CodeForces 814E An unavoidable detour for home

洛谷传送门

CF 传送门

考虑给图分层，一层的点一一对应上一层的一些点。设 \(f_{i, j}\) 为考虑了前 \(i\) 个点，最后一层有 \(j\) 个点，除了最后一层点的其他点度数限制已经满足的方案数。

转移系数是 \(g_{i, j, k}\) 表示这一层有 \(i\) 个点，上一层有 \(j\) 个 \(2\) 度点，\(k\) 个 \(3\) 度点（实际上因为上一层已经和上上层连边所以是 \(1, 2\) 度点）。这个的递推是容易的。

时间复杂度就是 \(O(n^3)\)。

code

// Problem: E. An unavoidable detour for home
// Contest: Codeforces - Codeforces Round 418 (Div. 2)
// URL: https://codeforces.com/problemset/problem/814/E
// Memory Limit: 256 MB
// Time Limit: 3000 ms
// 
// Powered by CP Editor (https://cpeditor.org)

#include <bits/stdc++.h>
#define pb emplace_back
#define fst first
#define scd second
#define mkp make_pair
#define mems(a, x) memset((a), (x), sizeof(a))

using namespace std;
typedef long long ll;
typedef double db;
typedef unsigned long long ull;
typedef long double ldb;
typedef pair<ll, ll> pii;

const int maxn = 55;
const ll mod = 1000000007;
const ll inv2 = (mod + 1) / 2;

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], fac[maxn], ifac[maxn], f[maxn][maxn], g[maxn][maxn][maxn];

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;
	}
}

void solve() {
	scanf("%lld", &n);
	for (int i = 1; i <= n; ++i) {
		scanf("%lld", &a[i]);
	}
	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;
	}
	g[0][0][0] = 1;
	for (int i = 3; i <= n; ++i) {
		for (int j = 3; j <= i; ++j) {
			g[0][0][i] = (g[0][0][i] + g[0][0][i - j] * C(i - 1, j - 1) % mod * fac[j - 1] % mod * inv2) % mod;
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int j = 0; j <= n; ++j) {
			if (i >= 2) {
				g[0][i][j] = g[0][i - 2][j] * (i - 1) % mod;
			}
			if (j) {
				g[0][i][j] = (g[0][i][j] + g[0][i][j - 1] * j) % mod;
			}
		}
	}
	for (int i = 1; i <= n; ++i) {
		for (int k = 0; k <= n; ++k) {
			for (int j = 0; j <= n; ++j) {
				if (j) {
					g[i][j][k] = g[i - 1][j - 1][k] * j % mod;
				}
				if (k) {
					g[i][j][k] = (g[i][j][k] + g[i - 1][j + 1][k - 1] * k) % mod;
				}
			}
		}
	}
	f[a[1] + 1][a[1]] = 1;
	for (int i = a[1] + 2; i <= n; ++i) {
		for (int j = 1; j < i; ++j) {
			int c2 = 0, c3 = 0;
			for (int k = 1; j + k < i; ++k) {
				c2 += (a[i - j - k + 1] == 2);
				c3 += (a[i - j - k + 1] == 3);
				f[i][j] = (f[i][j] + f[i - j][k] * g[j][c2][c3]) % mod;
			}
		}
	}
	int c2 = 0, c3 = 0;
	ll ans = 0;
	for (int i = 1; i < n; ++i) {
		c2 += (a[n - i + 1] == 2);
		c3 += (a[n - i + 1] == 3);
		ans = (ans + f[n][i] * g[0][c2][c3]) % mod;
	}
	printf("%lld\n", ans);
}

int main() {
	int T = 1;
	// scanf("%d", &T);
	while (T--) {
		solve();
	}
	return 0;
}