POJ 2186 Popular Cows 受欢迎的奶牛

Popular Cows

Time Limit: 2000MS Memory Limit: 65536K
Total Submissions: 45992 Accepted: 18779
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

题目大意：
给出一些奶牛之间受欢迎的关系，（关系具有传递性，A欢迎B，B欢迎C，那么A也欢迎C）问有几头奶牛是受所有奶牛欢迎的
题目分析：
缩点求强连通分量，然后统计出度，最多只能有一个出度为0的点

// 这么多年不写Tarjan 居然是1A的  好哇塞啊
// 本来刚度这个题的时候 我以为是无向边 那我一想 Tarjan只要所有的点都在一起答案就是n  只要不是都在一起就是0呗
// 想了想才知道 是有向边
#include 
#include 
#include 
#include 
#include 
using namespace std;
#define Maxn 10005
struct Edge {
    int to,next;
}e[Maxn << 3];
int dfn[Maxn],low[Maxn],head[Maxn],tot,visx,sum[Maxn],cnt,tar[Maxn],out[Maxn];
bool exist[Maxn];
inline void Add_Edge(int u,int v) {
    e[++tot].to = v; e[tot].next = head[u]; head[u] = tot;
}
stack<int> st;
void Tarjan(int u) {
    low[u] = dfn[u] = ++visx;
    st.push(u); exist[u] = 1;
    for(int i=head[u]; i; i=e[i].next) {
        int v = e[i].to;
        if(!dfn[v]) {
            Tarjan(v);
            low[u] = min(low[u],low[v]);
        }
        else low[u] = min(low[u],dfn[v]);
    }
    if(low[u] == dfn[u]) {
        cnt++;// 缩点加一
        while(st.top() != u ) {
            tar[st.top()] = cnt;
            sum[cnt]++;
            st.pop();
        }
        tar[u] = cnt;
        sum[cnt]++;
        st.pop();// 弹出u
    }
    return ;
}

int main() {
    int n,m; scanf("%d %d",&n,&m);
    for(int u,v,i=1; i<=m; i++) {
        scanf("%d %d",&u,&v);
        Add_Edge(u,v);
    }
    memset(dfn,0,sizeof(dfn));
    for(int i=1; i<=n; i++)
        if(!dfn[i]) Tarjan(i);

    for(int i=1; i<=n; i++) {
        for(int j=head[i]; j; j=e[j].next) {
            int v = e[j].to;
            if(tar[i] != tar[v]) out[tar[i]]++;// u,v 缩点之后不再同一个强连通分量 并且存在从u指向v的边
        }
    }
    int Ans = 0;
    for(int i=1; i<=cnt; i++)
        if(out[i] == 0) {
            if(Ans != 0) { printf("0\n"); return 0; }
            Ans = sum[i];
        }
    printf("%d\n",Ans);
    return 0;
}