看不懂题，蒟蒻中文英文都太差了。。。
于是Orz itwiiioi巨巨！
结果终于理解了：就是求有向图非单点的强连通分量个数。
tarjan妥妥的。。。（板子*1 get√）
1 /************************************************************** 2 Problem: 1654 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:32 ms 7 Memory:1476 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <cstring> 12 #include <algorithm> 13 14 using namespace std; 15 const int N = 10005; 16 const int M = 50005; 17 struct edges{ 18 int next, to; 19 }e[M]; 20 int n, m; 21 int cnt, top, num, tot, ans, sz; 22 int dfn[N], low[N], vis[N], s[N], first[N]; 23 24 inline int read(){ 25 int x = 0, sgn = 1; 26 char ch = getchar(); 27 while (ch < '0' || ch > '9'){ 28 if (ch == '-') sgn = -1; 29 ch = getchar(); 30 } 31 while (ch >= '0' && ch <= '9'){ 32 x = x * 10 + ch - '0'; 33 ch = getchar(); 34 } 35 return sgn * x; 36 } 37 38 inline void add_edge(int x, int y){ 39 e[++tot].next = first[x]; 40 first[x] = tot; 41 e[tot].to = y; 42 } 43 44 void DFS(int p){ 45 dfn[p] = low[p] = ++cnt; 46 s[++top] = p, vis[p] = 1; 47 int x, y; 48 for (x = first[p]; x; x = e[x].next){ 49 y = e[x].to; 50 if (!vis[y]) DFS(y); 51 if (vis[y] < 2) 52 low[p] = min(low[p], low[y]); 53 } 54 if (dfn[p] == low[p]){ 55 ++num, sz = 0; 56 while (s[top + 1] != p){ 57 y = s[top--]; 58 vis[y] = 2; 59 ++sz; 60 } 61 if (sz != 1) ++ ans; 62 } 63 } 64 65 inline void tarjan(){ 66 cnt = top = num = 0; 67 memset(vis, sizeof(vis), 0); 68 for (int i = 1; i <= n; ++i) 69 if (!vis[i]) DFS(i); 70 } 71 72 int main(){ 73 n = read(), m = read(); 74 int x, y; 75 for (int i = 1; i <= m; ++i){ 76 x = read(), y = read(); 77 add_edge(x, y); 78 } 79 tarjan(); 80 printf("%d ", ans); 81 return 0; 82 }
（p.s. 为什么改进版比不改进版要慢。。。这不科学的T T）