poj 2186 Popular Cows tarjan

Popular Cows

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

Hint

Cow 3 is the only cow of high popularity. 

low数组的理解：对于一个点，能从未遍历的边到达的最低的深度优先级

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll __int64
#define inf 2000000001
int scan()
{
    int res = 0 , ch ;
    while( !( ( ch = getchar() ) >= '0' && ch <= '9' ) )
    {
        if( ch == EOF )  return 1 << 30 ;
    }
    res = ch - '0' ;
    while( ( ch = getchar() ) >= '0' && ch <= '9' )
        res = res * 10 + ( ch - '0' ) ;
    return res ;
}
struct is
{
    int u,v;
    int next;
}edge[50010];
int head[50010];
int belong[50010];
int dfn[50010];
int low[50010];
int stackk[50010];
int instack[50010];
int number[50010];
int du[100010];
int n,m,jiedge,lu,bel,top;
void update(int u,int v)
{
    jiedge++;
    edge[jiedge].u=u;
    edge[jiedge].v=v;
    edge[jiedge].next=head[u];
    head[u]=jiedge;
}
void dfs(int x)
{
    dfn[x]=low[x]=++lu;
    stackk[++top]=x;
    instack[x]=1;
    for(int i=head[x];i;i=edge[i].next)
    {
        if(!dfn[edge[i].v])
        {
            dfs(edge[i].v);
            low[x]=min(low[x],low[edge[i].v]);
        }
        else if(instack[edge[i].v])
        low[x]=min(low[x],dfn[edge[i].v]);
    }
    if(low[x]==dfn[x])
    {
        int sum=0;
        bel++;
        int ne;
        do
        {
            sum++;
            ne=stackk[top--];
            belong[ne]=bel;
            instack[ne]=0;
        }while(x!=ne);
        number[bel]=sum;
    }
}
void tarjan()
{
    memset(dfn,0,sizeof(dfn));
    bel=lu=top=0;
    for(int i=1;i<=n;i++)
    if(!dfn[i])
    dfs(i);
}
int main()
{
    int i,t;
    while(~scanf("%d%d",&n,&m))
    {
        memset(head,0,sizeof(head));
        jiedge=0;
        for(i=1;i<=m;i++)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            update(u,v);
        }
        tarjan();
        for(i=1;i<=jiedge;i++)
        if(belong[edge[i].v]!=belong[edge[i].u])
        du[belong[edge[i].u]]++;
        int flag=0,pos;
        /*for(i=1;i<=n;i++)
        cout<<belong[i]<<endl;*/
        for(i=1;i<=bel;i++)
        {
            if(!du[i])
            {
                flag++;
                pos=i;
            }
        }
        if(flag!=1)
        printf("0
");
        else
        printf("%d
",number[pos]);
    }
    return 0;
}