• Java实现最小费用最大流问题


    1 问题描述
    在最大流有多组解时,给每条边在附上一个单位费用的量,问在满足最大流时的最小费用是多少?

    2 解决方案
    下面代码所使用的测试数据如下图:

    在这里插入图片描述

    package com.liuzhen.practice;
    
    import java.util.ArrayList;
    import java.util.Scanner;
    
    public class Main {
        public static int MAX = 1000;
        public static int n;   //图中顶点数目
        public static boolean[] used = new boolean[MAX];   //判断顶点是否在队列中
        public static int[] pre = new int[MAX];   //记录最短增广路径中相应节点的前节点
        public static int[] distance = new int[MAX];   //记录源点到图中其他所有顶点的最短距离
        public static int[] capacity = new int[MAX];  //用于记录遍历图每一次得到增广路径的流量
        public static ArrayList<edge>[] map;   //图的邻接表
        //表示图中边信息内部类
        static class edge {
            public int from;   //边的起点
            public int to;     //边的终点
            public int cap;    //边的容量
            public int cost;   //边的费用
            
            public edge(int from, int to, int cap, int cost) {
                this.from = from;
                this.to = to;
                this.cap = cap;
                this.cost = cost;
            }
        }
        //输入给定图数据
        @SuppressWarnings("unchecked")
        public void init() {
            Scanner in = new Scanner(System.in);
            n = in.nextInt();
            int k = in.nextInt();  //给定图的边数目
            map = new ArrayList[n];
            for(int i = 0;i < n;i++)
                map[i] = new ArrayList<edge>();
            for(int i = 0;i < k;i++) {
                int from = in.nextInt();
                int to = in.nextInt();
                int cap = in.nextInt();
                int cost = in.nextInt();
                map[from].add(new edge(from, to, cap, cost));  //正向边
                map[to].add(new edge(to, from, 0, -cost));     //反向边
            }
        }
        
        //寻找顶点start到顶点end的最短路径(PS:即费用最少的一条增广路径)
        public boolean spfa(int start, int end) {  
            int[] count = new int[n];
            for(int i = 0;i < n;i++) {
                used[i] = false;
                pre[i] = -1;
                distance[i] = Integer.MAX_VALUE;
                capacity[i] = Integer.MAX_VALUE;
            }
            used[start] = true;
            pre[start] = start;
            distance[start] = 0;
            count[start]++;
            ArrayList<Integer> list = new ArrayList<Integer>();
            list.add(start);
            while(!list.isEmpty()) {
                int index = list.get(0);
                list.remove(0);
                used[index] = false;
                for(int i = 0;i < map[index].size();i++) {
                    edge temp = map[index].get(i);
                    if(temp.cap > 0 && distance[temp.to] > distance[index] + temp.cost) {
                        //记录顶点start到图中其它顶点之间的最短费用距离
                        distance[temp.to] = distance[index] + temp.cost;
                        pre[temp.to] = index;
                        //记录增广路径能够流通的最大流量
                        capacity[temp.to] = Math.min(capacity[index], temp.cap);
                        if(!used[temp.to]) {
                            used[temp.to] = true;
                            list.add(temp.to);
                            count[temp.to]++;
                            if(count[temp.to] > n)   //用于判断图中是否有负环
                                return false;
                        }
                    }
                }
            }
            if(distance[end] != Integer.MAX_VALUE && pre[end] != -1)
                return true;
            return false;
        }
        
        public int getResult() {
            init();   //输入给定图数据
            int minCost = 0;
            int start = 0;   //把源点设置为顶点0
            int end = n - 1;  //把汇点设置为顶点n - 1
            while(true) {
                if(spfa(start, end) == false)
                    break;
                System.out.println("增广路径增量:"+capacity[end]+", 费用流:"+distance[end]);
                minCost += distance[end] * capacity[end];
                int last = end;
                int begin = end;
                System.out.print("汇点出发");
                while(begin != start) {
                    last = begin;
                    begin = pre[last];
                    int i = 0, j = 0;
                    System.out.print("——>"+last);
                    for(;i < map[begin].size();i++) {
                        if(map[begin].get(i).to == last)
                            break;
                    }
                    map[begin].get(i).cap -= capacity[end];  //正向边剩余流量减少
                    for(;j < map[last].size();j++) {
                        if(map[last].get(j).to == begin)
                            break;
                    }
                    map[last].get(j).cap += capacity[end];  //反向边剩余流量增加
                }    
                System.out.println("——>"+begin);
            }
            return minCost;
        }
        
        public static void main(String[] args) {
            Main test = new Main();
            int result = test.getResult();
            System.out.println(result);
        }
    }
    

    运行结果:

    7
    1 2 1
    3 3 2
    2 5 5
    4 3 4
    5 2 10
    2 1 3
    5 4 7
    增广路径增量:2, 费用流:12
    汇点出发——>5——>4——>1——>0
    增广路径增量:1, 费用流:15
    汇点出发——>5——>2——>3——>0
    
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  • 原文地址:https://www.cnblogs.com/a1439775520/p/12947897.html
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