Solution -「洛谷 P5827」边双连通图计数

$\mathcal{Description}$

  link.
  求包含 $n$ 个点的边双连通图的个数。
  $n\le10^5$。

$\mathcal{Solution}$

  类似于这道题，仍令 $D(x)$ 为有根无向连通图的 $\text{EGF}$，$B(x)$ 为边双连通图的 $\text{EGF}$，考虑用 $B$ 表示 $D$。显然，根仅存在于一个边双连通分量中。我们可以在这个连通分量的任意一个点上挂一个有根连通图，这显然不会影响当前的边双，$\text{EGF}$ 为 $nD(x)$（$n$ 为根所在边双大小。）事实上可以挂许多图，所以处根所在边双以外的 $\text{EGF}$ 为 $\exp\left(nD(x)\right)$。枚举根所在边双大小，就有：

$$D(x)=\sum_{i=1}^{+\infty}\frac{b_ix^i\exp\left(iD(x)\right)}{i!}=B\left(x\exp D(x)\right)$$

  令 $F(x)=x\exp D(x)$，$F^{-1}(x)$ 为其复合逆。将 $F^{-1}(x)$ 代入上式得：

$$B(x)=D\left(F^{-1}(x)\right)$$

  扩展拉格朗日反演：

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

  重新展开 $F$，化简得：

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

  求出 $D$，继而求出 $[x^n]B(x)$。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];
int F[MAXN + 5], G[MAXN + 5], H[MAXN + 5], T[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 polyDer ( 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] {};
	polyDer ( 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 () {
	int len = MAXN >> 1;
	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;
	}
	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 * G[i] * i % MOD;
	Poly::polyDer ( len, G, H ), Poly::NTT ( MAXN, H, 1 );
}

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

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