POJ 2186 Popular Cows

题目链接

题意：一个奶牛，之间有互相仰慕的关系，如果A仰慕B，B仰慕C，意味着A也仰慕C，问有多少奶牛被所有奶牛仰慕

思路：强连通缩点，点权为连通集合个数，如果出度为0的点个数大于1个，答案就是0，如果等于1个，答案就是那个点的权值

代码：

#include <cstdio>
#include <cstring>
#include <vector>
#include <algorithm>
#include <stack>
using namespace std;

const int N = 10005;

int n, m;
vector<int> g[N];

int pre[N], dfn[N], dfs_clock, sccn, sccno[N], val[N];
stack<int> S;

void dfs_scc(int u) {
	pre[u] = dfn[u] = ++dfs_clock;
	S.push(u);
	for (int i = 0; i < g[u].size(); i++) {
		int v = g[u][i];
		if (!pre[v]) {
			dfs_scc(v);
			dfn[u] = min(dfn[u], dfn[v]);
		} else if (!sccno[v]) dfn[u] = min(dfn[u], pre[v]);
	}
	if (dfn[u] == pre[u]) {
		sccn++;
		int cnt = 0;
		while (1) {
			cnt++;
			int x = S.top(); S.pop();
			sccno[x] = sccn;
			if (x == u) break;
		}
		val[sccn] = cnt;
	}
}

void find_scc() {
	dfs_clock = sccn = 0;
	memset(pre, 0, sizeof(pre));
	memset(sccno, 0, sizeof(sccno));
	for (int i = 1; i <= n; i++)
		if (!pre[i]) dfs_scc(i);
}

int out[N];

int main() {
	while (~scanf("%d%d", &n, &m)) {
		int u, v;
		while (m--) {
			scanf("%d%d", &u, &v);
			g[u].push_back(v);
		}
		find_scc();
		memset(out, 0, sizeof(out));
		for (int i = 1; i <= n; i++) {
			for (int j = 0; j < g[i].size(); j++) {
				int v = g[i][j];
				if (sccno[i] != sccno[v])
					out[sccno[i]]++;
			}
		}
		int cnt = 0, ans;
		for (int i = 1; i <= sccn; i++) {
			if (!out[i]) {
				cnt++;
				ans = val[i];
			}
		}
		if (cnt > 1) ans = 0;
		printf("%d\n", ans);
	}
	return 0;
}

POJ 2186 Popular Cows（强连通）