P2863 牛的舞会

题面:https://www.luogu.org/problemnew/show/P2863

本题直接用tarjan求出图中节点数大于一的强联通分量个数,然后输出即可。

Code:
#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<ctime>
using namespace std;
const int N=50005;
int n,m,head[N],dfn[N],q[N],low[N],cnt,t,top,ans;
bool vis[N];
struct node{
    int u,v,next;
}edge[N];
void push(int u,int v){
    edge[cnt].u=u;
    edge[cnt].v=v;
    edge[cnt].next=head[u];
    head[u]=cnt++;
}
void tarjan(int u){
    dfn[u]=low[u]=++t;
    q[++top]=u;
    vis[q[top]]=true;
    for(int i=head[u];i!=-1;i=edge[i].next){
        int v=edge[i].v;
        if(dfn[v]==-1){
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(vis[v]){
            low[u]=min(low[u],dfn[v]);
        }
    }
    if(low[u]==dfn[u]){
        int num=0;
        while(q[top+1]!=u&&top>0){
            ++num;
            vis[q[top--]]=false;
        }
        if(num>1){
            ++ans;
        }
    }
}
int main(){
    int u,v;
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    memset(dfn,-1,sizeof(dfn));
    memset(vis,false,sizeof(false));
    for(int i=1;i<=m;i++){
        scanf("%d%d",&u,&v);
        push(u,v);
    }
    for(int i=1;i<=n;i++){
        if(dfn[i]==-1){
            tarjan(i);
        }
    }
    printf("%d\n",ans);
    return 0;
}