【HDU5279】YJC plays Minecraft

【题目链接】
点击打开链接
【思路要点】
令 $f_i,g_i,h_i$ 表示 $i$ 个点的有标号树、森林、 $1$ 号与 $i$ 号点不连通的森林的个数。

那么答案应为 $2^N\prod_{i=1}^{N}g_{a_i}-\prod_{i=1}^{N}(g_{a_i}-h_{a_i})$

令 $F(x),G(x),H(x)$ 为 $f_i,g_i,h_i$ 的指数型生成函数。

那么首先有 $F(x)=\sum\frac{i^{i-2}}{i!}x^i,G(x)=e^{F(x)}$

其次，考虑计算 $h_i$ ，有 $h_i=\sum_{j=1}^{i-1}\binom{i-2}{j-1}f_jg_{i-j}$

因此 $H''(x)=F'(x)G'(x)$

进而 $H(x)=\int\int F'(x)G'(x)$

时间复杂度 $O(NLogN)$ 。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int MAXN = 262144;
const int P = 998244353;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); } 
template <typename T> void read(T &x) {
	x = 0; int f = 1;
	char c = getchar();
	for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
	for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
	x *= f;
}
template <typename T> void write(T x) {
	if (x < 0) x = -x, putchar('-');
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
	write(x);
	puts("");
}
namespace Poly {
	const int MAXN = 262144;
	const int P = 998244353;
	const int LOG = 25;
	const int G = 3;
	int power(int x, int y) {
		if (y == 0) return 1;
		int tmp = power(x, y / 2);
		if (y % 2 == 0) return 1ll * tmp * tmp % P;
		else return 1ll * tmp * tmp % P * x % P;
	}
	int invn[MAXN], tmpa[MAXN], tmpb[MAXN];
	int N, Log, home[MAXN]; bool initialized;
	int forward[MAXN], bckward[MAXN], inv[LOG];
	void init() {
		initialized = true;
		forward[0] = bckward[0] = inv[0] = invn[1] = 1;
		for (int len = 2, lg = 1; len <= MAXN; len <<= 1, lg++)
			inv[lg] = power(len, P - 2);
		for (int i = 2; i < MAXN; i++)
			invn[i] = (P - 1ll * (P / i) * invn[P % i] % P) % P;
		int delta = power(G, (P - 1) / MAXN);
		for (int i = 1; i < MAXN; i++)
			forward[i] = bckward[MAXN - i] = 1ll * forward[i - 1] * delta % P;
	}
	void NTTinit() {
		for (int i = 0; i < N; i++) {
			int ans = 0, tmp = i;
			for (int j = 1; j <= Log; j++) {
				ans <<= 1;
				ans += tmp & 1;
				tmp >>= 1;
			}
			home[i] = ans;
		}
	}
	void NTT(int *a, int mode) {
		assert(initialized);
		for (int i = 0; i < N; i++)
			if (home[i] < i) swap(a[i], a[home[i]]);
		int *g;
		if (mode == 1) g = forward;
		else g = bckward;
		for (int len = 2, lg = 1; len <= N; len <<= 1, lg++) {
			for (int i = 0; i < N; i += len) {
				for (int j = i, k = i + len / 2; k < i + len; j++, k++) {
					int tmp = a[j];
					int tnp = 1ll * a[k] * g[MAXN / len * (j - i)] % P;
					a[j] = (tmp + tnp > P) ? (tmp + tnp - P) : (tmp + tnp);
					a[k] = (tmp - tnp < 0) ? (tmp - tnp + P) : (tmp - tnp);
				}
			}
		}
		if (mode == -1) {
			for (int i = 0; i < N; i++)
				a[i] = 1ll * a[i] * inv[Log] % P;
		}
	}
	void times(vector <int> &a, vector <int> &b, vector <int> &c) {
		assert(a.size() >= 1), assert(b.size() >= 1);
		int goal = a.size() + b.size() - 1;
		N = 1, Log = 0;
		while (N < goal) {
			N <<= 1;
			Log++;
		}
		for (unsigned i = 0; i < a.size(); i++)
			tmpa[i] = a[i];
		for (int i = a.size(); i < N; i++)
			tmpa[i] = 0;
		for (unsigned i = 0; i < b.size(); i++)
			tmpb[i] = b[i];
		for (int i = b.size(); i < N; i++)
			tmpb[i] = 0;
		NTTinit();
		NTT(tmpa, 1);
		NTT(tmpb, 1);
		for (int i = 0; i < N; i++)
			tmpa[i] = 1ll * tmpa[i] * tmpb[i] % P;
		NTT(tmpa, -1);
		c.resize(goal);
		for (int i = 0; i < goal; i++)
			c[i] = tmpa[i];
	}
	void timesabb(vector <int> &a, vector <int> &b, vector <int> &c) {
		assert(a.size() >= 1), assert(b.size() >= 1);
		int goal = a.size() + b.size() * 2 - 2;
		N = 1, Log = 0;
		while (N < goal) {
			N <<= 1;
			Log++;
		}
		for (unsigned i = 0; i < a.size(); i++)
			tmpa[i] = a[i];
		for (int i = a.size(); i < N; i++)
			tmpa[i] = 0;
		for (unsigned i = 0; i < b.size(); i++)
			tmpb[i] = b[i];
		for (int i = b.size(); i < N; i++)
			tmpb[i] = 0;
		NTTinit();
		NTT(tmpa, 1);
		NTT(tmpb, 1);
		for (int i = 0; i < N; i++)
			tmpa[i] = 1ll * tmpa[i] * tmpb[i] % P * tmpb[i] % P;
		NTT(tmpa, -1);
		c.resize(goal);
		for (int i = 0; i < goal; i++)
			c[i] = tmpa[i];
	}
	void getinv(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1), assert(a[0] != 0);
		b.clear(), b.push_back(power(a[0], P - 2));
		while (b.size() < a.size()) {
			vector <int> c, ta = a;
			ta.resize(b.size() * 2);
			timesabb(ta, b, c);
			b.resize(b.size() * 2);
			for (unsigned i = 0; i < b.size(); i++)
				b[i] = (2ll * b[i] - c[i] + P) % P;
		}
		b.resize(a.size());
	}
	void getder(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1);
		if (a.size() == 1) {
			b.clear();
			b.resize(1);
		} else {
			b.resize(a.size() - 1);
			for (unsigned i = 0; i < b.size(); i++)
				b[i] = (i + 1ll) * a[i + 1] % P;
		}
	}
	void getint(vector <int> &a, vector <int> &b) {
		b.resize(a.size() + 1), b[0] = 0;
		for (unsigned i = 0; i < a.size(); i++)
			b[i + 1] = 1ll * invn[i + 1] * a[i] % P;
	}
	void getlog(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1), assert(a[0] == 1);
		vector <int> da, inva, db;
		getder(a, da), getinv(a, inva);
		times(da, inva, db), getint(db, b);
		b.resize(a.size());
	}
	void getexp(vector <int> &a, vector <int> &b) {
		assert(a.size() >= 1), assert(a[0] == 0);
		b.clear(), b.push_back(1);
		while (b.size() < a.size()) {
			vector <int> lnb, res;
			b.resize(b.size() * 2), getlog(b, lnb);
			for (unsigned i = 0; i < lnb.size(); i++)
				if (i == 0) lnb[i] = (P + 1 + a[i] - lnb[i]) % P;
				else if (i < a.size()) lnb[i] = (P + a[i] - lnb[i]) % P;
				else lnb[i] = (P - lnb[i]) % P;
			times(lnb, b, res);
			res.resize(b.size());
			swap(res, b);
		}
		b.resize(a.size());
	}
	void getshl(vector <int> &a, vector <int> &b, ull bits) {
		if (a.size() < bits) bits = a.size();
		b.clear(), b.resize(bits);
		for (unsigned i = 0; b.size() < a.size(); i++)
			b.push_back(a[i]);
	}
	void getshr(vector <int> &a, vector <int> &b, ull bits) {
		if (a.size() < bits) bits = a.size(); b.clear();
		for (unsigned i = bits; i < a.size(); i++)
			b.push_back(a[i]);
		b.resize(a.size());
	}
	void getpowk(vector <int> &a, vector <int> &b, int k) {
		assert(k >= 1);
		unsigned pos = a.size();
		for (unsigned i = 0; i < a.size(); i++)
			if (a[i]) {
				pos = i;
				break;
			}
		if (pos == a.size()) {
			b = a;
			return;
		}
		int val = power(a[pos], k), inv = power(a[pos], P - 2);
		vector <int> lntmp, tmp;
		getshr(a, tmp, pos);
		for (unsigned i = 0; i < tmp.size(); i++)
			tmp[i] = 1ll * tmp[i] * inv % P;
		getlog(tmp, lntmp);
		for (unsigned i = 0; i < lntmp.size(); i++)
			lntmp[i] = 1ll * lntmp[i] * k % P;
		getexp(lntmp, tmp);
		for (unsigned i = 0; i < tmp.size(); i++)
			tmp[i] = 1ll * tmp[i] * val % P;
		getshl(tmp, b, 1ull * pos * k);
	}
	int getinvfunc(vector <int> &f, unsigned n) {
		assert(f[0] == 0 && f[1] != 0);
		assert(n >= 1 && n <= f.size());
		int inv = power(n, P - 2);
		vector <int> tmp;
		getshr(f, tmp, 1);
		vector <int> invf;
		getinv(tmp, invf);
		vector <int> res;
		getpowk(invf, res, n);
		return 1ll * inv * res[n - 1] % P;
	}
}
int fac[MAXN], inv[MAXN];
int power(int x, int y) {
	if (y == 0) return 1;
	int tmp = power(x, y / 2);
	if (y % 2 == 0) return 1ll * tmp * tmp % P;
	else return 1ll * tmp * tmp % P * x % P;
}
vector <int> tree, forest, divorce;
void init(int n) {
	fac[0] = 1;
	for (int i = 1; i <= n; i++)
		fac[i] = 1ll * fac[i - 1] * i % P;
	inv[n] = power(fac[n], P - 2);
	for (int i = n - 1; i >= 0; i--)
		inv[i] = inv[i + 1] * (i + 1ll) % P;
	Poly :: init();
	tree.push_back(0), tree.push_back(1);
	for (int i = 2; i <= n; i++)
		tree.push_back(1ll * power(i, i - 2) * inv[i] % P);
	Poly :: getexp(tree, forest);
	vector <int> dtree, dforest, dddivorce, ddivorce;
	Poly :: getder(tree, dtree);
	Poly :: getder(forest, dforest);
	Poly :: times(dtree, dforest, dddivorce);
	Poly :: getint(dddivorce, ddivorce);
	Poly :: getint(ddivorce, divorce);
	for (int i = 0; i <= n; i++) {
		tree[i] = 1ll * tree[i] * fac[i] % P;
		forest[i] = 1ll * forest[i] * fac[i] % P;
		divorce[i] = 1ll * divorce[i] * fac[i] % P;
	}
}
void update(int &x, int y) {
	x += y;
	if (x >= P) x -= P;
}
int n, a[MAXN];
int main() {
	init(1e5);
	int T; read(T);
	while (T--) {
		read(n);
		for (int i = 1; i <= n; i++)
			read(a[i]);
		a[0] = a[n];
		int ans = power(2, n), now = 1;
		for (int i = 1; i <= n; i++) {
			ans = 1ll * ans * forest[a[i]] % P;
			now = 1ll * now * (forest[a[i]] - divorce[a[i]] + P) % P;
		}
		printf("%d\n", (ans - now + P) % P);
	}
	return 0;
}
【HDU5279】YJC plays Minecraft

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