经典拓扑排序

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Problem Description

ACM-DIY is a large QQ group where many excellent acmers get together. It is so harmonious that just like a big family. Every day,many “holy cows” like HH, hh, AC, ZT, lcc, BF, Qinz and so on chat on-line to exchange their ideas. When someone has questions, many warm-hearted cows like Lost will come to help. Then the one being helped will call Lost “master”, and Lost will have a nice “prentice”. By and by, there are many pairs of “master and prentice”. But then problem occurs: there are too many masters and too many prentices, how can we know whether it is legal or not?

We all know a master can have many prentices and a prentice may have a lot of masters too, it’s legal. Nevertheless，some cows are not so honest, they hold illegal relationship. Take HH and 3xian for instant, HH is 3xian’s master and, at the same time, 3xian is HH’s master,which is quite illegal! To avoid this,please help us to judge whether their relationship is legal or not.

Please note that the “master and prentice” relation is transitive. It means that if A is B’s master ans B is C’s master, then A is C’s master.  

Input

The input consists of several test cases. For each case, the first line contains two integers, N (members to be tested) and M (relationships to be tested)(2 <= N, M <= 100). Then M lines follow, each contains a pair of (x, y) which means x is y’s master and y is x’s prentice. The input is terminated by N = 0.

TO MAKE IT SIMPLE, we give every one a number (0, 1, 2,…, N-1). We use their numbers instead of their names.  

Output

For each test case, print in one line the judgement of the messy relationship.

If it is legal, output “YES”, otherwise “NO”.  

Sample Input

3 2 0 1 1 2 2 2 0 1 1 0 0 0  

Sample Output

YES NO

题意：给你一组数据，n表示有多少个人，m表示这n人中间有多少种关系，然后就是n个人中m个关系，a b，a表示师傅，b表示徒弟，而题目要求是判断给你这组数据中是否存在即是某个人的师傅，又是其徒弟的关系（也就是 a b，b a，那么a就即是b的徒弟又是师傅）

结题思路：这是一个典型的拓扑排序的例题，很适合学习拓扑排序。可以将每个人看作一个结点，而每个人的师傅的个数则表示这个结点对应的入度。然后通过对每个结点进行遍历，依次找出入度为0的结点，将其拿出，并将其对应指向的结点的入度减1，也就是相当于如果没有这个人后，还有人可以只做师傅，依次循环，直到将所有结点都遍历完，都没有出现找不到这样的人（只可以做师傅的人）则表示没有出现不符合条件的关系，否则有（即图中有回路）

import java.util.Scanner;

public class P3342 {
	static int n,m;
	static int degree[],visit[];//degree记录每个结点的入度（也就是师傅的个数），visit标记已经访问的结点
	static int arc[][];//记录结点之间的指向（谁是谁的师傅，前者是师傅，后者是徒弟）
	static Scanner sc=new Scanner(System.in);
	public static void main(String[] args) {
		while(sc.hasNext()){
			n=sc.nextInt();
			m=sc.nextInt();
			if(n==0){
				break;
			}
			init();//初始化
//			for(int i=0;i<n;i++){
//				System.out.println(i+" degree:"+degree[i]);
//			}
			if(topoSort()){
				System.out.println("YES");
			}else{
				System.out.println("NO");
			}
		}
	}
	private static boolean topoSort() {//拓扑排序
		int s=0;
		while(s<n){
			int i=0;
			for(;i<n;i++){//找入度为零且没有访问过的结点
				if(degree[i]==0&&visit[i]==0){
					break;
				}
			}
			if(i==n){//没找到这样的结点，则表示图中有回路，也就是即是师傅又是徒弟的人
				return false;
			}
			s++;
			visit[i]=1;//若找到这样的结点，这标记为访问过
			for(int j=0;j<n;j++){
				if(arc[i][j]==1){//然后除去这个点，也就是对应指向的结点入度减1，继续循环遍历其他所有结点
					degree[j]--;
//					System.out.println(i+"--->"+j+":"+degree[j]);
				}
			}
		}
		return true;
	}
	private static void init() {
		degree=new int[n];
		visit=new int[n];
		arc=new int[n][n];
		for(int i=0;i<m;i++){
			int a=sc.nextInt();
			int b=sc.nextInt();
			if(arc[a][b]==0){//这里注意，一定要防止重边的出现（就是防止给的数据会重复）
				arc[a][b]=1;
				degree[b]++;	
			}
		}
	}
}