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