这道题挺难的，可以加深对割点的理解，还有，排列组合好重要了，分连通块，然后乘法原理（加法原理计数什么的）

传送门   https://www.luogu.org/problem/P3225

省选oi题好难啊QAQ

 

 

 

 

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<vector>
#define maxn 10000
using namespace std;
vector<int>G[maxn];
void insert(int be, int en) {
	G[be].push_back(en);
}
int low[maxn], dfn[maxn], vis[maxn], df;
int cnt[maxn];
void tarjan(int root ,int x) {
	low[x] = dfn[x] = ++df;
	int chal = 0;
	for (int i = 0; i < G[x].size(); i++) {
		int p = G[x][i];
		if (!dfn[p]) {
			tarjan(root, p);
			chal++;
			low[x] = min(low[x], low[p]);
			if (root != x) {
				if (low[p] >= dfn[x]) {
					cnt[x] = 1;
				}
			}
			else {
				if (chal >= 2) {
					cnt[root] = 1;
				}
			}
		}
		else {
			low[x] = min(low[x], dfn[p]);
		}
	}
}
int num, cn;
int gp;
void dfs(int x) {
	if (cnt[x]) return;
	if (vis[x]) return;
	num++;
	vis[x] = gp;
	for (int i = 0; i < G[x].size(); i++) {
		int p = G[x][i];
		if (cnt[p] && vis[p] != gp) {
			cn++;
			vis[p] = gp;
		}
		dfs(p);
	}
}
int main() {
	int m;
	int ccc = 0;
	while (~scanf("%d", &m) && m) {
		int n = -1;
		int be, en;
		ccc++;
		memset(dfn, 0, sizeof(dfn));
		memset(vis, 0, sizeof(vis));
		memset(cnt, 0, sizeof(cnt));
		for (int i = 0; i < m; i++) {
			scanf("%d %d", &be, &en);
			insert(be, en);
			insert(en, be);
			n = max(be, n);
			n = max(en, n);
		}
		for (int i = 1; i <= n; i++) {
			if (!dfn[i]) tarjan(i, i);
		}
		long long ans1 = 0, ans2 = 1;
		for (int i = 1; i <= n; i++) {

			if (!vis[i] && !cnt[i]) {
				cn = num = 0;
				gp++;
				dfs(i);

				if (cn == 0) {
					ans1 += 2;
					ans2 *= num * (num - 1) / 2;
				}
				if (cn == 1) {
					ans1 += 1;
					ans2 *= num;
				}
			}
		}
		printf("Case %d: %lld %lld\n",ccc, ans1, ans2);
		for (int i = 0; i <= n; i++) G[i].clear();
	}
	return 0;
}