比较平凡的一个容斥。
考虑把问题转化成，求 \(\forall i \in [1, n], r_i \ne i \land r_i \ne p_i\) 的 \(r\) 方案数。考虑到不弱于错排，所以容斥。设钦定 \(i\) 个 \(r_i\) 取了 \(i, p_i\) 中的一个的方案数为 \(f_i\)，其余任意，那么：
\[ans = \sum\limits_{i = 0}^n (-1)^i f_i (n - i)! \]考虑求 \(f_i\)。连边 \(i \to p_i\)，对每个环单独考虑。设第 \(i\) 个环点数为 \(s_i\)。这个东西抽象到环上就相当于，每一个点，可以不选，可以选择选它自己，也可以选择选它在环上的下一个点。设 \(h_{i, j, 0/1}\) 表示当前考虑到环上第 \(i\) 个点，有 \(j\) 个点选了，这个点是否选择第 \(i + 1\) 个点。枚举第一个点的状态，然后直接做即可。合并到 \(f_i\)，就是做一个加法卷积，暴力即可。
总时间复杂度 \(O(\sum s_i^2 + n \sum s_i) = O(n^2)\)。
code
// Problem: G - Three Permutations
// Contest: AtCoder - AtCoder Beginner Contest 214
// URL: https://atcoder.jp/contests/abc214/tasks/abc214_g
// Memory Limit: 1024 MB
// Time Limit: 2000 ms
//
// Powered by CP Editor (https://cpeditor.org)
#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mems(a, x) memset((a), (x), sizeof(a))
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ldb;
typedef pair<ll, ll> pii;
const int maxn = 3030;
const ll mod = 1000000007;
inline ll qpow(ll b, ll p) {
ll res = 1;
while (p) {
if (p & 1) {
res = res * b % mod;
}
b = b * b % mod;
p >>= 1;
}
return res;
}
ll n, a[maxn], b[maxn], c[maxn], fac[maxn], ifac[maxn], fa[maxn], sz[maxn], f[maxn], g[maxn], h[maxn][maxn][2];
inline ll C(ll n, ll m) {
if (n < m || n < 0 || m < 0) {
return 0;
} else {
return fac[n] * ifac[m] % mod * ifac[n - m] % mod;
}
}
int find(int x) {
return fa[x] == x ? x : fa[x] = find(fa[x]);
}
inline void merge(int x, int y) {
x = find(x);
y = find(y);
if (x != y) {
fa[x] = y;
sz[y] += sz[x];
}
}
void solve() {
scanf("%lld", &n);
for (int i = 1; i <= n; ++i) {
fa[i] = i;
sz[i] = 1;
}
fac[0] = 1;
for (int i = 1; i <= n; ++i) {
fac[i] = fac[i - 1] * i % mod;
}
ifac[n] = qpow(fac[n], mod - 2);
for (int i = n - 1; ~i; --i) {
ifac[i] = ifac[i + 1] * (i + 1) % mod;
}
for (int i = 1; i <= n; ++i) {
scanf("%lld", &a[i]);
c[a[i]] = i;
}
for (int i = 1; i <= n; ++i) {
scanf("%lld", &b[i]);
b[i] = c[b[i]];
merge(i, b[i]);
}
int s = 0;
f[0] = 1;
for (int _ = 1; _ <= n; ++_) {
if (fa[_] == _) {
int m = sz[_];
for (int i = 0; i <= n; ++i) {
g[i] = f[i];
f[i] = 0;
}
if (m == 1) {
f[0] = g[0];
for (int i = 1; i <= n; ++i) {
f[i] = (g[i] + g[i - 1]) % mod;
}
++s;
continue;
}
for (int x = 0; x <= 2; ++x) {
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= i; ++j) {
for (int k = 0; k < 2; ++k) {
h[i][j][k] = 0;
}
}
}
h[1][x >= 1][x == 2] = 1;
for (int i = 2; i <= m; ++i) {
for (int j = 0; j < i; ++j) {
for (int p = 0; p <= 1; ++p) {
for (int q = 0; q <= 2; ++q) {
if (p && q == 1) {
continue;
}
int nj = j + (q >= 1), np = (q == 2);
h[i][nj][np] = (h[i][nj][np] + h[i - 1][j][p]) % mod;
}
}
}
}
for (int i = 0; i <= m; ++i) {
for (int j = 0; j <= s; ++j) {
ll val = h[m][i][0];
if (x != 1) {
val = (val + h[m][i][1]) % mod;
}
f[i + j] = (f[i + j] + val * g[j] % mod) % mod;
}
}
}
s += m;
}
}
ll ans = 0;
for (int i = 0; i <= n; ++i) {
ans = (ans + ((i & 1) ? mod - 1 : 1) * f[i] % mod * fac[n - i] % mod) % mod;
}
printf("%lld\n", ans);
}
int main() {
int T = 1;
// scanf("%d", &T);
while (T--) {
solve();
}
return 0;
}