LuoGu P2863 [USACO06JAN]牛的舞会The Cow Prom

Posted 2021-01-06 equinox-flower

题目传送门
这个题还是个缩点的板子题......
答案就是size大于1的强连通分量的个数
加一个size来统计就好了

#include <iostream>
#include <cstdlib>
#include <cstdio>

using namespace std;

const int N=1e5+5;
const int M=5e5+5;

struct edge{
    int to,next;
}e[M];

int n,m,dfn[N],low[N],cnt,head[N];
int idx[N],s[N],top,tot,sum,siz[N];
bool ins[N];int ind[N],ans;

inline void build(int u,int v){
    e[++tot].next=head[u];
    head[u]=tot;e[tot].to=v;
    return ;
}

inline void tarjan(int cur){
    s[++top]=cur;ins[cur]=true;
    dfn[cur]=low[cur]=++cnt;
    for(int i=head[cur];i;i=e[i].next){
        int k=e[i].to;
        if(!dfn[k]){
            tarjan(k);
            low[cur]=min(low[cur],low[k]);
        }else if(ins[k]) low[cur]=min(low[cur],dfn[k]);
    }
    if(low[cur]==dfn[cur]){
        register int ss=0;
        ++sum;
        while(s[top+1]!=cur){
            ++ss;
            idx[s[top]]=sum;
            ins[s[top--]]=false;
        }
        siz[sum]=ss;
    }
    return ;
}

int main(){
    scanf("%d%d",&n,&m);
    for(int i=1;i<=m;++i){
        register int u,v;
        scanf("%d%d",&u,&v);
        build(u,v);
    }
    for(int i=1;i<=n;++i) if(!dfn[i]) tarjan(i);
    for(int i=1;i<=n;++i)
        for(int j=head[i];j;j=e[j].next){
            int k=e[j].to;
            if(idx[i]!=idx[k]) ++ind[idx[k]];
        }
    for(int i=1;i<=sum;++i) if(!ind[i]&&siz[i]>1) ++ans;
    printf("%d
",ans);
    return 0;
}