Popular Cows - POJ 2186 Tarjan

最新推荐文章于 2021-04-14 23:52:08 发布

原创最新推荐文章于 2021-04-14 23:52:08 发布 · 431 阅读

0 ·

CC 4.0 BY-SA版权

POJ 同时被 2 个专栏收录

206 篇文章

订阅专栏

图论

76 篇文章

订阅专栏

本文介绍了一种通过Tarjan算法解决计算图中被所有其他点可达的超级节点数量的问题。利用Tarjan缩点法，文章详细展示了如何找出图中唯一的无出度超级节点，并以此确定最终答案。

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 24359		Accepted: 9999

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

Sample Output

题意：给定一些点和单向边，问有多少点使得其他所有的点到这个点都存在路径。

思路：Tarjan缩点后，如果有且仅有一个点的出度为0，那么这个点内的点的数目就是最后的答案。

AC代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    int v,next;
}edge[50010];
int n,m,Head[10010],DFN[10010],Low[10010],stack[10010],scnt,cnt,tot,top,Belong[10010];
bool instack[10010],flag;
int num[10010],in[10010],out[10010];
void add(int u,int v)
{
    edge[tot].v=v;
    edge[tot].next=Head[u];
    Head[u]=tot++;
}
void Tarjan(int u)
{
    int i,j,k,v;
    DFN[u]=Low[u]=++cnt;
    stack[top++]=u;
    instack[u]=1;
    for(i=Head[u];i!=-1;i=edge[i].next)
    {
        v=edge[i].v;
        if(!DFN[v])
        {
            Tarjan(v);
            Low[u]=min(Low[u],Low[v]);
        }
        else if(instack[v])
          Low[u]=min(Low[u],DFN[v]);
    }
    if(DFN[u]==Low[u])
    {
        scnt++;
        do
        {
            v=stack[--top];
            instack[v]=0;
            Belong[v]=scnt;
            num[scnt]++;
        }while(u!=v);
    }
}
int main()
{
    int i,j,k,u,v,ans;
    memset(Head,-1,sizeof(Head));
    scanf("%d%d",&n,&m);
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&u,&v);
        add(u,v);
    }
    for(i=1;i<=n;i++)
       if(!DFN[i])
         Tarjan(i);
    for(u=1;u<=n;u++)
       for(j=Head[u];j!=-1;j=edge[j].next)
       {
           v=edge[j].v;
           if(Belong[u]!=Belong[v])
           {
               out[Belong[u]]++;
               in[Belong[v]]++;
           }
       }
    k=0;
    for(i=1;i<=scnt;i++)
       if(out[i]==0)
       {
           k++;
           ans=num[i];
       }
    if(k!=1)
      ans=0;
    printf("%d\n",ans);
}