2186 Popular Cows 图的连通性

Popular Cows

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 9234 Accepted: 3690

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 31 22 12 3

Sample Output

1

Hint

Cow 3 is the only cow of high popularity.

Source

USACO 2003 Fall

缩点判断出度为0的点的数目，当为1时即为那个联通分支里的数目

当大于1时无解，输出0

如果只有一个联通分支，嘿嘿，很简单

下面为关键代码

 

void Tarjan(int u) //Tarjan

{

    int v;

    node[u].DFN=node[u].LOW=(++idx);

    instack[u]=true;

    stack[++top]=u;

    for(Edge*p=node[u].first; p; p=p->next)

    {

        v=p->adj;

        if(!node[v].DFN)

        {

            Tarjan(v);

            if(node[v].LOW<node[u].LOW)

                node[u].LOW=node[v].LOW;

        }

        else if(instack[v]&&node[v].DFN<node[u].LOW)

            node[u].LOW=node[v].DFN;

    }

    if(node[u].DFN==node[u].LOW)

    {

        b_cnt++;

        do

        {

            v=stack[top--];

            instack[v]=false;

            node[v].belongs=b_cnt;

        }

        while(u!=v);

    }

}

 

 

 

        for(i=1; i<=N; i++)

        {

            flag[i]=true;

            for(Edge*p=node[i].first; p; p=p->next)

            if(node[i].belongs!=node[p->adj].belongs

               &&hash[node[i].belongs]!=node[p->adj].belongs)

            {

                    hash[node[i].belongs]=node[p->adj].belongs;

                    InsertEdge2(node[i].belongs,node[p->adj].belongs);

            }

            else

            if(node[i].belongs==node[p->adj].belongs&&!flag[p->adj])

            {

                    now[node[i].belongs].sum++;

                    flag[p->adj]=true;

            }

        }