POJ2186 Popular Cows 强连通分量Tarjan算法

今天听见某大佬说POJ2186是连通图的入门题，然而我这个菜鸡什么都不会，因此决定记录下来...

首先贴大佬博客Orz：https://www.cnblogs.com/void/articles/2048209.html

                                   https://blog.csdn.net/chang_mu/article/details/38709047

附上AC代码：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<algorithm>
#include<stack>
using namespace std;
#define ll long long

const int MAXN=10005;
const int MAXM=50005;
int n,m;
struct Edge
{
    int to,next;
}edge[MAXM];
int head[MAXN],tol;
int low[MAXN],dfn[MAXN],belong[MAXN];
stack<int>s;
int index;
int scc;
bool vis[MAXN];
int num[MAXN];

void Tarjan(int u)
{
    int v;
    low[u]=dfn[u]=++index;
    s.push(u);
    vis[u]=true;
    for(int i=head[u];i!=-1;i=edge[i].next)
    {
        v=edge[i].to;
        if(!dfn[v])
        {
            Tarjan(v);
            if(low[u]>low[v])
                low[u]=low[v];
        }
        else if(vis[v]&&low[u]>dfn[v])
            low[u]=dfn[v];
    }
    if(low[u]==dfn[u])
    {
        scc++;
        do{
            v=s.top();
            s.pop();
            vis[v]=false;
            belong[v]=scc;
            num[scc]++;
        }while(v!=u);
    }
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(head,-1,sizeof(head));
        memset(dfn,0,sizeof(dfn));
        memset(vis,false,sizeof(vis));
        memset(num,0,sizeof(num));
        tol=0;
        index=scc=0;
        int u,v;
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&u,&v);
            edge[tol].to=v;
            edge[tol].next=head[u];
            head[u]=tol++;
        }
        for(int i=1;i<=n;i++)
            if(!dfn[i])
                Tarjan(i);
        if(scc==1)
        {
            printf("%d\n",n);
            continue;
        }
        int out[MAXN];
        memset(out,0,sizeof(out));
        for(int i=1;i<=n;i++)
            for(int j=head[i];j!=-1;j=edge[j].next)
                if(belong[i]!=belong[edge[j].to])
                    out[belong[i]]++;
        int cnt=0,ans=0;
        for(int i=1;i<=scc;i++)
            if(out[i]==0)
            {
                cnt++;
                ans=num[i];
            }
        if(cnt!=1)
            printf("0\n");
        else
            printf("%d\n",ans);
    }
	return 0;
}