题意：判断是否成环。

策略：如题。

这道题就是简单的拓扑排序题，但是要注意一点要去重复的数据。我用了两种结构体：链式前向星和邻接矩阵。

代码1：（用链式前向星）（不用增加去重）

#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 0x3f3f3f3f
#define MAXN 105
struct EdgeNode{
	int to;
	int next;
}edges[MAXN];
int head[MAXN], n, in[MAXN], queue[MAXN];
int toposort()
{
	int i, iq = 0, j;
	for(i = 0; i < n; i ++){
		if(!in[i]){
			queue[iq++] = i;
		}
	}
	for(i = 0; i < iq; i ++){
		int temp = queue[i];
		for(j = head[temp]; j != -1; j = edges[j].next){
			if(!--in[edges[j].to]){
				queue[iq++] = edges[j].to;
			}
		}
	}
	return iq == n;
}
int main()
{
	int m, a, b, i;
	while(scanf("%d%d", &n, &m), n||m){
		memset(head, -1, sizeof(head));
		memset(in, 0, sizeof(in));
		for(i = 0; i < m; i ++){
			scanf("%d%d", &a, &b);
			edges[i].to = b;
			edges[i].next = head[a];
			head[a] = i;
			++in[b];
		}
		printf("%s\n", toposort()?"YES":"NO");
	}
}

#include<stdio.h>
#include<string.h>
#include<queue>
#define INF 0x3f3f3f3f
#define MAXN 105
int map[105][105];
int queue[MAXN], in[MAXN];
int n;
int toposort()
{
	int i, j;
	int iq = 0;
	for(i = 0; i < n; i ++){
		if(!in[i]){
			queue[iq++] = i;
		}
	}
	for(i = 0; i < iq; i ++){
		int temp = queue[i];
		for(j = 0; j < n; j ++){
			if(map[temp][j]){
				--in[j];
				if(!in[j]){
					queue[iq++] = j;
				}
			}
		}
	}
	return iq == n;
}
int main()
{
	int m, i, j, a, b;
	while(scanf("%d%d", &n, &m), n||m){
		memset(map, 0, sizeof(map));
		memset(in, 0, sizeof(in));
		for(i = 0; i < m; i ++){
			scanf("%d%d", &a, &b);
			if(!map[a][b]){ //去重
			map[a][b] = 1;
			in[b]++;
		}
	}
		printf("%s\n", toposort()?"YES":"NO");
	}
	return 0;
}