Proving Equivalences UVA - 12167

题文：https://vjudge.net/problem/UVA-12167

 

题解：
　　很明显，先要缩点。然后画一下图就会发现是入度为0的点和出度为0的点取max。

 

代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#include <iostream>
#include <stack>
#define MAXN 50100
using namespace std;
struct edge{
    int first;
    int next;
    int to;
}a[MAXN*2];
int low[MAXN],dfn[MAXN],have[MAXN],fa[MAXN];
int in[MAXN],out[MAXN];
int num=0,num1=0,num2=0,n,m;
stack<int> s;

void cl(){
    memset(low,0,sizeof(low));
    memset(dfn,0,sizeof(dfn));
    memset(have,0,sizeof(have));
    memset(fa,0,sizeof(fa));
    memset(in,0,sizeof(in));
    memset(out,0,sizeof(out));
    memset(a,0,sizeof(a));
    num1=num2=num=0;
}

void addedge(int from,int to){
    a[++num].to=to;
    a[num].next=a[from].first;
    a[from].first=num;
}

void tarjian(int now){
    low[now]=dfn[now]=++num1;
    have[now]=1;s.push(now);
    for(int i=a[now].first;i;i=a[i].next){
        int to=a[i].to;
        if(!dfn[to]){
            tarjian(to);
            low[now]=min(low[now],low[to]);
        }
        else if(have[to]) low[now]=min(low[now],dfn[to]);
    }
    if(low[now]==dfn[now]){
        int u=-1;
        num2++;
        while(u!=now){
            u=s.top();s.pop();
            have[u]=0;
            fa[u]=num2;
        }
    }
}

void make(){
    for(int now=1;now<=n;now++){
        for(int i=a[now].first;i;i=a[i].next){
            int to=a[i].to;
            if(fa[now]!=fa[to]) out[fa[now]]++,in[fa[to]]++;
        }
    }
}

int main()
{
    int t;cin>>t;
    while(t--){
        cl();
        scanf("%d%d",&n,&m);
        for(int i=1;i<=m;i++){
            int x,y;scanf("%d%d",&x,&y);
            addedge(x,y);
        }
        while(!s.empty()) s.pop();
        for(int i=1;i<=n;i++) if(!dfn[i]) tarjian(i);
        make();
        int x=0,y=0;
        for(int i=1;i<=num2;i++) if(in[i]==0) x++;
        for(int i=1;i<=num2;i++) if(out[i]==0) y++;
        if(num2==1) printf("0\n");
        else printf("%d\n",max(x,y));
    }
    return 0;
}