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40 lines
1.2 KiB

2 years ago
#include <bits/stdc++.h>
using namespace std;
const int N = 1e3 + 10, M = N << 1;
int ind[N];
int n, m;
//邻接表
int e[M], h[N], idx, ne[M];
void add(int a, int b) {
e[idx] = b, ne[idx] = h[a], h[a] = idx++;
}
int q[N];
int main() {
while (~scanf("%d %d", &n, &m) && n) {
memset(ind, 0, sizeof(ind));
memset(h, -1, sizeof h);
idx = 0;
for (int i = 1; i <= m; i++) {
int a, b;
scanf("%d %d", &a, &b);
add(a, b);
ind[b]++; //记录入度
}
int hh = 0, tt = -1;
for (int i = 0; i < n; i++)
if (ind[i] == 0) q[++tt] = i; //入度为0的入队列
while (hh <= tt) {
int u = q[hh++];
for (int i = h[u]; ~i; i = ne[i]) {
int j = e[i];
ind[j]--; //拓扑模拟减去边,入度-1
if (ind[j] == 0) q[++tt] = j; //如果入度为0则此节点入队列
}
}
//检查入过队列的节点个数是不是n个相等则无环
hh == n ? puts("YES") : puts("NO");
}
return 0;
}