目录

1 问题描述

2 解决方案

 
1 问题描述

Description

In order to make their sons brave, Jiajia and Wind take them to a big cave. The cave has n rooms, and one-way corridors connecting some rooms. Each time, Wind choose two rooms x and y, and ask one of their little sons go from one to the other. The son can either go from x to y, or from y to x. Wind promised that her tasks are all possible, but she actually doesn't know how to decide if a task is possible. To make her life easier, Jiajia decided to choose a cave in which every pair of rooms is a possible task. Given a cave, can you tell Jiajia whether Wind can randomly choose two rooms without worrying about anything?

Input

The first line contains a single integer T, the number of test cases. And followed T cases. 

The first line for each case contains two integers n, m(0 < n < 1001,m < 6000), the number of rooms and corridors in the cave. The next m lines each contains two integers u and v, indicating that there is a corridor connecting room u and room v directly. 

Output

The output should contain T lines. Write 'Yes' if the cave has the property stated above, or 'No' otherwise.

Sample Input

1
3 3
1 2
2 3
3 1

Sample Output

Yes

Source

 

 

 

2 解决方案

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;  //顶点数
    public static int count;
    public static int[] DFN;
    public static int[] Low;
    public static boolean[] inStack;
    public static int group;  //强连通分量组
    public static int[] belong;
    public static Stack<Integer> stack;
    public static ArrayList<edge>[] map;
    public static ArrayList<String> result = new ArrayList<String>();
    
    static class edge {
        public int a;
        public int b;
        
        public edge(int a, int b) {
            this.a = a;
            this.b = b;
        }
    }
    
    @SuppressWarnings("unchecked")
    public void init() {
        count = 1;
        DFN = new int[n + 1];
        Low = new int[n + 1];
        inStack = new boolean[n + 1];
        group = 0;
        belong = new int[n + 1];
        stack = new Stack<Integer>();
        map = new ArrayList[n + 1];
        for(int i = 1;i <= n;i++) {
            DFN[i] = -1;
            Low[i] = -1;
            inStack[i] = false;
            belong[i] = -1;
            map[i] = new ArrayList<edge>();
        }
    }
    
    public void TarJan(int start) {
        DFN[start] = count++;
        Low[start] = DFN[start];
        inStack[start] = true;
        stack.push(start);
        int j = start;
        for(int i = 0;i < map[start].size();i++) {
            j = map[start].get(i).b;
            if(DFN[j] == -1) {
                TarJan(j);
                Low[start] = Math.min(Low[start], Low[j]);
            } else if(inStack[j]) {
                Low[start] = Math.min(Low[start], DFN[j]);
            }
        }
        if(DFN[start] == Low[start]) {
            group++;
            do {
                j = stack.pop();
                belong[j] = group;
                inStack[j] = false;
            } while(j != start);
        }
    }
    
    public boolean TopSort(ArrayList<edge>[] lessMap, int[] degree) {
        int count = 0;
        Stack<Integer> s = new Stack<Integer>();
        for(int i = 1;i < degree.length;i++) {
            if(degree[i] == 0) {
                count++;
                s.push(i);
            }
        }
        if(count > 1)
            return false;
        while(!s.empty()) {
            int start = s.pop();
            count = 0;
            for(int i = 0;i < lessMap[start].size();i++) {
                int j = lessMap[start].get(i).b;
                degree[j]--;
                if(degree[j] == 0) {
                    count++;
                    s.push(j);
                }
            }
            if(count > 1)
                return false;
        }
        return true;
    }
    
    @SuppressWarnings("unchecked")
    public void getResult() {
        for(int i = 1;i <= n;i++) {
            if(DFN[i] == -1)
                TarJan(i);
        }
        ArrayList<edge>[] lessMap = new ArrayList[group + 1];
        int[] degree = new int[group + 1];
        for(int i = 1;i <= group;i++)
            lessMap[i] = new ArrayList<edge>();
        for(int i = 1;i < map.length;i++) {
            for(int j = 0;j < map[i].size();j++) {
                int a = map[i].get(j).a;
                int b = map[i].get(j).b;
                if(belong[a] != belong [b]) {
                    lessMap[belong[a]].add(new edge(belong[a], belong[b]));
                    degree[belong[b]]++;
                }
            }
        }
        if(TopSort(lessMap, degree)) {
            result.add("Yes");
        } else {
            result.add("No");
        }
        return;
    }
    
    public static void main(String[] args) {
        Main test = new Main();
        Scanner in = new Scanner(System.in);
        int t = in.nextInt();
        while(t > 0) {
            t--;
            n = in.nextInt();
            int k = in.nextInt();
            test.init();
            for(int i = 0;i < k;i++) {
                int a = in.nextInt();
                int b = in.nextInt();
                map[a].add(new edge(a, b));
            }
            test.getResult();
        }
        for(int i = 0;i < result.size();i++)
            System.out.println(result.get(i));
    }
}

 

运行结果：

1
3 3
1 2
2 3
3 1
Yes

 

 

 

 

 

参考资料：

   1. 欧拉回路

    2. POJ2762 Going from u to v or from v to u?（强连通分量缩点+拓扑排序）

 

每天一小步，成就一大步