poj2186 Popular Cows 题解——S.B.S.

Popular Cows

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 29642		Accepted: 11996

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. 

Source

USACO 2003 Fall

——————————我是分割线————————————————————————————————

图的强联通分量。

题目大意：

有n只牛，牛A认为牛B很牛，牛B认为牛C很牛。

给你M个关系（谁认为谁牛），求大家都认为它很牛的牛有几只。

p.s.如果牛A认为牛B很牛，牛B认为牛C很牛。那么我们就认为牛A认为牛C很牛。

/*
    Problem:   poj 2186
    OJ:         POJ
    User:     S.B.S.
    Time:     297 ms
    Memory:     1468 kb
    Length:     2420 b
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<cstdlib>
#include<iomanip>
#include<cassert>
#include<climits>
#include<functional>
#include<bitset>
#include<vector>
#include<list>
#include<map>
#define maxn 10001
#define F(i,j,k) for(int i=j;i<=k;i++)
#define M(a,b) memset(a,b,sizeof(a))
#define FF(i,j,k) for(int i=j;i>=k;i--)
#define inf 0x3f3f3f3f
#define maxm 50001
#define mod 998244353
//#define LOCAL
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,m;
struct EDGE  
{  
    int from;
    int to;
    int next;  
}edge[maxm];  
int head[maxn];
int dfn[maxn],low[maxn],stack1[maxn];
int num[maxn],du[maxn],vis[maxn];  
int tim,tp,dcnt,sum;
int cnt,time=1,top,cut,tot;
inline void addedge(int u,int v)  
{
    edge[tot].from=u;  
    edge[tot].to=v;  
    edge[tot].next=head[u];  
    head[u]=tot;  
    tot++;  
}
inline void dfs(int u,int fa)  
{  
    dfn[u]=time;  
    low[u]=time;  
    time++;  
    vis[u]=1;  
    stack1[top]=u;  
    top++;  
    for(int i=head[u]; i!=-1; i=edge[i].next)  
    {  
        int v=edge[i].to;  
        if(!vis[v]){  
            dfs(v,u);  
            low[u]=min(low[u],low[v]);  
        }  
        else if(vis[v]){  
            low[u]=min(low[u],dfn[v]);  
        }  
    }  
    if(low[u]==dfn[u]){  
        cut++;  
        while(top>0&&stack1[top]!=u)  
        {  
            top--;  
            vis[stack1[top]]=2;  
            num[stack1[top]]=cut;  
        }  
    }  
}  
int main()
{
    std::ios::sync_with_stdio(false);//cout<<setiosflags(ios::fixed)<<setprecision(1)<<y;
    #ifdef LOCAL
    freopen("data.in","r",stdin);
    freopen("data.out","w",stdout);
    #endif
    cin>>n>>m;
    M(head,-1);
    F(i,0,m-1){
        int a,b;
        cin>>a>>b;
        addedge(a,b);
    }
    F(i,1,n) if(!vis[i]) dfs(i,0);
    F(i,1,n)for(int j=head[i];j!=-1;j=edge[j].next){
        if(num[i]!=num[edge[j].to]){
            du[num[i]]++;
        }
    }
    int x;sum=0;
    F(i,1,cut) 
     if(!du[i]){
            sum++;
            x=i;
    }
    if(sum==1){
        sum=0;
        F(i,1,n) if(num[i]==x) sum++;
        cout<<sum<<endl;
    }
    else cout<<"0"<<endl;
    return 0;
}