HDu 3342 Legal or Not(トポロジーソート)

Legal or Not
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 3151    Accepted Submission(s): 1432
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
   
   
   
   

師匠と弟子の関係を判断し、出会った人の関係が乱れていないかを判断する.
問題を解く構想:実はトポロジーの順序付けを行うかどうかを判断して、トポロジーの順序付けの性質を考察して、有向無環、DFSで手がかりを収めて判断します.

#include<stdio.h>
#include<string.h>
#define N 105
int n, m;
int vis[N], map[N][N];

int DFS(int k){
	vis[k] = 1;
	for (int i = 0; i < n; i++){
		if (map[k][i]){
			if(vis[i])
				return 1;
			else if(DFS(i))
				return 1;
		}
	}
	vis[k] = 0;
	return 0;
}

int main(){
	while (scanf("%d%d", &n, &m), n || m){
		// Init.
		memset(map, 0, sizeof(map));
		int ok = 0;

		// Read.
		for (int i = 0; i < m; i++){
			int a, b;
			scanf("%d%d", &a, &b);
			map[b][a] = 1;
		}

		// DFS.
		for (int i = 0; i < n; i++){
			memset(vis, 0, sizeof(vis));
			if (DFS(i)){
				ok = 1;
				break;
			}
		}

		if (!ok)
			printf("YES
");
		else
			printf("NO
");
	}
	return 0;}