Solution -「洛谷 P5827」点双连通图计数

$\mathcal{Description}$

  link.
  求有 $n$ 个结点的点双连通图的个数，对 $998244353$ 取模。
  $n\le10^5$。

$\mathcal{Solution}$

  奇怪的 GF 增加了 w！
  对于带标号简单无向图，其 $\text{EGF}$ 为 $F(x)=\displaystyle\sum_{i=0}^{+\infty}\frac{2^{i\choose2}x^i}{i!}$（任意两点间有连与不连两种情况。）在此基础上，我们要求图连通（注意这里不是点双连通），即对于带标号简单连通无向图，$\text{EGF}$ 有 $G(x)=\ln F(x)$。
  以下规定所有的图都是带标号的简单无向图。设有根连通图的 $\text{EGF}$ 为 $D(x)$，显然 $D(x)=nG(x)$。并设 $i$ 个结点的点双连通图个数为 $b_i$。考虑任意一个简单无向图的根，它可能被包含在多个点双中。首先特判掉 $n=1$ ——单点的情况。现在对于每一个不是根且在点双连通分量上的点，我们都可以在上面插上一个以其为根的无向连通图，并且不会影响到包含原来的根的任何点双连通分量的大小。所以每一个点双块的 $\text{EGF}$ 是：

$$\sum_{i=1}\frac{b_{i+1}D^i(x)}{i!}$$

  令 $B(x)=\sum_{i=0}^{+\infty}b_{i+1}\frac{x^i}{i!}$，我们反过来表示 $D(x)$，则：

$$D(x)=x\exp B\left(D(x)\right)$$

  开始推式子，先对上式变形：

$$B\left(D(x)\right)=\ln\frac{D(x)}x$$

  令 $D^{-1}$ 是 $D$ 的复合逆，代入得：

$$B(x)=\ln\frac{x}{D^{-1}(x)}$$

  令 $H(x)=\ln\frac{D(x)}x$，那么有 $B(x)=H\left(D^{-1}(x)\right)$。利用扩展拉格朗日反演，有：

$$[x^n]B(x)=\frac{1}n[x^{n-1}]H'(x)\left(\frac{x}{D(x)}\right)^n$$

  把后面的多项式幂变形，并交换分子分母以便运算，得：

$$[x^n]B(x)=\frac{1}n[x^{n-1}]H'(x)\exp\left(-n\ln\frac{D(x)}x\right)$$

  $D$ 易求，那么整个式子都能用亿堆多项式模板算出来。复杂度 $\mathcal O(n\log n)$。

$\mathcal{Code}$

#include <cmath>
#include <cstdio>

const int MAXN = 1 << 18, MOD = 998244353;
int n, fac[MAXN + 5], ifac[MAXN + 5], inv[MAXN + 5], F[MAXN + 5], G[MAXN + 5];
int H[MAXN + 5], lH[MAXN + 5], dH[MAXN + 5];

inline int qkpow ( int a, int b, const int p = MOD ) {
	int ret = 1;
	for ( ; b; a = 1ll * a * a % p, b >>= 1 ) ret = 1ll * ret * ( b & 1 ? a : 1 ) % p;
	return ret;
}

namespace Poly {

const int G = 3;

inline void NTT ( const int n, int* A, const int tp ) {
	static int lstn = -1, rev[MAXN + 5] {};
	if ( lstn ^ n ) {
		int lgn = log ( n ) / log ( 2 ) + 0.5;
		for ( int i = 0; i < n; ++ i ) rev[i] = ( rev[i >> 1] >> 1 ) | ( ( i & 1 ) << lgn >> 1 );
		lstn = n;
	}
	for ( int i = 0; i < n; ++ i ) if ( i < rev[i] ) A[i] ^= A[rev[i]] ^= A[i] ^= A[rev[i]];
	for ( int i = 2, stp = 1; i <= n; i <<= 1, stp <<= 1 ) {
		int w = qkpow ( G, ( MOD - 1 ) / i );
		if ( ! ~ tp ) w = qkpow ( w, MOD - 2 );
		for ( int j = 0; j < n; j += i ) {
			for ( int k = j, r = 1; k < j + stp; ++ k, r = 1ll * r * w % MOD ) {
				int ev = A[k], ov = 1ll * r * A[k + stp] % MOD;
				A[k] = ( ev + ov ) % MOD, A[k + stp] = ( ev - ov + MOD ) % MOD;
			}
		}
	}
	if ( ! ~ tp ) for ( int i = 0; i < n; ++ i ) A[i] = 1ll * A[i] * inv[n] % MOD;
}

inline void polyDir ( const int n, const int* A, int* R ) {
	for ( int i = 1; i < n; ++ i ) R[i - 1] = 1ll * i * A[i] % MOD;
	R[n - 1] = 0;
}

inline void polyInt ( const int n, const int* A, int* R ) {
	for ( int i = n - 1; ~ i; -- i ) R[i + 1] = 1ll * inv[i + 1] * A[i] % MOD;
	R[0] = 0;
}

inline void polyInv ( const int n, const int* A, int* R ) {
	static int tmp[2][MAXN + 5] {};
	if ( n == 1 ) return void ( R[0] = qkpow ( A[0], MOD - 2 ) );
	polyInv ( n >> 1, A, R );
	for ( int i = 0; i < n; ++ i ) tmp[0][i] = A[i], tmp[1][i] = R[i];
	NTT ( n << 1, tmp[0], 1 ), NTT ( n << 1, tmp[1], 1 );
	for ( int i = 0; i < n << 1; ++ i ) tmp[0][i] = 1ll * tmp[0][i] * tmp[1][i] % MOD * tmp[1][i] % MOD;
	NTT ( n << 1, tmp[0], -1 );
	for ( int i = 0; i < n; ++ i ) R[i] = ( 2ll * R[i] % MOD - tmp[0][i] + MOD ) % MOD;
	for ( int i = 0; i < n << 1; ++ i ) tmp[0][i] = tmp[1][i] = 0;
}

inline void polyLn ( const int n, const int* A, int* R ) {
	static int tmp[2][MAXN + 5] {};
	polyDir ( n, A, tmp[0] ), polyInv ( n, A, tmp[1] );
	NTT ( n << 1, tmp[0], 1 ), NTT ( n << 1, tmp[1], 1 );
	for ( int i = 0; i < n << 1; ++ i ) tmp[0][i] = 1ll * tmp[0][i] * tmp[1][i] % MOD;
	NTT ( n << 1, tmp[0], -1 ), polyInt ( n << 1, tmp[0], R );
	for ( int i = 0; i < n << 1; ++ i ) tmp[0][i] = tmp[1][i] = 0;
}

inline void polyExp ( const int n, const int* A, int* R ) {
	static int tmp[MAXN + 5] {};
	if ( n == 1 ) return void ( R[0] = 1 );
	polyExp ( n >> 1, A, R ), polyLn ( n, R, tmp );
	tmp[0] = ( A[0] + 1 - tmp[0] + MOD ) % MOD;
	for ( int i = 1; i < n; ++ i ) tmp[i] = ( A[i] - tmp[i] + MOD ) % MOD;
	NTT ( n << 1, tmp, 1 ), NTT ( n << 1, R, 1 );
	for ( int i = 0; i < n << 1; ++ i ) R[i] = 1ll * R[i] * tmp[i] % MOD;
	NTT ( n << 1, R, -1 );
	for ( int i = n; i < n << 1; ++ i ) R[i] = tmp[i] = 0;
}

} // namespace Poly.

inline void init () {
	inv[1] = fac[0] = ifac[0] = fac[1] = ifac[1] = 1;
	for ( int i = 2; i <= MAXN; ++ i ) {
		fac[i] = 1ll * i * fac[i - 1] % MOD;
		inv[i] = 1ll * ( MOD - MOD / i ) * inv[MOD % i] % MOD;
		ifac[i] = 1ll * inv[i] * ifac[i - 1] % MOD;
	}
	int len = MAXN >> 1;
	for ( int i = 0; i < len; ++ i ) F[i] = 1ll * qkpow ( 2, ( i * ( i - 1ll ) >> 1 ) % ( MOD - 1 ) ) * ifac[i] % MOD;
	Poly::polyLn ( len, F, G );
	for ( int i = 0; i < len; ++ i ) G[i] = 1ll * i * G[i] % MOD;
	for ( int i = 0; i < len - 1; ++ i ) G[i] = G[i + 1];
	G[len - 1] = 0;
	Poly::polyLn ( len, G, H ), Poly::polyDir ( len, H, dH );
	Poly::NTT ( MAXN, dH, 1 );
}

inline void solve () {
	int len = MAXN >> 1;
	if ( ! -- n ) return void ( puts ( "1" ) );
	for ( int i = 0; i < MAXN; ++ i ) F[i] = G[i] = 0;
	for ( int i = 0; i < len; ++ i ) F[i] = 1ll * ( MOD - n ) % MOD * H[i] % MOD;
	Poly::polyExp ( len, F, G ), Poly::NTT ( MAXN, G, 1 );
	for ( int i = 0; i < MAXN; ++ i ) G[i] = 1ll * dH[i] * G[i] % MOD;
	Poly::NTT ( MAXN, G, -1 );
	printf ( "%d\n", int ( 1ll * inv[n] * fac[n] % MOD * G[n - 1] % MOD ) );
}

int main () {
	init ();
	for ( int i = 1; i <= 5; ++ i ) scanf ( "%d", &n ), solve ();
	return 0;
}