18.11.08 POJ 2186 Popular Cows（有向图+缩点）

描述

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.
输入

* 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.
输出* Line 1: A single integer that is the number of cows who are considered popular by every other cow.
样例输入

3 3
1 2
2 1
2 3

样例输出

1

提示

Cow 3 is the only cow of high popularity.
来源

USACO 2003 Fall

#include <iostream>
#include <string.h>
#include <algorithm>
#include <stack>
#include <string>
#include <math.h>
#include <queue>
#include <stdio.h>
#include <string.h>
#include <set>
#include <vector>
#define maxn 10005
#define inf 999999
#define EPS 1e-10
#define lowbit(x) (int)(x&(-x))
using namespace std;

int n, m, tm = 1, colorid;
int dfn[maxn], low[maxn], color[maxn];
bool vis[maxn], to[maxn];
vector<vector<int>>G(maxn);
stack<int>q;

void tarjan(int u) {
    vis[u] = true;
    q.push(u);
    low[u]=dfn[u] = tm++;
    int size = G[u].size();
    for (int i = 0; i < size; i++) {
        int v = G[u][i];
        if (!dfn[v]) {
            tarjan(v);
            low[u] = min(low[u], low[v]);
        }
        else if (vis[v])
            low[u] = min(low[u], dfn[v]);
    }
    if (low[u] == dfn[u]) {
        int tmp = 0;
        ++colorid;
        while (tmp != u) {
            tmp = q.top(); q.pop();
            color[tmp] = colorid;
            vis[tmp] = false;
        }
    }
}

void solve() {
    for (int i = 1; i <= n; i++) {
        if (to[color[i]])continue;
        int size = G[i].size();
        for(int j=0;j<size;j++)
            if (color[i] != color[G[i][j]]) {
                to[color[i]] = true;
                break;
            }
    }
    int cnt=0, ans;
    for(int i=1;i<=colorid;i++)
        if (!to[i]) {
            cnt++;
            ans = i;
        }
    if (cnt != 1) {
        printf("0
");
        return;
    }
    cnt = 0;
    for (int i = 1; i <= n; i++)
        if (color[i] == ans)
            cnt++;
    printf("%d
", cnt);
}

void init() {
    scanf("%d%d", &n, &m);
    while (m--) {
        int x, y;
        scanf("%d%d", &x, &y);
        G[x].push_back(y);
    }
    for(int i=1;i<=n;i++)
        if(!dfn[i])
            tarjan(i);
    solve();
}

int main() {
    init();
    return 0;
}