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;
}