[poj] 3180 the cow prom

原题

这是一道强连通分量板子题。
我们只用输出点数大于1的强连通分量的个数！

#include<cstdio>
#include<algorithm>
#include<stack>
#define N 10010
#define M 50010
using namespace std;
int n,m,head[N],dfn[N],low[N],cnt=1,t,sum,bel[N],num[N],out[N],ans;
bool instk[N];
stack <int> stk;
struct hhh
{
    int to,next;
}edge[M];

int read()
{
    int ans=0,fu=1;
    char j=getchar();
    for (;(j<'0' || j>'9') && j!='-';j=getchar()) ;
    if (j=='-') fu=-1,j=getchar();
    for (;j>='0' && j<='9';j=getchar()) ans*=10,ans+=j-'0';
    return ans*fu;
}

void add(int u,int v)
{
    edge[cnt].to=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}

void Tarjan(int x)
{
    dfn[x]=low[x]=++t;
    stk.push(x);
    instk[x]=1;
    int v;
    for (int i=head[x];i;i=edge[i].next)
    {
	v=edge[i].to;
	if (!dfn[v])
	{
	    Tarjan(v);
	    low[x]=min(low[x],low[v]);
	}
	else if (instk[v]) low[x]=min(low[x],dfn[v]);
    }
    if (dfn[x]==low[x])
    {
	sum++;
	do
	{
	    v=stk.top();
	    stk.pop();
	    instk[v]=0;
	    bel[v]=sum;
	    num[sum]++;
	}while(v!=x);
    }
}

int main()
{
    n=read();
    m=read();
    for (int i=1,a,b;i<=m;i++)
    {
	a=read();
	b=read();
	add(a,b);
    }
    for (int i=1;i<=n;i++)
	if (!dfn[i]) Tarjan(i);
    for (int i=1;i<=sum;i++)
	if (num[i]>1) ans++;
    printf("%d",ans);
    return 0;
}