luoguP4778 Counting swaps

题目链接

题解

首先,对于每个(i)向(a[i])连边.

这样会连出许多独立的环.

可以证明,交换操作不会跨越环.

每个环内的点到最终状态最少交换步数是 (环的大小-1)

那么设(f[i])表示环大小为(i)的方案数

则

[f[i] = sum_{x+y=i}f[x] * f[y] * g(x,y) * (^{i-1}_{x-1}) ]

其中

[g(x,y)={^{frac{x+y}{2}{(x+y为偶数且x=y)}}_{x+y(else)} ]

打标可以发现(f[n] = n^{n-2}(n≠1))

那么假设有(k)个环,第(i)个环大小为(a[i])

则

[ans = prod f[a[i]] *T ]

(T)是把(n-k)步分进每个环的方案数

(T=frac{(n-k)!}{prod(a[i]-1)!})

还有另一种方法算(T)(具体看代码)

Code

#include<bits/stdc++.h>
using namespace std;
const int N = 1e5 + 10, Mod = 1e9 + 9;
int fpow(int a, int b) {
	if (b <= 0) return 1;
	int res = 1;
	for (; b; b >>= 1, a = 1ll * a * a % Mod) if (b & 1) res = 1ll * res * a % Mod;
	return res;
}
bool vis[N];
int to[N], a[N];
int dfs(int x) {
	vis[x] = 1;
	if (vis[to[x]]) return 1;
	else return 1 + dfs(to[x]);
}
int fac[N], ifac[N]; 
int C(int n, int m) {
	if (n < m) return 0;
	return 1ll * fac[n] * ifac[n - m] % Mod * ifac[m] % Mod;
}
void solve() {
	int n, ans = 1;
	scanf("%d", &n);
	for (int i = 1; i <= n; i++) scanf("%d", &to[i]);
	memset(vis, 0, sizeof(vis));
	int len = 0;
	for (int i = 1; i <= n; i++) 
		if (!vis[i])
			a[++len] = dfs(i);
//	for (int i = 1; i <= len; i++) ans = 1ll * ans * fpow(a[i], a[i] - 2) % Mod * ifac[a[i] - 1] % Mod;
	for (int i = 1, sum = 0; i <= len; sum += a[i++] - 1)
		ans = 1ll * ans * fpow(a[i], a[i] - 2) % Mod * C(n - sum - len, a[i] - 1) % Mod;
	printf("%d
", /*1ll * ans * fac[n - len] % Mod*/ans);
	return ;
}
int main() {
	int T;
	scanf("%d", &T);
	fac[0] = 1;
	for (int i = 1; i <= 100000; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
	ifac[100000] = fpow(fac[100000], Mod - 2);
	for (int i = 100000; i >= 1; i--) ifac[i - 1] = 1ll * ifac[i] * i % Mod; 
	while (T--) solve();
	return 0;
}