最大流

在以下几个部分中，您将获得一个最大流示例 （最大数据流）问题。

最大流示例

下图定义了该问题，该图表示交通 网络：

网络流图

您想要将材料从节点 0（源）传输到节点 4（ sink)。弧形旁边的数字是各自的容量，即 弧形容积是指 固定期限容量是问题的限制条件。

流是为每条弧（ flow amount），且该规则满足以下数据流保守规则：

最大流问题是找到一个流量总和 尽可能扩大整个广告网络

以下各部分介绍了一些程序，它们可从 数据来源 (0) 发送到接收器 (4)。

导入库

以下代码会导入所需的库。

Python

import numpy as np

from ortools.graph.python import max_flow

C++

#include <cstdint>
#include <vector>

#include "ortools/graph/max_flow.h"

Java

import com.google.ortools.Loader;
import com.google.ortools.graph.MaxFlow;

C#

using System;
using Google.OrTools.Graph;

声明求解器

要解决此问题，您可以使用 SimpleMaxFlow 求解器使用。

Python

# Instantiate a SimpleMaxFlow solver.
smf = max_flow.SimpleMaxFlow()

C++

// Instantiate a SimpleMaxFlow solver.
SimpleMaxFlow max_flow;

Java

// Instantiate a SimpleMaxFlow solver.
MaxFlow maxFlow = new MaxFlow();

C#

// Instantiate a SimpleMaxFlow solver.
MaxFlow maxFlow = new MaxFlow();

定义数据

您使用三个数组（起始节点、结束节点）定义问题图表 节点和弧形的容量每个数组的长度等于 曲线中的弧线

对于每个 i，弧形 i 从 start_nodes[i] 到 end_nodes[i]，及其容量 由 capacities[i] 提供。下一部分将介绍如何使用 这些数据。

Python

# Define three parallel arrays: start_nodes, end_nodes, and the capacities
# between each pair. For instance, the arc from node 0 to node 1 has a
# capacity of 20.
start_nodes = np.array([0, 0, 0, 1, 1, 2, 2, 3, 3])
end_nodes = np.array([1, 2, 3, 2, 4, 3, 4, 2, 4])
capacities = np.array([20, 30, 10, 40, 30, 10, 20, 5, 20])

C++

// Define three parallel arrays: start_nodes, end_nodes, and the capacities
// between each pair. For instance, the arc from node 0 to node 1 has a
// capacity of 20.
std::vector<int64_t> start_nodes = {0, 0, 0, 1, 1, 2, 2, 3, 3};
std::vector<int64_t> end_nodes = {1, 2, 3, 2, 4, 3, 4, 2, 4};
std::vector<int64_t> capacities = {20, 30, 10, 40, 30, 10, 20, 5, 20};

Java

// Define three parallel arrays: start_nodes, end_nodes, and the capacities
// between each pair. For instance, the arc from node 0 to node 1 has a
// capacity of 20.
// From Taha's 'Introduction to Operations Research',
// example 6.4-2.
int[] startNodes = new int[] {0, 0, 0, 1, 1, 2, 2, 3, 3};
int[] endNodes = new int[] {1, 2, 3, 2, 4, 3, 4, 2, 4};
int[] capacities = new int[] {20, 30, 10, 40, 30, 10, 20, 5, 20};

C#

// Define three parallel arrays: start_nodes, end_nodes, and the capacities
// between each pair. For instance, the arc from node 0 to node 1 has a
// capacity of 20.
// From Taha's 'Introduction to Operations Research',
// example 6.4-2.
int[] startNodes = { 0, 0, 0, 1, 1, 2, 2, 3, 3 };
int[] endNodes = { 1, 2, 3, 2, 4, 3, 4, 2, 4 };
int[] capacities = { 20, 30, 10, 40, 30, 10, 20, 5, 20 };

添加弧形

对于每个起始节点和结束节点，您需要创建从起始节点到结束节点的弧形 使用 方法， AddArcWithCapacity。容量是 来解决问题。

Python

# Add arcs in bulk.
#   note: we could have used add_arc_with_capacity(start, end, capacity)
all_arcs = smf.add_arcs_with_capacity(start_nodes, end_nodes, capacities)

C++

// Add each arc.
for (int i = 0; i < start_nodes.size(); ++i) {
  max_flow.AddArcWithCapacity(start_nodes[i], end_nodes[i], capacities[i]);
}

Java

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

C#

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

调用求解器

现在，所有弧线均已定义，剩下的工作就是调用 求解器并显示结果。您可以调用 Solve() 方法，并提供 数据来源 (0) 和接收器 (4)。

Python

# Find the maximum flow between node 0 and node 4.
status = smf.solve(0, 4)

C++

// Find the maximum flow between node 0 and node 4.
int status = max_flow.Solve(0, 4);

Java

// Find the maximum flow between node 0 and node 4.
MaxFlow.Status status = maxFlow.solve(0, 4);

C#

// Find the maximum flow between node 0 and node 4.
MaxFlow.Status status = maxFlow.Solve(0, 4);

显示结果

现在，您可以显示横跨每个弧形的流。

Python

if status != smf.OPTIMAL:
    print("There was an issue with the max flow input.")
    print(f"Status: {status}")
    exit(1)
print("Max flow:", smf.optimal_flow())
print("")
print(" Arc    Flow / Capacity")
solution_flows = smf.flows(all_arcs)
for arc, flow, capacity in zip(all_arcs, solution_flows, capacities):
    print(f"{smf.tail(arc)} / {smf.head(arc)}   {flow:3}  / {capacity:3}")
print("Source side min-cut:", smf.get_source_side_min_cut())
print("Sink side min-cut:", smf.get_sink_side_min_cut())

C++

if (status == MaxFlow::OPTIMAL) {
  LOG(INFO) << "Max flow: " << max_flow.OptimalFlow();
  LOG(INFO) << "";
  LOG(INFO) << "  Arc    Flow / Capacity";
  for (std::size_t i = 0; i < max_flow.NumArcs(); ++i) {
    LOG(INFO) << max_flow.Tail(i) << " -> " << max_flow.Head(i) << "  "
              << max_flow.Flow(i) << "  / " << max_flow.Capacity(i);
  }
} else {
  LOG(INFO) << "Solving the max flow problem failed. Solver status: "
            << status;
}

Java

if (status == MaxFlow.Status.OPTIMAL) {
  System.out.println("Max. flow: " + maxFlow.getOptimalFlow());
  System.out.println();
  System.out.println("  Arc     Flow / Capacity");
  for (int i = 0; i < maxFlow.getNumArcs(); ++i) {
    System.out.println(maxFlow.getTail(i) + " -> " + maxFlow.getHead(i) + "    "
        + maxFlow.getFlow(i) + "  /  " + maxFlow.getCapacity(i));
  }
} else {
  System.out.println("Solving the max flow problem failed. Solver status: " + status);
}

C#

if (status == MaxFlow.Status.OPTIMAL)
{
    Console.WriteLine("Max. flow: " + maxFlow.OptimalFlow());
    Console.WriteLine("");
    Console.WriteLine("  Arc     Flow / Capacity");
    for (int i = 0; i < maxFlow.NumArcs(); ++i)
    {
        Console.WriteLine(maxFlow.Tail(i) + " -> " + maxFlow.Head(i) + "    " +
                          string.Format("{0,3}", maxFlow.Flow(i)) + "  /  " +
                          string.Format("{0,3}", maxFlow.Capacity(i)));
    }
}
else
{
    Console.WriteLine("Solving the max flow problem failed. Solver status: " + status);
}

程序的输出如下：

Max flow: 60

  Arc    Flow / Capacity
0 -> 1    20  /  20
0 -> 2    30  /  30
0 -> 3    10  /  10
1 -> 2     0  /  40
1 -> 4    20  /  30
2 -> 3    10  /  10
2 -> 4    20  /  20
3 -> 2     0  /   5
3 -> 4    20  /  20
Source side min-cut: [0]
Sink side min-cut: [4, 1]

每条弧上的流量都显示在 Flow 下。

完成计划

综上，以下是完整的计划。

Python

"""From Taha 'Introduction to Operations Research', example 6.4-2."""
import numpy as np

from ortools.graph.python import max_flow


def main():
    """MaxFlow simple interface example."""
    # Instantiate a SimpleMaxFlow solver.
    smf = max_flow.SimpleMaxFlow()

    # Define three parallel arrays: start_nodes, end_nodes, and the capacities
    # between each pair. For instance, the arc from node 0 to node 1 has a
    # capacity of 20.
    start_nodes = np.array([0, 0, 0, 1, 1, 2, 2, 3, 3])
    end_nodes = np.array([1, 2, 3, 2, 4, 3, 4, 2, 4])
    capacities = np.array([20, 30, 10, 40, 30, 10, 20, 5, 20])

    # Add arcs in bulk.
    #   note: we could have used add_arc_with_capacity(start, end, capacity)
    all_arcs = smf.add_arcs_with_capacity(start_nodes, end_nodes, capacities)

    # Find the maximum flow between node 0 and node 4.
    status = smf.solve(0, 4)

    if status != smf.OPTIMAL:
        print("There was an issue with the max flow input.")
        print(f"Status: {status}")
        exit(1)
    print("Max flow:", smf.optimal_flow())
    print("")
    print(" Arc    Flow / Capacity")
    solution_flows = smf.flows(all_arcs)
    for arc, flow, capacity in zip(all_arcs, solution_flows, capacities):
        print(f"{smf.tail(arc)} / {smf.head(arc)}   {flow:3}  / {capacity:3}")
    print("Source side min-cut:", smf.get_source_side_min_cut())
    print("Sink side min-cut:", smf.get_sink_side_min_cut())


if __name__ == "__main__":
    main()

C++

// From Taha 'Introduction to Operations Research', example 6.4-2."""
#include <cstdint>
#include <vector>

#include "ortools/graph/max_flow.h"

namespace operations_research {
// MaxFlow simple interface example.
void SimpleMaxFlowProgram() {
  // Instantiate a SimpleMaxFlow solver.
  SimpleMaxFlow max_flow;

  // Define three parallel arrays: start_nodes, end_nodes, and the capacities
  // between each pair. For instance, the arc from node 0 to node 1 has a
  // capacity of 20.
  std::vector<int64_t> start_nodes = {0, 0, 0, 1, 1, 2, 2, 3, 3};
  std::vector<int64_t> end_nodes = {1, 2, 3, 2, 4, 3, 4, 2, 4};
  std::vector<int64_t> capacities = {20, 30, 10, 40, 30, 10, 20, 5, 20};

  // Add each arc.
  for (int i = 0; i < start_nodes.size(); ++i) {
    max_flow.AddArcWithCapacity(start_nodes[i], end_nodes[i], capacities[i]);
  }

  // Find the maximum flow between node 0 and node 4.
  int status = max_flow.Solve(0, 4);

  if (status == MaxFlow::OPTIMAL) {
    LOG(INFO) << "Max flow: " << max_flow.OptimalFlow();
    LOG(INFO) << "";
    LOG(INFO) << "  Arc    Flow / Capacity";
    for (std::size_t i = 0; i < max_flow.NumArcs(); ++i) {
      LOG(INFO) << max_flow.Tail(i) << " -> " << max_flow.Head(i) << "  "
                << max_flow.Flow(i) << "  / " << max_flow.Capacity(i);
    }
  } else {
    LOG(INFO) << "Solving the max flow problem failed. Solver status: "
              << status;
  }
}

}  // namespace operations_research

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

Java

package com.google.ortools.graph.samples;
import com.google.ortools.Loader;
import com.google.ortools.graph.MaxFlow;

/** Minimal MaxFlow program. */
public final class SimpleMaxFlowProgram {
  public static void main(String[] args) throws Exception {
    Loader.loadNativeLibraries();
    // Instantiate a SimpleMaxFlow solver.
    MaxFlow maxFlow = new MaxFlow();

    // Define three parallel arrays: start_nodes, end_nodes, and the capacities
    // between each pair. For instance, the arc from node 0 to node 1 has a
    // capacity of 20.
    // From Taha's 'Introduction to Operations Research',
    // example 6.4-2.
    int[] startNodes = new int[] {0, 0, 0, 1, 1, 2, 2, 3, 3};
    int[] endNodes = new int[] {1, 2, 3, 2, 4, 3, 4, 2, 4};
    int[] capacities = new int[] {20, 30, 10, 40, 30, 10, 20, 5, 20};

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

    // Find the maximum flow between node 0 and node 4.
    MaxFlow.Status status = maxFlow.solve(0, 4);

    if (status == MaxFlow.Status.OPTIMAL) {
      System.out.println("Max. flow: " + maxFlow.getOptimalFlow());
      System.out.println();
      System.out.println("  Arc     Flow / Capacity");
      for (int i = 0; i < maxFlow.getNumArcs(); ++i) {
        System.out.println(maxFlow.getTail(i) + " -> " + maxFlow.getHead(i) + "    "
            + maxFlow.getFlow(i) + "  /  " + maxFlow.getCapacity(i));
      }
    } else {
      System.out.println("Solving the max flow problem failed. Solver status: " + status);
    }
  }

  private SimpleMaxFlowProgram() {}
}

C#

// From Taha 'Introduction to Operations Research', example 6.4-2.
using System;
using Google.OrTools.Graph;

public class SimpleMaxFlowProgram
{
    static void Main()
    {
        // Instantiate a SimpleMaxFlow solver.
        MaxFlow maxFlow = new MaxFlow();

        // Define three parallel arrays: start_nodes, end_nodes, and the capacities
        // between each pair. For instance, the arc from node 0 to node 1 has a
        // capacity of 20.
        // From Taha's 'Introduction to Operations Research',
        // example 6.4-2.
        int[] startNodes = { 0, 0, 0, 1, 1, 2, 2, 3, 3 };
        int[] endNodes = { 1, 2, 3, 2, 4, 3, 4, 2, 4 };
        int[] capacities = { 20, 30, 10, 40, 30, 10, 20, 5, 20 };

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

        // Find the maximum flow between node 0 and node 4.
        MaxFlow.Status status = maxFlow.Solve(0, 4);

        if (status == MaxFlow.Status.OPTIMAL)
        {
            Console.WriteLine("Max. flow: " + maxFlow.OptimalFlow());
            Console.WriteLine("");
            Console.WriteLine("  Arc     Flow / Capacity");
            for (int i = 0; i < maxFlow.NumArcs(); ++i)
            {
                Console.WriteLine(maxFlow.Tail(i) + " -> " + maxFlow.Head(i) + "    " +
                                  string.Format("{0,3}", maxFlow.Flow(i)) + "  /  " +
                                  string.Format("{0,3}", maxFlow.Capacity(i)));
            }
        }
        else
        {
            Console.WriteLine("Solving the max flow problem failed. Solver status: " + status);
        }
    }
}