这是道强连通分量的裸题．

#include <bits/stdc++.h>
using namespace std;
const int N = 10005;
int dfn[N], low[N], dfs_clock, scc_cnt, sccno[N];
int n, m;
int cnt;
struct edge {
    int to, next;
} e[N * 10];
int head[N];
void add_edge(int u, int v)
{
    e[++cnt].to = v, e[cnt].next = head[u], head[u] = cnt;
}
stack<int> s;
void dfs(int u)
{
    dfn[u] = low[u] = ++dfs_clock;
    s.push(u);
    for (int i = head[u]; i; i = e[i].next) {
        if (!dfn[e[i].to]) {
            dfs(e[i].to);
            low[u] = min(low[u], low[e[i].to]);
        } else {
            low[u] = min(low[u], dfn[e[i].to]);
        }
    }
    if (dfn[u] == low[u]) {
        scc_cnt++;
        for (;;) {
            int x = s.top();
            s.pop();
            sccno[x] = scc_cnt;
            if (x == u)
                break;
        }
    }
}
int main()
{
    while (~scanf("%d %d", &n, &m) && (m + n)) {
        cnt = 0;
        memset(head, 0, sizeof(head));
        for (int i = 0; i < m; i++) {
            int u, v;
            scanf("%d %d", &u, &v);
            add_edge(u, v);
        }
        scc_cnt = dfs_clock = 0;
        memset(dfn, 0, sizeof(dfn));
        memset(sccno, 0, sizeof(sccno));
        for (int i = 1; i <= n; i++) {
            if (!dfn[i])
                dfs(i);
        }
        if (scc_cnt == 1) {
            printf("Yes\n");
        } else {
            printf("No\n");
        }
    }
    return 0;
}