Hdu 5279 YJC plays Minecraft

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多项式 $\exp$ .

规定每一块内部不能有环.连接块与块的边随便选.

这样只会多算一种情况,就是块与块的边都被选,并且每块的第一个点和最后一个点都连通.

把这种情况减掉即可,答案为
$$
ans=2^n\prod_{i=1}^n (g_{a_i})-\prod_{i=1}^n (g_{a_i}-h_{a_i})
$$

其中 $g_n,h_n$ 分别表示 $n$ 个点的完全图生成森林的数目, $n$ 个点的完全图, $n$ 与 $n$ 不连通的生成森林数目.

记 $f_n=n^{n-2}$ 表示 $n$ 个点的完全图生成树的数目, $F,G,H$ 分别为三者的 EGF.

易知 $G=\exp F$ ,只需要考虑如何计算 $H$ .

考虑 $h_n$ 如何计算,枚举 $1​$ 所在的生成树大小即可.
$$
h_n=\sum_{i=1}^{n-1}\binom{n-2}{i-1} f_i\cdot g_{n-i} \\
\frac{h_n}{(n-2)!}=\sum_{i=1}^{n-1} \frac{f_i}{(i-1)!}\cdot \frac{g_{n-i}}{(n-i-1)!}
$$
做一次卷积即可求出所有 $h$ ,时间复杂度 $O(n\log n)$ .

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//%std
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
inline int read()
{
	int out = 0, fh = 1;
	char jp = getchar();
	while ((jp > '9' || jp < '0') && jp != '-')
		jp = getchar();
	if (jp == '-')
		fh = -1, jp = getchar();
	while (jp >= '0' && jp <= '9')
		out = out * 10 + jp - '0', jp = getchar();
	return out * fh;
}
void print(int x)
{
	if (x >= 10)
		print(x / 10);
	putchar('0' + x % 10);
}
void write(int x, char c)
{
	if (x < 0)
		putchar('-'), x = -x;
	print(x);
	putchar(c);
}
const int P = 998244353, G = 3;
int add(int a, int b)
{
	return a + b >= P ? a + b - P : a + b;
}
void inc(int &a, int b)
{
	a = add(a, b);
}
int mul(int a, int b)
{
	return 1LL * a * b % P;
}
int fpow(int a, ll b)
{
	int res = 1;
	while (b)
	{
		if (b & 1LL)
			res = mul(res, a);
		a = mul(a, a);
		b >>= 1;
	}
	return res;
}
const int N = 1 << 18 | 10;
int rev[N], omega[N], inv[N], curn = 0;
void init(int n)
{
	if (curn == n)
		return;
	for (int i = 0; i < n; ++i)
		rev[i] = (rev[i >> 1] >> 1) | ((i & 1) * (n >> 1));
	for (int l = 2; l <= n; l <<= 1)
	{
		omega[l] = fpow(G, (P - 1) / l);
		inv[l] = fpow(omega[l], P - 2);
	}
	curn = n;
}
void DFT(int *a, int n, bool invflag)
{
	init(n);
	for (int i = 0; i < n; ++i)
		if(i < rev[i])
			swap(a[i], a[rev[i]]);
	for (int l = 2; l <= n; l <<= 1)
	{
		int m = l >> 1;
		int gi = omega[l];
		if (invflag)
			gi = inv[l];
		for (int *p = a; p != a + n; p += l)
		{
			int g = 1;
			for (int i = 0; i < m; ++i)
			{
				int t = mul(g, p[i + m]);
				p[i + m] = add(p[i], P - t);
				p[i] = add(p[i], t);
				g = mul(g, gi);
			}
		}
	}
	if(invflag)
	{
		int invn = fpow(n, P - 2);
		for (int i = 0; i < n; ++i)
			a[i] = mul(a[i], invn);
	}
}
void NTT(int *A, int *B, int *C, int lenA, int lenB)
{
	int lenC = lenA + lenB - 1, n = 1;
	while (n < lenC)
		n <<= 1;
	static int a[N], b[N];
	copy(A, A + lenA, a);
	fill(a + lenA, a + n, 0);
	copy(B, B + lenB, b);
	fill(b + lenB, b + n, 0);
	DFT(a, n, false);
	DFT(b, n, false);
	for (int i = 0; i < n; ++i)
		C[i] = mul(a[i], b[i]);
	DFT(C, n, true);
}
void Inverse(int *A, int *B, int n)
{
	int len = 1;
	while (len < n)
		len <<= 1;
	static int res[N], tmp[N];
	res[0] = fpow(A[0], P - 2);
	for (int i = 2; i <= len; i <<= 1)
	{
		NTT(A, res, tmp, i, i);
		NTT(tmp, res, tmp, i, i);
		for (int j = 0; j < i; ++j)
			res[j] = add(mul(2, res[j]), P - tmp[j]);
	}
	copy(res, res + n, B);
}
void Diff(int *A, int n)
{
	for (int i = 0; i < n - 1; ++i)
		A[i] = mul(i + 1, A[i + 1]);
	A[n - 1] = 0;
}
int Inv[N];
void Inte(int *A, int n)
{
	for (int i = n + 1; i >= 1; --i)
		A[i] = mul(Inv[i], A[i - 1]);
	A[0] = 0;
}
void Log(int *A, int *B, int n)
{
	static int invA[N], tmp[N];
	Inverse(A, invA, n);
	copy(A, A + n, tmp);
	Diff(tmp, n);
	NTT(tmp, invA, tmp, n, n);
	Inte(tmp, n);
	copy(tmp, tmp + n, B);
}
void Exp(int *A, int *B, int n)
{
	int len = 1;
	while (len < n)
		len <<= 1;
	static int res[N], tmp[N];
	res[0] = 1;
	for (int i = 2; i <= len; i <<= 1)
	{
		Log(res, tmp, i);
		for (int j = 0; j < i; ++j)
			tmp[j] = add(A[j], P - tmp[j]);
		inc(tmp[0], 1);
		NTT(tmp, res, res, i, i);
	}
	copy(res, res + n, B);
}
int bin[N], fac[N], invfac[N], f[N], g[N], tf[N], tg[N], h[N];
int a[N];
void solve()
{
	int n = read(), ans = 0, prod = bin[n];
	for (int i = 1; i <= n; ++i)
		a[i] = read(), prod = mul(prod, g[a[i]]);
	ans = prod;
	prod = 1;
	for (int i = 1; i <= n; ++i)
		prod = mul(prod, add(g[a[i]], P - h[a[i]]));
	inc(ans, P - prod);
	write(ans, '\n');
}
int main()
{
	int n = 100000 + 1;
	bin[0] = fac[0] = fac[1] = invfac[0] = invfac[1] = Inv[1] = 1;
	bin[1] = 2;
	for (int i = 2; i < n; ++i)
	{
		bin[i] = add(bin[i - 1], bin[i - 1]);
		fac[i] = mul(fac[i - 1], i);
		Inv[i] = mul(P - P / i, Inv[P % i]);
		invfac[i] = mul(invfac[i - 1], Inv[i]);
	}
	f[0] = 0, f[1] = 1;
	for (int i = 2; i < n; ++i)
		f[i] = mul(fpow(i, i - 2), invfac[i]);
	Exp(f, g, n);
	g[0] = 0;
	for (int i = 1; i < n; ++i)
	{
		tf[i] = mul(f[i], i);
		tg[i] = mul(g[i], i);
		f[i] = mul(f[i], fac[i]);
		g[i] = mul(g[i], fac[i]);
	}
	NTT(tf, tg, h, n, n);
	h[0] = h[1] = 0;
	for (int i = 2; i < n; ++i)
		h[i] = mul(h[i], fac[i - 2]);
	int T = read();
	while (T--)
		solve();
	return 0;
}