POJ 2186 Popular Cows(缩点+出度判断）

最新推荐文章于 2020-05-18 14:45:19 发布

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本文介绍了一种使用强连通分量缩点成为有向无环图的方法来解决特定类型的图论问题。通过此方法可以有效地找出哪些节点被认为是“最热门”的节点，即被所有其他节点认为是受欢迎的。文章提供了详细的实现步骤和完整的代码示例。

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Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions:36908		Accepted: 15026

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

题解：
强连通分量缩点。成为一个有向无环图。若有两个以上的点
出度为0，则这两个点一定不会产生关系，若有一个点出度
为0，则这个点所包含的连通分量的个数便是答案的解。

代码：

#include<cstdio>
#include<string.h>
#include<stack>
using namespace std;
const int maxn=10007;
int head[maxn],dfn[maxn],low[maxn],tol;
int from[maxn*5],to[maxn*5],vis[maxn];
int out[maxn],par[maxn],num,sum[maxn];
stack<int>P;
struct node
{
    int to,next;
}rode[maxn*5];
void add(int a,int b)
{
    rode[tol].to=b;
    rode[tol].next=head[a];
    head[a]=tol++;
}
void tarjan(int x)
{
    P.push(x);
    vis[x]=1;
    dfn[x]=low[x]=++tol;
    for(int i=head[x];i!=-1;i=rode[i].next)
    {
        node e=rode[i];
        if(!vis[e.to])tarjan(e.to);
        if(vis[e.to])low[x]=min(low[x],low[e.to]);
    }
    if(dfn[x]==low[x])
    {
        num++;int v;
        sum[num]=0;
        do
        {
            v=P.top();P.pop();
            sum[num]++;
            par[v]=num;
        }while(v!=x);
    }
}
int main()
{
    int n,m,i,j;tol=0;
    scanf("%d%d",&n,&m);
    memset(head,-1,sizeof(head));
    memset(vis,0,sizeof(vis));
    memset(out,0,sizeof(out));
    memset(in,0,sizeof(in));
    for(i=1;i<=m;i++)
    {
        scanf("%d%d",&from[i],&to[i]);
        add(from[i],to[i]);
    }
    tol=num=0;
    for(i=1;i<=n;i++)
        if(!vis[i])tarjan(i);
    for(i=1;i<=m;i++)
    {
        int x=from[i],y=to[i];
        x=par[x];y=par[y];
        if(x!=y)
        {
            out[x]++;
        }
    }
    int ans=0,cnt=0;
    for(i=1;i<=num;i++)
    {
        if(out[i]==0)cnt++,ans=sum[i];
    }
    if(cnt>1)ans=0;
    printf("%d\n",ans);
    return 0;
}