这道题实质就是缩点，缩完点后一定是$DAG$，统计每个缩点的大小。如果有且只有一个缩点的出度为$0$，说明其他所有点都会直接或间接到达这个点。

1 #include 
     
       2  3 using namespace std; 4  5 #define re register 6 #define rep(i, a, b) for (re int i = a; i <= b; ++i) 7 #define repd(i, a, b) for (re int i = a; i >= b; --i) 8 #define maxx(a, b) a = max(a, b); 9 #define minn(a, b) a = min(a, b);10 #define LL long long11 #define inf (1 << 30)12 13 inline int read() {14     int w = 0, f = 1; char c = getchar();15     while (!isdigit(c)) f = c == '-' ? -1 : f, c = getchar();16     while (isdigit(c)) w = (w << 3) + (w << 1) + (c ^ '0'), c = getchar();17     return w * f;18 }19 20 const int maxn  = 1e4 + 5, maxm = 5e4 + 5;21 22 struct Edge {23     int u, v, pre;24 };25 26 struct Graph {27     Edge edges[maxm << 1];28     int n, m;29     int G[maxn];30     int dfs_clock, sccno[maxn], scc_cnt, cnt[maxn], pre[maxn], low[maxn];31     void init(int n) {32         this->n = n;33         m = 0;34         memset(G, 0, sizeof(G));35         dfs_clock = scc_cnt = 0;36     }37     void Add(int u, int v) {38         edges[++m] = (Edge){u, v, G[u]};39         G[u] = m;40     }41     stack
      
        s;42     void dfs(int u) {43         pre[u] = low[u] = ++dfs_clock;44         s.push(u);45         for (int i = G[u]; i; i = edges[i].pre) {46             register int v = edges[i].v;47             if (!pre[v]) {48                 dfs(v); minn(low[u], low[v]);49             }50             else if (!sccno[v]) minn(low[u], pre[v]);51         }52         if (pre[u] == low[u]) {53             int x = 0; scc_cnt++;54             while (x != u) {55                 x = s.top(); s.pop();56                 sccno[x] = scc_cnt;57                 cnt[scc_cnt]++;58             }59         }60     }61 } G;62 63 int n, m, b[maxn];64 65 int main() {66     n = read(), m = read();67     G.init(n);68 69     rep(i, 1, m) {70         int u = read(), v = read();71         G.Add(u, v);72     }73     rep(i, 1, n) if (!G.sccno[i]) G.dfs(i);74 75     rep(i, 1, G.m) {76         int u = G.edges[i].u, v = G.edges[i].v;77         if (G.sccno[u] != G.sccno[v])78             b[G.sccno[u]] = 1;79     }80 81     int ans = 0;82     rep(i, 1, G.scc_cnt) 83         if (!b[i]) {84             if (ans) { ans = 0; break; }85             ans = i;86         }87 88     printf("%d", G.cnt[ans]);89 90     return 0;91 }