hdoj 3342 Legal or Not

Legal or Not

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：有N个人编号从0到N-1，给出M组师徒关系即a是b师徒。问你 这N个人之间 存不存在 师徒辈分混乱的情况。

这是一道简单的拓扑排序问题，考察就是排序中出现成环的情况，题目不难，看到题，很快就写出来了，但是有一点坑，就是要提前跳出加一个break；哎考虑问题不深入，这就是教训呀，不过终于发现了这个错误。。。

#include<cstdio>
#include<cstring>
int map[110][110],indegree[110];
void tuop(int n)
{
	int i,j,ans,flag=1;
	for(i=0;i<n;i++)
	{
		ans=-1;
		for(j=0;j<n;j++)
		{	
			if(indegree[j]==0)
			{	
				ans=j;
				break;
			}
		}	
		indegree[ans]=-1;
		if(ans==-1)
			break;
		for(j=0;j<n;j++)
		{
			if(map[ans][j])
				indegree[j]--;	
		}
	}
	if(ans>=0)
			printf("YES\n");
		else
			printf("NO\n");
}
int main()
{
	int m,n,a,b,i;
	while(scanf("%d%d",&n,&m),n)
	{
		memset(map,0,sizeof(map));
		memset(indegree,0,sizeof(indegree));
		for(i=0;i<m;i++)
		{
			scanf("%d%d",&a,&b);
			if(map[a][b]==0)
			{
				map[a][b]=1;
				indegree[b]++;
			}
		}
		tuop(n);
	}
	return 0;
}