在以下各節中，您將找到最大流程 (流程上限) 問題的示例。
最大流程示例
問題由下方代表運輸網路的圖形所定義：
您想要將內容從節點 0 (「來源」) 傳輸到節點 4 (接收器)。弧形旁邊的數字是其「容量」，弧形的容量是指在固定時間內可傳輸的最大數量。容量是問題的限制條件。
「資料流」是為符合以下「資料流保」規則的每個弧形 (「流程金額」) 指派非負數的弧度：
最大流程問題是找出流程中整個網路的流程金額總和最大。
以下各節會介紹尋找從來源 (0) 到接收器的最大資料流 (4) 的程式。
匯入程式庫
下列程式碼會匯入必要的程式庫。
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;
宣告解題工具
如要解決這個問題，您可以使用 SimpleMaxFlow 解題工具。
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();
定義資料
您將使用三個陣列定義問題的圖表，用於起始節點、結束節點和弧形的容量。每個陣列的長度等於圖形中的陣列數量。
每個 i 的弧形都會從 start_nodes[i] 到 end_nodes[i]，容量則由 capacities[i] 提供。下一節將說明如何使用這項資料建立弧形。
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 };
新增弧形
針對每個起始節點和結束節點，您可以使用 AddArcWithCapacity 方法，從起始節點到具有指定容量的結束節點建立弧形。容量是問題的限制
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");
}
叫用求解工具
現在所有弧形都已定義完畢，剩下的就是叫用解數工具並顯示結果您必須叫用 Solve() 方法，並提供來源 (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);
顯示結果
您現在可以顯示每個弧形的流程。
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);
}
以下是程式的輸出內容：
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]
每個弧形的流動量會顯示在 Flow 下方。
完成計畫
總整理如下：
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);
}
}
}