POJ 3280 The Cow Prom (强连通分量)

http://poj.org/problem?id=3180
题意：n个点，m条边，求节点数大于1的强连通分量数。

强连通分量的模板题，只需要计数每个强连通分量的点数，找出大于1的即可。

#include<iostream>
#include<cstdio>
#include<stack>
#include<algorithm>
#include<cstring>using namespace std;
const int maxn=20010,maxm=200100;
int dfn[maxn],low[maxn],head[maxn],color[maxn],sum[maxn];
int n,m,k,cnt,deep,ans;
bool vis[maxn];
struct edge{
    int v,next;
}e[maxm];stack<int> s;void addedge(int u,int v){
    e[k].v=v;e[k].next=head[u];head[u]=k++;}void tarjan(int u)
{
    deep++;dfn[u]=deep;low[u]=deep;vis[u]=1;s.push(u);for(int i=head[u];~i;i=e[i].next){
    int v=e[i].v;if(!dfn[v]){
    tarjan(v);low[u]=min(low[u],low[v]);}else{
    if(vis[v]){
    low[u]=min(low[u],low[v]);}}}if(dfn[u]==low[u]){
    cnt++;color[u]=cnt;vis[u]=0;while(s.top()!=u){
    color[s.top()]=cnt;vis[s.top()]=0;s.pop();}s.pop();}
}int main()
{
    int x,y;memset(head,-1,sizeof(head));scanf("%d%d",&n,&m);for(int i=1;i<=m;i++){
    scanf("%d%d",&x,&y);addedge(x,y);}for(int i=1;i<=n;i++){
    if(!dfn[i])tarjan(i);}for(int i=1;i<=n;i++)sum[color[i]]++;for(int i=1;i<=cnt;i++)if(sum[i]>1)ans++;printf("%d\n",ans);return 0;
}