codeforces 156D Clues（prufer序列）

codeforces 156D Clues

题意

给定一个无向图，不保证联通。求添加最少的边使它联通的方案数。

题解

根据prufer序列，带标号无根树的方案数是(n^{n-2})
依这个思想构建树即可。

代码

#include<bits/stdc++.h>
using namespace std;
#define fi first
#define se second
#define mp make_pair
#define pb push_back
#define rep(i, a, b) for(int i=(a); i<(b); i++)
#define sz(a) (int)a.size()
#define de(a) cout << #a << " = " << a << endl
#define dd(a) cout << #a << " = " << a << " "
#define all(a) a.begin(), a.end()
#define endl "
"
typedef long long ll;
typedef pair<int, int> pii;
typedef vector<int> vi;
//---

const int N = 101010;

int n, m, p;
int pre[N], cnt[N];

int find(int x) {
	if(x == pre[x]) return x;
	return pre[x] = find(pre[x]);
}
void join(int x, int y) {
	x = find(x);
	y = find(y);
	pre[x] = y;
}

int main() {
	std::ios::sync_with_stdio(false);
	std::cin.tie(0);
	cin >> n >> m >> p;
	rep(i, 1, n+1) pre[i] = i;
	rep(i, 0, m) {
		int u, v;
		cin >> u >> v;
		join(u, v);
	}
	rep(i, 1, n+1) ++cnt[find(i)];
	ll ans = 1;
	int c = 0;
	rep(i, 1, n+1) if(cnt[i]) {
		ans = ans * cnt[i] % p;
		++c;
	}
	rep(i, 0, c-2) ans = ans * n % p;
	if(c == 1) ans = 1 % p;
	cout << ans << endl;
	return 0;
}