POJ 2186 Popular Cows (强连通分量）

题意：—给定一个有向图，求有多少个顶点是由任何顶点出发都可达的。
题解：1. 求出所有强连通分量  2. 每个强连通分量缩成一点，则形成一个有向无环图DAG。—3. DAG上面如果有唯一的出度为0的点,则该点能被所有的点可达。那么该点所代表的连通分量上的所有的原图中的点，都能被原图中的所有点可达，则该连通分量的点数，就是答案。—4. DAG上面如果有不止一个出度为0的点，则这些点互相不可达，原问题无解，答案为0

PS:有向无环图中唯一出度为0的点，一定可以由任何点出发均可达(由于无环，所以从任何点出发往前走，必然终止于一个出度为0的点)

#include <iostream>using namespace std;#define N 10009#define min(a,b) (a<b?a:b)#define max(a,b) (a>b?a:b)int n, m;int size, cnt, top, id;int instack[N], stack[N];int head[N], low[N], dfn[N];int out[N], block[N], num[N];struct Edge { int v, next; } edge[N*5];void Tarjan ( int u ){  int v;instack[u] = 1;  stack[++top] = u;  dfn[u] = low[u] = ++id;  for ( int i = head[u]; i; i = edge[i].next ){  v = edge[i].v;if ( ! dfn[v] ){  Tarjan(v);  low[u] = min ( low[u], low[v] );  }  else if ( instack[v] ) //instack保证不是横叉边low[u] = min ( low[u], dfn[v] ); //注意不是 min(low[u],low[v])！}  if ( low[u] == dfn[u] ){  cnt++;  do {    v = stack[top--];  instack[v] = 0;  block[v] = cnt; num[cnt]++;        } while ( u != v );  }  }  int shrink (){int u, v, i, ans, flag = 0;for ( u = 1; u <= n; u++ ){for ( i = head[u]; i; i = edge[i].next ){v = edge[i].v;if ( block[u] != block[v] )out[block[u]]++;}}for ( i = 1; i <= cnt; i++ ){if ( out[i] == 0 ){ ans = num[i];flag++;}if ( flag > 1 ){ans = 0; break;}}return ans;}void Initial(){size = cnt = id = top = 0;for ( int i = 1; i <= n; i++ ){out[i] = block[i] = num[i] = 0;head[i] = instack[i] = dfn[i] = 0;}}void add ( int u, int v ){size++;edge[size].v = v;edge[size].next = head[u];head[u] = size;}int main(){int u, v;Initial();scanf("%d%d",&n,&m);while ( m-- ){scanf("%d%d",&u,&v);add(u,v);}for ( v = 1; v <= n; v++ )    if ( ! dfn[v] ) Tarjan ( v ); printf("%d\n", shrink() );return 0;}