poj 3180 The Cow Prom（强连通分量 Tarjan）

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;
}