصفوفة التكلفة (التي تستخدمها أداة حلّ المهام الخطية)، وتكون مسطّحة في شكل متجه.
• وتتجاوب الصفيفة الثالثة مع الأقواس المؤدية إلى الحوض.

تتضمن البيانات أيضًا الخط المتجه supplies، الذي يعطي المعلومات في كل عقدة.

كيف تمثل مشكلة تدفق الحد الأدنى للتكلفة مشكلة مهمة

كيف تمثل مشكلة تدفق الحد الأدنى للتكلفة أعلاه مشكلة مهمة؟ أولاً، نظرًا لأن قدرة كل قوس تبلغ 1، فإن توفير 4 في المصدر يدفع كل من الأقواس الأربعة المؤدية إلى العمّال إلى الحصول على تدفق بقيمة 1.

بعد ذلك، تفرض حالة "التدفق في تساوي التدفق للخارج" أن يكون التدفق خارج كل عامل هو 1. وإذا أمكن، ستوجّه أداة الحلّ هذه التدفق إلى الحدّ الأدنى للتكلفة والذي يخرج من كل عامل. ومع ذلك، لا يمكن للأداة الحل توجيه التدفقات من عاملين مختلفين إلى مهمة واحدة. وإذا كان الأمر كذلك، فسيكون هناك تدفق مجمّع 2 في تلك المهمة، والذي لا يمكن إرساله عبر القوس الواحد بالسعة 1 من المهمة إلى الحوض. هذا يعني أنّ أداة الحلّ لا يمكنها تعيين مهمة إلا لعامل واحد، على النحو الذي تتطلبه مشكلة المهمة.

أخيرًا، يفرض شرط "التدفق في-يساوي التدفق والخارج" أن يكون لكل مهمة تدفق خارجي 1، لذلك يتم أداء كل مهمة بواسطة بعض العمال.

إنشاء الرسم البياني والقيود

يُنشئ الرمز التالي الرسم البياني والقيود.

Python

# Add each arc.
for i in range(len(start_nodes)):
    smcf.add_arc_with_capacity_and_unit_cost(
        start_nodes[i], end_nodes[i], capacities[i], costs[i]
    )
# Add node supplies.
for i in range(len(supplies)):
    smcf.set_node_supply(i, supplies[i])

C++

// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
  int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
      start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
  if (arc != i) LOG(FATAL) << "Internal error";
}

// Add node supplies.
for (int i = 0; i < supplies.size(); ++i) {
  min_cost_flow.SetNodeSupply(i, supplies[i]);
}

Java

// Add each arc.
for (int i = 0; i < startNodes.length; ++i) {
  int arc = minCostFlow.addArcWithCapacityAndUnitCost(
      startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
  if (arc != i) {
    throw new Exception("Internal error");
  }
}

// Add node supplies.
for (int i = 0; i < supplies.length; ++i) {
  minCostFlow.setNodeSupply(i, supplies[i]);
}

C#

// Add each arc.
for (int i = 0; i < startNodes.Length; ++i)
{
    int arc =
        minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
    if (arc != i)
        throw new Exception("Internal error");
}

// Add node supplies.
for (int i = 0; i < supplies.Length; ++i)
{
    minCostFlow.SetNodeSupply(i, supplies[i]);
}

استدعاء أداة الحلّ

يستدعي الرمز التالي أداة الحلّ ويعرض الحلّ.

Python

# Find the minimum cost flow between node 0 and node 10.
status = smcf.solve()

C++

// Find the min cost flow.
int status = min_cost_flow.Solve();

Java

// Find the min cost flow.
MinCostFlowBase.Status status = minCostFlow.solve();

C#

// Find the min cost flow.
MinCostFlow.Status status = minCostFlow.Solve();

يتكوّن الحل من أقواس بين العاملين والمهام التي تم تخصيص تدفقها بقيمة 1 بواسطة أداة الحلّ. (الأقواس المتصلة بالمصدر أو الحوض ليست جزءًا من الحل).

يفحص البرنامج كل قوس لمعرفة ما إذا كان يحتوي على التدفق 1، وإذا كان كذلك، يطبع Tail (عقدة البداية) وHead (عقدة النهاية) للقوس، والتي تقابل العامل والمهمة في المهمة.

نتيجة البرنامج

Python

if status == smcf.OPTIMAL:
    print("Total cost = ", smcf.optimal_cost())
    print()
    for arc in range(smcf.num_arcs()):
        # Can ignore arcs leading out of source or into sink.
        if smcf.tail(arc) != source and smcf.head(arc) != sink:
            # Arcs in the solution have a flow value of 1. Their start and end nodes
            # give an assignment of worker to task.
            if smcf.flow(arc) > 0:
                print(
                    "Worker %d assigned to task %d.  Cost = %d"
                    % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
                )
else:
    print("There was an issue with the min cost flow input.")
    print(f"Status: {status}")

C++

if (status == MinCostFlow::OPTIMAL) {
  LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
  LOG(INFO) << "";
  for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
    // Can ignore arcs leading out of source or into sink.
    if (min_cost_flow.Tail(i) != source && min_cost_flow.Head(i) != sink) {
      // Arcs in the solution have a flow value of 1. Their start and end
      // nodes give an assignment of worker to task.
      if (min_cost_flow.Flow(i) > 0) {
        LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
                  << " assigned to task " << min_cost_flow.Head(i)
                  << " Cost: " << min_cost_flow.UnitCost(i);
      }
    }
  }
} else {
  LOG(INFO) << "Solving the min cost flow problem failed.";
  LOG(INFO) << "Solver status: " << status;
}

Java

if (status == MinCostFlow.Status.OPTIMAL) {
  System.out.println("Total cost: " + minCostFlow.getOptimalCost());
  System.out.println();
  for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
    // Can ignore arcs leading out of source or into sink.
    if (minCostFlow.getTail(i) != source && minCostFlow.getHead(i) != sink) {
      // Arcs in the solution have a flow value of 1. Their start and end nodes
      // give an assignment of worker to task.
      if (minCostFlow.getFlow(i) > 0) {
        System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
            + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
      }
    }
  }
} else {
  System.out.println("Solving the min cost flow problem failed.");
  System.out.println("Solver status: " + status);
}

C#

if (status == MinCostFlow.Status.OPTIMAL)
{
    Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
    Console.WriteLine("");
    for (int i = 0; i < minCostFlow.NumArcs(); ++i)
    {
        // Can ignore arcs leading out of source or into sink.
        if (minCostFlow.Tail(i) != source && minCostFlow.Head(i) != sink)
        {
            // Arcs in the solution have a flow value of 1. Their start and end nodes
            // give an assignment of worker to task.
            if (minCostFlow.Flow(i) > 0)
            {
                Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
                                  " Cost: " + minCostFlow.UnitCost(i));
            }
        }
    }
}
else
{
    Console.WriteLine("Solving the min cost flow problem failed.");
    Console.WriteLine("Solver status: " + status);
}

إليك ناتج البرنامج.

Total cost = 265

Worker 1 assigned to task 8.  Cost = 70
Worker 2 assigned to task 7.  Cost = 55
Worker 3 assigned to task 6.  Cost = 95
Worker 4 assigned to task 5.  Cost = 45

Time = 0.000245 seconds

وتكون النتيجة مماثلة للنتائج الخاصة بأداة حلّ المهام الخطية (باستثناء ترقيم العاملين والتكاليف). تعد أداة حل التعيينات الخطية أسرع قليلاً من تدفق الحد الأدنى للتكلفة - 0.000147 ثانية مقابل 0.000458 ثانية.

البرنامج بأكمله

يظهر البرنامج بأكمله أدناه.

Python

"""Linear assignment example."""
from ortools.graph.python import min_cost_flow


def main():
    """Solving an Assignment Problem with MinCostFlow."""
    # Instantiate a SimpleMinCostFlow solver.
    smcf = min_cost_flow.SimpleMinCostFlow()

    # Define the directed graph for the flow.
    start_nodes = (
        [0, 0, 0, 0] + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4] + [5, 6, 7, 8]
    )
    end_nodes = (
        [1, 2, 3, 4] + [5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8] + [9, 9, 9, 9]
    )
    capacities = (
        [1, 1, 1, 1] + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1] + [1, 1, 1, 1]
    )
    costs = (
        [0, 0, 0, 0]
        + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115]
        + [0, 0, 0, 0]
    )

    source = 0
    sink = 9
    tasks = 4
    supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]

    # Add each arc.
    for i in range(len(start_nodes)):
        smcf.add_arc_with_capacity_and_unit_cost(
            start_nodes[i], end_nodes[i], capacities[i], costs[i]
        )
    # Add node supplies.
    for i in range(len(supplies)):
        smcf.set_node_supply(i, supplies[i])

    # Find the minimum cost flow between node 0 and node 10.
    status = smcf.solve()

    if status == smcf.OPTIMAL:
        print("Total cost = ", smcf.optimal_cost())
        print()
        for arc in range(smcf.num_arcs()):
            # Can ignore arcs leading out of source or into sink.
            if smcf.tail(arc) != source and smcf.head(arc) != sink:
                # Arcs in the solution have a flow value of 1. Their start and end nodes
                # give an assignment of worker to task.
                if smcf.flow(arc) > 0:
                    print(
                        "Worker %d assigned to task %d.  Cost = %d"
                        % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
                    )
    else:
        print("There was an issue with the min cost flow input.")
        print(f"Status: {status}")


if __name__ == "__main__":
    main()

C++

#include <cstdint>
#include <vector>

#include "ortools/graph/min_cost_flow.h"

namespace operations_research {
// MinCostFlow simple interface example.
void AssignmentMinFlow() {
  // Instantiate a SimpleMinCostFlow solver.
  SimpleMinCostFlow min_cost_flow;

  // Define four parallel arrays: sources, destinations, capacities,
  // and unit costs between each pair. For instance, the arc from node 0
  // to node 1 has a capacity of 15.
  const std::vector<int64_t> start_nodes = {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2,
                                            3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
  const std::vector<int64_t> end_nodes = {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8,
                                          5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
  const std::vector<int64_t> capacities = {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                                           1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
  const std::vector<int64_t> unit_costs = {0,  0,   0,  0,   90,  76, 75, 70,
                                           35, 85,  55, 65,  125, 95, 90, 105,
                                           45, 110, 95, 115, 0,   0,  0,  0};

  const int64_t source = 0;
  const int64_t sink = 9;
  const int64_t tasks = 4;
  // Define an array of supplies at each node.
  const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};

  // Add each arc.
  for (int i = 0; i < start_nodes.size(); ++i) {
    int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
        start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
    if (arc != i) LOG(FATAL) << "Internal error";
  }

  // Add node supplies.
  for (int i = 0; i < supplies.size(); ++i) {
    min_cost_flow.SetNodeSupply(i, supplies[i]);
  }

  // Find the min cost flow.
  int status = min_cost_flow.Solve();

  if (status == MinCostFlow::OPTIMAL) {
    LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
    LOG(INFO) << "";
    for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
      // Can ignore arcs leading out of source or into sink.
      if (min_cost_flow.Tail(i) != source && min_cost_flow.Head(i) != sink) {
        // Arcs in the solution have a flow value of 1. Their start and end
        // nodes give an assignment of worker to task.
        if (min_cost_flow.Flow(i) > 0) {
          LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
                    << " assigned to task " << min_cost_flow.Head(i)
                    << " Cost: " << min_cost_flow.UnitCost(i);
        }
      }
    }
  } else {
    LOG(INFO) << "Solving the min cost flow problem failed.";
    LOG(INFO) << "Solver status: " << status;
  }
}

}  // namespace operations_research

int main() {
  operations_research::AssignmentMinFlow();
  return EXIT_SUCCESS;
}

Java

package com.google.ortools.graph.samples;
import com.google.ortools.Loader;
import com.google.ortools.graph.MinCostFlow;
import com.google.ortools.graph.MinCostFlowBase;

/** Minimal Assignment Min Flow. */
public class AssignmentMinFlow {
  public static void main(String[] args) throws Exception {
    Loader.loadNativeLibraries();
    // Instantiate a SimpleMinCostFlow solver.
    MinCostFlow minCostFlow = new MinCostFlow();

    // Define four parallel arrays: sources, destinations, capacities, and unit costs
    // between each pair.
    int[] startNodes =
        new int[] {0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8};
    int[] endNodes =
        new int[] {1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9};
    int[] capacities =
        new int[] {1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
    int[] unitCosts = new int[] {
        0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 0, 0, 0, 0};

    int source = 0;
    int sink = 9;
    int tasks = 4;
    // Define an array of supplies at each node.
    int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};

    // Add each arc.
    for (int i = 0; i < startNodes.length; ++i) {
      int arc = minCostFlow.addArcWithCapacityAndUnitCost(
          startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
      if (arc != i) {
        throw new Exception("Internal error");
      }
    }

    // Add node supplies.
    for (int i = 0; i < supplies.length; ++i) {
      minCostFlow.setNodeSupply(i, supplies[i]);
    }

    // Find the min cost flow.
    MinCostFlowBase.Status status = minCostFlow.solve();

    if (status == MinCostFlow.Status.OPTIMAL) {
      System.out.println("Total cost: " + minCostFlow.getOptimalCost());
      System.out.println();
      for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
        // Can ignore arcs leading out of source or into sink.
        if (minCostFlow.getTail(i) != source && minCostFlow.getHead(i) != sink) {
          // Arcs in the solution have a flow value of 1. Their start and end nodes
          // give an assignment of worker to task.
          if (minCostFlow.getFlow(i) > 0) {
            System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
                + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
          }
        }
      }
    } else {
      System.out.println("Solving the min cost flow problem failed.");
      System.out.println("Solver status: " + status);
    }
  }

  private AssignmentMinFlow() {}
}

C#

using System;
using Google.OrTools.Graph;

public class AssignmentMinFlow
{
    static void Main()
    {
        // Instantiate a SimpleMinCostFlow solver.
        MinCostFlow minCostFlow = new MinCostFlow();

        // Define four parallel arrays: sources, destinations, capacities, and unit costs
        // between each pair.
        int[] startNodes = { 0, 0, 0, 0, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 6, 7, 8 };
        int[] endNodes = { 1, 2, 3, 4, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 5, 6, 7, 8, 9, 9, 9, 9 };
        int[] capacities = { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
        int[] unitCosts = { 0,   0,  0,  0,   90, 76,  75, 70,  35, 85, 55, 65,
                            125, 95, 90, 105, 45, 110, 95, 115, 0,  0,  0,  0 };

        int source = 0;
        int sink = 9;
        int tasks = 4;
        // Define an array of supplies at each node.
        int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };

        // Add each arc.
        for (int i = 0; i < startNodes.Length; ++i)
        {
            int arc =
                minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
            if (arc != i)
                throw new Exception("Internal error");
        }

        // Add node supplies.
        for (int i = 0; i < supplies.Length; ++i)
        {
            minCostFlow.SetNodeSupply(i, supplies[i]);
        }

        // Find the min cost flow.
        MinCostFlow.Status status = minCostFlow.Solve();

        if (status == MinCostFlow.Status.OPTIMAL)
        {
            Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
            Console.WriteLine("");
            for (int i = 0; i < minCostFlow.NumArcs(); ++i)
            {
                // Can ignore arcs leading out of source or into sink.
                if (minCostFlow.Tail(i) != source && minCostFlow.Head(i) != sink)
                {
                    // Arcs in the solution have a flow value of 1. Their start and end nodes
                    // give an assignment of worker to task.
                    if (minCostFlow.Flow(i) > 0)
                    {
                        Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
                                          " Cost: " + minCostFlow.UnitCost(i));
                    }
                }
            }
        }
        else
        {
            Console.WriteLine("Solving the min cost flow problem failed.");
            Console.WriteLine("Solver status: " + status);
        }
    }
}

مهمة مع فرق من العاملين

يعرض هذا القسم مشكلة مهمة أكثر عمومية. في هذه المشكلة، تم تقسيم ستة عاملين إلى فريقين. تكمن المشكلة في تعيين أربع مهام للعمال بحيث يتم موازنة عبء العمل بالتساوي بين الفرق - أي، يجري كل فريق اثنتين من المهام.

للحصول على حلّ لهذه المسألة ضمن MIP، يمكنك الاطّلاع على مقالة المهام الدراسية مع فِرق العاملين.

تصف الأقسام التالية برنامجًا يحل المشكلة باستخدام أداة حل مشكلة الحد الأدنى للتكلفة.

استيراد المكتبات

يستورد الرمز التالي المكتبة المطلوبة.

Python

from ortools.graph.python import min_cost_flow

C++

#include <cstdint>
#include <vector>

#include "ortools/graph/min_cost_flow.h"

Java

import com.google.ortools.Loader;
import com.google.ortools.graph.MinCostFlow;
import com.google.ortools.graph.MinCostFlowBase;

C#

using System;
using Google.OrTools.Graph;

تعريف أداة الحلّ

تنشئ التعليمة البرمجية التالية أداة حل الحد الأدنى لتدفق التكلفة.

Python

smcf = min_cost_flow.SimpleMinCostFlow()

C++

// Instantiate a SimpleMinCostFlow solver.
SimpleMinCostFlow min_cost_flow;

Java

// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();

C#

// Instantiate a SimpleMinCostFlow solver.
MinCostFlow minCostFlow = new MinCostFlow();

إنشاء البيانات

ينشئ الرمز التالي البيانات للبرنامج.

Python

# Define the directed graph for the flow.
team_a = [1, 3, 5]
team_b = [2, 4, 6]

start_nodes = (
    # fmt: off
  [0, 0]
  + [11, 11, 11]
  + [12, 12, 12]
  + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6]
  + [7, 8, 9, 10]
    # fmt: on
)
end_nodes = (
    # fmt: off
  [11, 12]
  + team_a
  + team_b
  + [7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10]
  + [13, 13, 13, 13]
    # fmt: on
)
capacities = (
    # fmt: off
  [2, 2]
  + [1, 1, 1]
  + [1, 1, 1]
  + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
  + [1, 1, 1, 1]
    # fmt: on
)
costs = (
    # fmt: off
  [0, 0]
  + [0, 0, 0]
  + [0, 0, 0]
  + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95]
  + [0, 0, 0, 0]
    # fmt: on
)

source = 0
sink = 13
tasks = 4
# Define an array of supplies at each node.
supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]

C++

// Define the directed graph for the flow.
const std::vector<int64_t> team_A = {1, 3, 5};
const std::vector<int64_t> team_B = {2, 4, 6};

const std::vector<int64_t> start_nodes = {
    0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
    3, 3, 4,  4,  4,  4,  5,  5,  5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
const std::vector<int64_t> end_nodes = {
    11, 12, 1, 3, 5, 2,  4, 6, 7, 8,  9, 10, 7, 8,  9,  10, 7,  8,
    9,  10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,  9, 10, 13, 13, 13, 13};
const std::vector<int64_t> capacities = {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                                         1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                                         1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
const std::vector<int64_t> unit_costs = {
    0,  0,   0,  0,  0,   0,  0,   0,   90, 76,  75, 70,
    35, 85,  55, 65, 125, 95, 90,  105, 45, 110, 95, 115,
    60, 105, 80, 75, 45,  65, 110, 95,  0,  0,   0,  0};

const int64_t source = 0;
const int64_t sink = 13;
const int64_t tasks = 4;
// Define an array of supplies at each node.
const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0,
                                       0,     0, 0, 0, 0, 0, -tasks};

Java

// Define the directed graph for the flow.
// int[] teamA = new int[] {1, 3, 5};
// int[] teamB = new int[] {2, 4, 6};

int[] startNodes = new int[] {0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3,
    4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
int[] endNodes = new int[] {11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7,
    8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13};
int[] capacities = new int[] {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
    1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
int[] unitCosts = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95,
    90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0};

int source = 0;
int sink = 13;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};

C#

// Define the directed graph for the flow.
int[] teamA = { 1, 3, 5 };
int[] teamB = { 2, 4, 6 };

// Define four parallel arrays: sources, destinations, capacities, and unit costs
// between each pair.
int[] startNodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
                     3, 3, 4,  4,  4,  4,  5,  5,  5, 5, 6, 6, 6, 6, 7, 8, 9, 10 };
int[] endNodes = { 11, 12, 1, 3, 5, 2,  4, 6, 7, 8,  9, 10, 7, 8,  9,  10, 7,  8,
                   9,  10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,  9, 10, 13, 13, 13, 13 };
int[] capacities = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                     1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
int[] unitCosts = { 0,  0,   0,  0,   0,  0,   0,  0,   90, 76, 75, 70, 35,  85, 55, 65, 125, 95,
                    90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0,  0,  0,   0 };

int source = 0;
int sink = 13;
int tasks = 4;
// Define an array of supplies at each node.
int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };

تتجاوب العوامل مع العقد 1 - 6. يتكون الفريق "أ" من العاملين 1 و3 و5، ويتكون الفريق "ب" من العاملين 2 و4 و6. يتم ترقيم المهام من 7 إلى 10.

هناك عُقدتان جديدتان، 11 و12، بين المصدر والعاملين. يتم توصيل العقدة 11 بالعُقد الخاصة بالفريق A، والعقدة 12 متصلة بالعُقد الخاصة بالفريق B، ذات أقواس ذات سعة 1. يُظهر الرسم البياني أدناه العُقد والأقواس فقط من المصدر إلى العاملين.

يكمن مفتاح موازنة عبء العمل في أن المصدر 0 متصل بالنقطتين 11 و12 من خلال أقواس السعة 2. هذا يعني أن العُقدتين 11 و12 (وبالتالي الفريقين "أ" و"ب") يمكن أن يحصلا على تدفق 2 كحد أقصى. نتيجة لذلك، يمكن لكل فريق أداء مهمتين على الأكثر.

وضع القيود

Python

# Add each arc.
for i in range(0, len(start_nodes)):
    smcf.add_arc_with_capacity_and_unit_cost(
        start_nodes[i], end_nodes[i], capacities[i], costs[i]
    )

# Add node supplies.
for i in range(0, len(supplies)):
    smcf.set_node_supply(i, supplies[i])

C++

// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
  int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
      start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
  if (arc != i) LOG(FATAL) << "Internal error";
}

// Add node supplies.
for (int i = 0; i < supplies.size(); ++i) {
  min_cost_flow.SetNodeSupply(i, supplies[i]);
}

Java

// Add each arc.
for (int i = 0; i < startNodes.length; ++i) {
  int arc = minCostFlow.addArcWithCapacityAndUnitCost(
      startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
  if (arc != i) {
    throw new Exception("Internal error");
  }
}

// Add node supplies.
for (int i = 0; i < supplies.length; ++i) {
  minCostFlow.setNodeSupply(i, supplies[i]);
}

C#

// Add each arc.
for (int i = 0; i < startNodes.Length; ++i)
{
    int arc =
        minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
    if (arc != i)
        throw new Exception("Internal error");
}

// Add node supplies.
for (int i = 0; i < supplies.Length; ++i)
{
    minCostFlow.SetNodeSupply(i, supplies[i]);
}

استدعاء أداة الحلّ

Python

# Find the minimum cost flow between node 0 and node 10.
status = smcf.solve()

C++

// Find the min cost flow.
int status = min_cost_flow.Solve();

Java

// Find the min cost flow.
MinCostFlowBase.Status status = minCostFlow.solve();

C#

// Find the min cost flow.
MinCostFlow.Status status = minCostFlow.Solve();

نتيجة البرنامج

Python

if status == smcf.OPTIMAL:
    print("Total cost = ", smcf.optimal_cost())
    print()
    for arc in range(smcf.num_arcs()):
        # Can ignore arcs leading out of source or intermediate, or into sink.
        if (
            smcf.tail(arc) != source
            and smcf.tail(arc) != 11
            and smcf.tail(arc) != 12
            and smcf.head(arc) != sink
        ):
            # Arcs in the solution will have a flow value of 1.
            # There start and end nodes give an assignment of worker to task.
            if smcf.flow(arc) > 0:
                print(
                    "Worker %d assigned to task %d.  Cost = %d"
                    % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
                )
else:
    print("There was an issue with the min cost flow input.")
    print(f"Status: {status}")

C++

if (status == MinCostFlow::OPTIMAL) {
  LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
  LOG(INFO) << "";
  for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
    // Can ignore arcs leading out of source or intermediate nodes, or into
    // sink.
    if (min_cost_flow.Tail(i) != source && min_cost_flow.Tail(i) != 11 &&
        min_cost_flow.Tail(i) != 12 && min_cost_flow.Head(i) != sink) {
      // Arcs in the solution have a flow value of 1. Their start and end
      // nodes give an assignment of worker to task.
      if (min_cost_flow.Flow(i) > 0) {
        LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
                  << " assigned to task " << min_cost_flow.Head(i)
                  << " Cost: " << min_cost_flow.UnitCost(i);
      }
    }
  }
} else {
  LOG(INFO) << "Solving the min cost flow problem failed.";
  LOG(INFO) << "Solver status: " << status;
}

Java

if (status == MinCostFlow.Status.OPTIMAL) {
  System.out.println("Total cost: " + minCostFlow.getOptimalCost());
  System.out.println();
  for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
    // Can ignore arcs leading out of source or intermediate nodes, or into sink.
    if (minCostFlow.getTail(i) != source && minCostFlow.getTail(i) != 11
        && minCostFlow.getTail(i) != 12 && minCostFlow.getHead(i) != sink) {
      // Arcs in the solution have a flow value of 1. Their start and end nodes
      // give an assignment of worker to task.
      if (minCostFlow.getFlow(i) > 0) {
        System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
            + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
      }
    }
  }
} else {
  System.out.println("Solving the min cost flow problem failed.");
  System.out.println("Solver status: " + status);
}

C#

if (status == MinCostFlow.Status.OPTIMAL)
{
    Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
    Console.WriteLine("");
    for (int i = 0; i < minCostFlow.NumArcs(); ++i)
    {
        // Can ignore arcs leading out of source or into sink.
        if (minCostFlow.Tail(i) != source && minCostFlow.Tail(i) != 11 && minCostFlow.Tail(i) != 12 &&
            minCostFlow.Head(i) != sink)
        {
            // Arcs in the solution have a flow value of 1. Their start and end nodes
            // give an assignment of worker to task.
            if (minCostFlow.Flow(i) > 0)
            {
                Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
                                  " Cost: " + minCostFlow.UnitCost(i));
            }
        }
    }
}
else
{
    Console.WriteLine("Solving the min cost flow problem failed.");
    Console.WriteLine("Solver status: " + status);
}

يوضح ما يلي نتائج البرنامج.

Total cost = 250

Worker 1 assigned to task 9.  Cost =  75
Worker 2 assigned to task 7.  Cost =  35
Worker 5 assigned to task 10.  Cost =  75
Worker 6 assigned to task 8.  Cost =  65

Time = 0.00031 seconds

تم تعيين المهمتين 9 و10 للفريق "أ"، بينما تم تعيين المهمتين 7 و8 للفريق "ب".

تجدر الإشارة إلى أنّ أداة حلّ الحدّ الأدنى للتكلفة لحلّ هذه المشكلة أسرع من أداة حلّ MIP التي تستغرق 0.006 ثانية تقريبًا.

البرنامج بأكمله

يظهر البرنامج بأكمله أدناه.

Python

"""Assignment with teams of workers."""
from ortools.graph.python import min_cost_flow


def main():
    """Solving an Assignment with teams of worker."""
    smcf = min_cost_flow.SimpleMinCostFlow()

    # Define the directed graph for the flow.
    team_a = [1, 3, 5]
    team_b = [2, 4, 6]

    start_nodes = (
        # fmt: off
      [0, 0]
      + [11, 11, 11]
      + [12, 12, 12]
      + [1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6]
      + [7, 8, 9, 10]
        # fmt: on
    )
    end_nodes = (
        # fmt: off
      [11, 12]
      + team_a
      + team_b
      + [7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10]
      + [13, 13, 13, 13]
        # fmt: on
    )
    capacities = (
        # fmt: off
      [2, 2]
      + [1, 1, 1]
      + [1, 1, 1]
      + [1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1]
      + [1, 1, 1, 1]
        # fmt: on
    )
    costs = (
        # fmt: off
      [0, 0]
      + [0, 0, 0]
      + [0, 0, 0]
      + [90, 76, 75, 70, 35, 85, 55, 65, 125, 95, 90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95]
      + [0, 0, 0, 0]
        # fmt: on
    )

    source = 0
    sink = 13
    tasks = 4
    # Define an array of supplies at each node.
    supplies = [tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks]

    # Add each arc.
    for i in range(0, len(start_nodes)):
        smcf.add_arc_with_capacity_and_unit_cost(
            start_nodes[i], end_nodes[i], capacities[i], costs[i]
        )

    # Add node supplies.
    for i in range(0, len(supplies)):
        smcf.set_node_supply(i, supplies[i])

    # Find the minimum cost flow between node 0 and node 10.
    status = smcf.solve()

    if status == smcf.OPTIMAL:
        print("Total cost = ", smcf.optimal_cost())
        print()
        for arc in range(smcf.num_arcs()):
            # Can ignore arcs leading out of source or intermediate, or into sink.
            if (
                smcf.tail(arc) != source
                and smcf.tail(arc) != 11
                and smcf.tail(arc) != 12
                and smcf.head(arc) != sink
            ):
                # Arcs in the solution will have a flow value of 1.
                # There start and end nodes give an assignment of worker to task.
                if smcf.flow(arc) > 0:
                    print(
                        "Worker %d assigned to task %d.  Cost = %d"
                        % (smcf.tail(arc), smcf.head(arc), smcf.unit_cost(arc))
                    )
    else:
        print("There was an issue with the min cost flow input.")
        print(f"Status: {status}")


if __name__ == "__main__":
    main()

C++

#include <cstdint>
#include <vector>

#include "ortools/graph/min_cost_flow.h"

namespace operations_research {
// MinCostFlow simple interface example.
void BalanceMinFlow() {
  // Instantiate a SimpleMinCostFlow solver.
  SimpleMinCostFlow min_cost_flow;

  // Define the directed graph for the flow.
  const std::vector<int64_t> team_A = {1, 3, 5};
  const std::vector<int64_t> team_B = {2, 4, 6};

  const std::vector<int64_t> start_nodes = {
      0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
      3, 3, 4,  4,  4,  4,  5,  5,  5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
  const std::vector<int64_t> end_nodes = {
      11, 12, 1, 3, 5, 2,  4, 6, 7, 8,  9, 10, 7, 8,  9,  10, 7,  8,
      9,  10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,  9, 10, 13, 13, 13, 13};
  const std::vector<int64_t> capacities = {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                                           1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                                           1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
  const std::vector<int64_t> unit_costs = {
      0,  0,   0,  0,  0,   0,  0,   0,   90, 76,  75, 70,
      35, 85,  55, 65, 125, 95, 90,  105, 45, 110, 95, 115,
      60, 105, 80, 75, 45,  65, 110, 95,  0,  0,   0,  0};

  const int64_t source = 0;
  const int64_t sink = 13;
  const int64_t tasks = 4;
  // Define an array of supplies at each node.
  const std::vector<int64_t> supplies = {tasks, 0, 0, 0, 0, 0, 0,
                                         0,     0, 0, 0, 0, 0, -tasks};

  // Add each arc.
  for (int i = 0; i < start_nodes.size(); ++i) {
    int arc = min_cost_flow.AddArcWithCapacityAndUnitCost(
        start_nodes[i], end_nodes[i], capacities[i], unit_costs[i]);
    if (arc != i) LOG(FATAL) << "Internal error";
  }

  // Add node supplies.
  for (int i = 0; i < supplies.size(); ++i) {
    min_cost_flow.SetNodeSupply(i, supplies[i]);
  }

  // Find the min cost flow.
  int status = min_cost_flow.Solve();

  if (status == MinCostFlow::OPTIMAL) {
    LOG(INFO) << "Total cost: " << min_cost_flow.OptimalCost();
    LOG(INFO) << "";
    for (std::size_t i = 0; i < min_cost_flow.NumArcs(); ++i) {
      // Can ignore arcs leading out of source or intermediate nodes, or into
      // sink.
      if (min_cost_flow.Tail(i) != source && min_cost_flow.Tail(i) != 11 &&
          min_cost_flow.Tail(i) != 12 && min_cost_flow.Head(i) != sink) {
        // Arcs in the solution have a flow value of 1. Their start and end
        // nodes give an assignment of worker to task.
        if (min_cost_flow.Flow(i) > 0) {
          LOG(INFO) << "Worker " << min_cost_flow.Tail(i)
                    << " assigned to task " << min_cost_flow.Head(i)
                    << " Cost: " << min_cost_flow.UnitCost(i);
        }
      }
    }
  } else {
    LOG(INFO) << "Solving the min cost flow problem failed.";
    LOG(INFO) << "Solver status: " << status;
  }
}

}  // namespace operations_research

int main() {
  operations_research::BalanceMinFlow();
  return EXIT_SUCCESS;
}

Java

package com.google.ortools.graph.samples;
import com.google.ortools.Loader;
import com.google.ortools.graph.MinCostFlow;
import com.google.ortools.graph.MinCostFlowBase;

/** Minimal Assignment Min Flow. */
public class BalanceMinFlow {
  public static void main(String[] args) throws Exception {
    Loader.loadNativeLibraries();
    // Instantiate a SimpleMinCostFlow solver.
    MinCostFlow minCostFlow = new MinCostFlow();

    // Define the directed graph for the flow.
    // int[] teamA = new int[] {1, 3, 5};
    // int[] teamB = new int[] {2, 4, 6};

    int[] startNodes = new int[] {0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 3,
        4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 6, 7, 8, 9, 10};
    int[] endNodes = new int[] {11, 12, 1, 3, 5, 2, 4, 6, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 7,
        8, 9, 10, 7, 8, 9, 10, 7, 8, 9, 10, 13, 13, 13, 13};
    int[] capacities = new int[] {2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
        1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1};
    int[] unitCosts = new int[] {0, 0, 0, 0, 0, 0, 0, 0, 90, 76, 75, 70, 35, 85, 55, 65, 125, 95,
        90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0, 0, 0, 0};

    int source = 0;
    int sink = 13;
    int tasks = 4;
    // Define an array of supplies at each node.
    int[] supplies = new int[] {tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks};

    // Add each arc.
    for (int i = 0; i < startNodes.length; ++i) {
      int arc = minCostFlow.addArcWithCapacityAndUnitCost(
          startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
      if (arc != i) {
        throw new Exception("Internal error");
      }
    }

    // Add node supplies.
    for (int i = 0; i < supplies.length; ++i) {
      minCostFlow.setNodeSupply(i, supplies[i]);
    }

    // Find the min cost flow.
    MinCostFlowBase.Status status = minCostFlow.solve();

    if (status == MinCostFlow.Status.OPTIMAL) {
      System.out.println("Total cost: " + minCostFlow.getOptimalCost());
      System.out.println();
      for (int i = 0; i < minCostFlow.getNumArcs(); ++i) {
        // Can ignore arcs leading out of source or intermediate nodes, or into sink.
        if (minCostFlow.getTail(i) != source && minCostFlow.getTail(i) != 11
            && minCostFlow.getTail(i) != 12 && minCostFlow.getHead(i) != sink) {
          // Arcs in the solution have a flow value of 1. Their start and end nodes
          // give an assignment of worker to task.
          if (minCostFlow.getFlow(i) > 0) {
            System.out.println("Worker " + minCostFlow.getTail(i) + " assigned to task "
                + minCostFlow.getHead(i) + " Cost: " + minCostFlow.getUnitCost(i));
          }
        }
      }
    } else {
      System.out.println("Solving the min cost flow problem failed.");
      System.out.println("Solver status: " + status);
    }
  }

  private BalanceMinFlow() {}
}

C#

using System;
using Google.OrTools.Graph;

public class BalanceMinFlow
{
    static void Main()
    {
        // Instantiate a SimpleMinCostFlow solver.
        MinCostFlow minCostFlow = new MinCostFlow();

        // Define the directed graph for the flow.
        int[] teamA = { 1, 3, 5 };
        int[] teamB = { 2, 4, 6 };

        // Define four parallel arrays: sources, destinations, capacities, and unit costs
        // between each pair.
        int[] startNodes = { 0, 0, 11, 11, 11, 12, 12, 12, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3,
                             3, 3, 4,  4,  4,  4,  5,  5,  5, 5, 6, 6, 6, 6, 7, 8, 9, 10 };
        int[] endNodes = { 11, 12, 1, 3, 5, 2,  4, 6, 7, 8,  9, 10, 7, 8,  9,  10, 7,  8,
                           9,  10, 7, 8, 9, 10, 7, 8, 9, 10, 7, 8,  9, 10, 13, 13, 13, 13 };
        int[] capacities = { 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
                             1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 };
        int[] unitCosts = { 0,  0,   0,  0,   0,  0,   0,  0,   90, 76, 75, 70, 35,  85, 55, 65, 125, 95,
                            90, 105, 45, 110, 95, 115, 60, 105, 80, 75, 45, 65, 110, 95, 0,  0,  0,   0 };

        int source = 0;
        int sink = 13;
        int tasks = 4;
        // Define an array of supplies at each node.
        int[] supplies = { tasks, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, -tasks };

        // Add each arc.
        for (int i = 0; i < startNodes.Length; ++i)
        {
            int arc =
                minCostFlow.AddArcWithCapacityAndUnitCost(startNodes[i], endNodes[i], capacities[i], unitCosts[i]);
            if (arc != i)
                throw new Exception("Internal error");
        }

        // Add node supplies.
        for (int i = 0; i < supplies.Length; ++i)
        {
            minCostFlow.SetNodeSupply(i, supplies[i]);
        }

        // Find the min cost flow.
        MinCostFlow.Status status = minCostFlow.Solve();

        if (status == MinCostFlow.Status.OPTIMAL)
        {
            Console.WriteLine("Total cost: " + minCostFlow.OptimalCost());
            Console.WriteLine("");
            for (int i = 0; i < minCostFlow.NumArcs(); ++i)
            {
                // Can ignore arcs leading out of source or into sink.
                if (minCostFlow.Tail(i) != source && minCostFlow.Tail(i) != 11 && minCostFlow.Tail(i) != 12 &&
                    minCostFlow.Head(i) != sink)
                {
                    // Arcs in the solution have a flow value of 1. Their start and end nodes
                    // give an assignment of worker to task.
                    if (minCostFlow.Flow(i) > 0)
                    {
                        Console.WriteLine("Worker " + minCostFlow.Tail(i) + " assigned to task " + minCostFlow.Head(i) +
                                          " Cost: " + minCostFlow.UnitCost(i));
                    }
                }
            }
        }
        else
        {
            Console.WriteLine("Solving the min cost flow problem failed.");
            Console.WriteLine("Solver status: " + status);
        }
    }
}