【Tarjan缩点】POJ2186 Popular Cows

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 35644		Accepted: 14532

Description

Every cow's dream is to become the most popular cow in the herd. In a herd of N (1 <= N <= 10,000) cows, you are given up to M (1 <= M <= 50,000) ordered pairs of the form (A, B) that tell you that cow A thinks that cow B is popular. Since popularity is transitive, if A thinks B is popular and B thinks C is popular, then A will also think that C is 
popular, even if this is not explicitly specified by an ordered pair in the input. Your task is to compute the number of cows that are considered popular by every other cow. 

Input

* Line 1: Two space-separated integers, N and M 

* Lines 2..1+M: Two space-separated numbers A and B, meaning that A thinks B is popular. 

Output

* Line 1: A single integer that is the number of cows who are considered popular by every other cow. 

Sample Input

3 3
1 2
2 1
2 3

Sample Output

1

题目大意：

　　给定一个有向图，求有多少个顶点是由任何顶 点出发都可达的。

 

题解：

　　Tarjan缩点构成DAG
　　若DAG上面如果有唯一的出度为0的点,则该点能被所有的点可达。那么该点所代表的连通分量上的所有的原图中的点，都能被原图中的所有点可达，则该连通分量的点数，就是答案。 
　　若DAG上面如果有不止一个出度为0的点，则这些点互相不可达，原问题无解，答案为0

 

代码：

//by 减维
#include<cstdio>
#include<iostream>
#include<cstring>
#include<queue>
#include<cstdlib>
#include<ctime>
#include<cmath>
#include<algorithm>
#define ll long long
#define maxn
using namespace std;

struct edge{
    int fr,to,ne;
}e[500005];

int n,m,ecnt,cnt,num,tot,ans,ans2;
int map[100005],dfn[100005],low[100005],chu[100005],shu[100005];
int head[100005],zhan[100005];
bool pd[100005];

void add(int x,int y)
{
    e[++ecnt].to=y;
    e[ecnt].fr=x;
    e[ecnt].ne=head[x];
    head[x]=ecnt;
}

void tarjan(int x)
{
    dfn[x]=low[x]=++num;
    zhan[++tot]=x;
    pd[x]=1;
    for(int i=head[x];i;i=e[i].ne)
    {
        int dd=e[i].to;
        if(!dfn[dd]){
            tarjan(dd);
            low[x]=min(low[x],low[dd]);
        }else if(pd[dd]){
            low[x]=min(dfn[dd],low[x]);
        }
    }
    if(dfn[x]==low[x]){
        cnt++;
        int t;
        do{
            t=zhan[tot--];
            map[t]=cnt;
            pd[t]=0;
            shu[cnt]++;
        }while(t!=x);
    }
}

void init()
{
    cnt=ecnt=num=tot=ans=0;
    memset(e,0,sizeof(e));
    memset(pd,0,sizeof(pd));
    memset(dfn,0,sizeof(dfn));
    memset(low,0,sizeof(low));
    memset(map,0,sizeof(map));
    memset(chu,0,sizeof(chu));
    memset(shu,0,sizeof(shu));
    memset(head,0,sizeof(head));
    memset(zhan,0,sizeof(zhan));
}

int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        init();
        for(int i=1;i<=m;++i){
            int x,y;
            scanf("%d%d",&x,&y);
            add(x,y);
        }
        for(int i=1;i<=n;++i)if(!dfn[i])tarjan(i);
        //for(int i=1;i<=n;++i)printf("%d ",dfn[i]);printf("
");
        //for(int i=1;i<=n;++i)printf("%d ",low[i]);printf("
");
        //for(int i=1;i<=n;++i)printf("%d ",map[i]);printf("
");
        for(int i=1;i<=ecnt;++i)
        {
            int x=e[i].fr,dd=e[i].to;
            if(map[x]!=map[dd])chu[map[x]]++;
        }
        memset(pd,0,sizeof(pd));
        for(int i=1;i<=n;++i)
            if(chu[map[i]]==0&&!pd[map[i]])
                pd[map[i]]=1,ans++,ans2=shu[map[i]];
        if(ans==1)printf("%d
",ans2);
        else printf("0
");
    }
}