算法笔记_149:图论之桥的应用(Java)

目录

1 问题描述

2 解决方案

1 问题描述

1310 One-way traffic
In a certain town there are n intersections connected by two- and one-way streets. The town is very
modern so a lot of streets run through tunnels or viaducts. Of course it is possible to travel between
any two intersections in both ways, i.e. it is possible to travel from an intersection a to an intersection
b as well as from b to a without violating traffic rules. Because one-way streets are safer, it has been
decided to create as much one-way traffic as possible. In order not to make too much confusion it has
also been decided that the direction of traffic in already existing one-way streets should not be changed.
Your job is to create a new traffic system in the town. You have to determine the direction of
traffic for as many two-way streets as possible and make sure that it is still possible to travel both ways
between any two intersections.
Write a program that:
? reads a description of the street system in the town from the standard input,
? for each two-way street determines one direction of traffic or decides that the street must remain
two-way,
? writes the answer to the standard output.
Input
The first line of the input contains two integers n and m, where 2 ≤ n ≤ 2000 and n?1 ≤ m ≤ n(n?1)/2.
Integer n is the number of intersections in the town and integer m is the number of streets.
Each of the next m lines contains three integers a, b and c, where 1 ≤ a ≤ n, 1 ≤ b ≤ n, a ?= b and
c belongs to {1, 2}. If c = 1 then intersections a and b are connected by an one-way street from a to
b. If c = 2 then intersections a and b are connected by a two-way street. There is at most one street
connecting any two intersections.
Output
The output contains exactly the same number of lines as the number of two-way streets in the input.
For each such street (in any order) the program should write three integers a, b and c meaning, the new
direction of the street from a to b (c = 1) or that the street connecting a and b remains two-way (c = 2).
If there are more than one solution with maximal number of one-way streets then your program should
output any of them but just one.
Sample Input
4 4
4 1 1
4 2 2
1 2 1
1 3 2

Sample Output
2 4 1
3 1 2

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=4056

2 解决方案

首先看看关于图论的割点和桥的相关概念定义：

引用文末参考资料1：

1.割点：若删掉某点后，原连通图分裂为多个子图，则称该点为割点。

2.割点集合：在一个无向连通图中，如果有一个顶点集合，删除这个顶点集合，以及这个集合中所有顶点相关联的边以后，原图变成多个连通块，就称这个点集为割点集合。

3.点连通度：最小割点集合中的顶点数。

4.割边(桥)：删掉它之后，图必然会分裂为两个或两个以上的子图。

5.割边集合：如果有一个边集合，删除这个边集合以后，原图变成多个连通块，就称这个点集为割边集合。

6.边连通度：一个图的边连通度的定义为，最小割边集合中的边数。

7.缩点：把没有割边的连通子图缩为一个点，此时满足任意两点之间都有两条路径可达。

注：求块<>求缩点。缩点后变成一棵k个点k-1条割边连接成的树。而割点可以存在于多个块中。

8.双连通分量：分为点双连通和边双连通。它的标准定义为：点连通度大于1的图称为点双连通图，边连通度大于1的图称为边双连通图。通俗地讲，满足任意两点之间，能通过两条或两条以上没有任何重复边的路到达的图称为双连通图。无向图G的极大双连通子图称为双连通分量。

具体代码如下：

package com.liuzhen.practice;

import java.util.ArrayList;
import java.util.Scanner;
import java.util.Stack;

public class Main {
    public static int n;  //给定图的顶点数
    public static int count;  //记录遍历次序
    public static int[] DFN;
    public static int[] Low;
    public static int[] parent;   //parent[i] = j，表示顶点i的直接父母顶点为j
    public static Stack<Integer> stack;
    public static ArrayList<edge>[] map;
    public static ArrayList<edge> ans;  //存储最终输出结果

    static class edge {
        public int a;  //边的起点
        public int b;  //边的终点
        public int c;  //c = 1表示单向边，c = 2表示双向边

        public edge(int a, int b, int c) {
            this.a = a;
            this.b = b;
            this.c = c;
        }
    }

    @SuppressWarnings("unchecked")
    public void init() {
        count = 0;
        DFN = new int[n + 1];
        Low = new int[n + 1];
        parent = new int[n + 1];
        stack = new Stack<Integer>();
        map = new ArrayList[n + 1];
        ans = new ArrayList<edge>();
        for(int i = 1;i <= n;i++) {
            DFN[i] = -1;
            Low[i] = -1;
            parent[i] = -1;
            map[i] = new ArrayList<edge>();
        }
    }

    public void TarJan(int start, int father) {
        DFN[start] = count++;
        Low[start] = DFN[start];
        parent[start] = father;
        stack.push(start);
        for(int i = 0;i < map[start].size();i++) {
            edge temp = map[start].get(i);
            int j = temp.b;
            if(DFN[j] == -1) {
                TarJan(j, start);
                Low[start] = Math.min(Low[start], Low[j]);
                if(temp.c == 2) {
                    if(Low[j] > DFN[start]) {   //当边temp为割边(或者桥)时
                        ans.add(temp);
                    } else {
                        ans.add(new edge(temp.a, temp.b, 1));
                    }
                }
            } else if(j != parent[start]) {  //当j不是start的直接父母节点时
                Low[start] = Math.min(Low[start], DFN[j]);
                if(temp.c == 2) {
                    ans.add(new edge(temp.a, temp.b, 1));
                }
            }
        }
    }

    public void getResult() {
        for(int i = 1;i <= n;i++) {
            if(parent[i] == -1)
                TarJan(i, 0);
        }
        for(int i = 0;i < ans.size();i++)
            System.out.println(ans.get(i).a+" "+ans.get(i).b+" "+ans.get(i).c);
    }

    public static void main(String[] args) {
        Main test = new Main();
        Scanner in = new Scanner(System.in);
        n = in.nextInt();
        int k = in.nextInt();
        test.init();
        for(int i = 0;i < k;i++) {
            int a = in.nextInt();
            int b = in.nextInt();
            int c = in.nextInt();
            map[a].add(new edge(a, b, c));
            if(c == 2)
                map[b].add(new edge(b, a, c));
        }
        test.getResult();
    }
}

运行结果：