Popular Cows+求强联通量简单题+tarjan算法+POJ

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 23066		Accepted: 9451

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

解决方案：先求强连通分量，缩点，反向建图，求入度为0的点。

code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<vector>
#include<stack>
#define MMAX 10003
using namespace std;
int low[MMAX],dfn[MMAX],instack[MMAX];
int in[MMAX];
int fa[MMAX],cnt[MMAX];
int index;
stack<int>S;
vector<int>Map[MMAX];
int N,M;
void init()
{
    index=0;
    memset(cnt,0,sizeof(cnt));
    memset(instack,0,sizeof(instack));
    memset(in,0,sizeof(in));
    for(int i=0; i<=N; i++)
    {
        fa[i]=i;
        Map[i].clear();
    }
    while(!S.empty())
        S.pop();

}
void tarjan(int u)
{
    low[u]=dfn[u]=index++;///时间戳
    S.push(u);
    instack[u]=1;
    int len=Map[u].size();
    for(int i=0; i<len; i++)
    {
        int v=Map[u][i];
        if(!instack[v])
        {
            tarjan(v);
            low[u]=min(low[u],low[v]);
        }
        else if(instack[v]==1)
        {
            low[u]=min(low[v],low[u]);
        }
    }
    if(dfn[u]==low[u])
    {
        while(!S.empty())
        {
            int temp=S.top();
            S.pop();
            fa[temp]=u;
            instack[temp]=2;
            if(temp==u) break;
        }

    }

}
int main()
{
    while(~scanf("%d%d",&N,&M))
    {
        init();
        for(int i=0; i<M; i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            Map[a].push_back(b);
        }
        for(int i=1; i<=N; i++)
        {
            if(!instack[i])
            {
                tarjan(i);
            }
        }
        for(int i=1; i<=N; i++)
        {
            int len=Map[i].size();
            for(int j=0; j<len; j++)
            {
                int v=Map[i][j];
                if(fa[v]!=fa[i])
                {
                    in[fa[i]]++;
                }
            }
        }
        int _sum=0,ss=0;
        int pre=-1;
        for(int i=1; i<=N; i++)
        {
            if(!in[fa[i]])
            {
                if(pre!=-1&&pre!=fa[i]) ss++;
                _sum++;
                pre=fa[i];
            }

        }
        if(!ss)
            printf("%d\n",_sum);
        else printf("0\n");
    }
    return 0;
}