学步园

题意：有N个牛，围在一水池边，它们用绳子互相绑着（有方向的）。如果绳子的方向一致，它们就能顺时针转，问有多少组牛可以跳舞。

思路：简单有向图的强连通分量。求出强连通分量，且强连通分量里的点数大于等于2的块就能跳舞。

//836K    79MS
#include
#include
#include
using namespace std;
const int VM = 10005;
const int EM = 50005;

struct Edge
{
    int to,nxt;
}edge[EM];

int head[VM],vis[VM],dfn[VM],low[VM];
int stack[VM+10],belong[VM];
int scc,cnt,top,ep;

void addedge (int cu,int cv)
{
    edge[ep].to = cv;
    edge[ep].nxt = head[cu];
    head[cu] = ep++;
}
int min (int a ,int b)
{
    return a > b ? b : a;
}
void Tarjan(int u)
{
    dfn[u] = low[u] = ++cnt;
    vis[u] = 1;
    stack[top++] = u;
    int v;
    for (int i = head[u];i != -1;i = edge[i].nxt)
    {
        v = edge[i].to;
        if (!dfn[v])
        {
            Tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else if (vis[v]) low[u] = min(low[u],dfn[v]);
    }
    if (dfn[u] == low[u])
    {
        ++scc;
        do{
            v = stack[--top];
            vis[v] = 0;
            belong[v] = scc;
        }while (u != v);
    }
}
void solve(int n)
{
    memset (vis,0,sizeof(vis));
    memset (dfn,0,sizeof(dfn));
    scc = cnt = top = 0;
    int u,v;
    for (u = 1;u <= n;u ++)
        if (!dfn[u])
            Tarjan(u);
    sort (belong+1,belong+n+1);
    belong[n+1] = -1;
   // for (int i = 0;i <= n;i ++)
   //     printf ("%d ",belong[i]);
    int pre = belong[1];
    cnt = 0;
    int ans = 0;
    for (int i = 1;i <= n+1;i ++)
    {
        if (pre == belong[i])
            ++cnt;
        else {
            pre = belong[i];
            if (cnt >= 2) ans ++;
            cnt = 1;
        }
    }
    printf ("%d\n",ans);
}
int main ()
{
    #ifdef LOCAL
        freopen("in.txt","r",stdin);
    #endif
    int n,m,u,v;
    while (~scanf ("%d%d",&n,&m))
    {
        memset (head,-1,sizeof(head));
        ep = 0;
        while (m --)
        {
            scanf ("%d%d",&u,&v);
            addedge (u,v);
        }
        solve(n);
    }
    return 0;
}