BZOJ3456 城市规划 【多项式求逆】

题目链接

BZOJ3456

题解

之前我们用分治\(ntt\)在\(O(nlog^2n)\)的复杂度下做了这题，今天我们使用多项式求逆

设\(f_n\)表示\(n\)个点带标号无向连通图数

设\(g_n\)表示\(n\)个点图的数量，显然\(g_n = 2^{{n \choose 2}}\)

枚举\(1\)号点所在联通块大小，我们有

\[g_n = \sum\limits_{i = 1}^{n} {n - 1 \choose i - 1}f_{i}g_{n - i}
\]

代入\(g_n\)

\[2^{{n \choose 2}} = \sum\limits_{i = 1}^{n} \frac{(n - 1)!}{(n - i)!(i - 1)!}f_{i}2^{{n - i \choose 2}}
\]

整理一下：

\[\frac{2^{{n \choose 2}}}{(n - 1)!} = \sum\limits_{i = 1}^{n} \frac{f_i}{(i - 1)!} * \frac{2^{{n - i \choose 2}}}{(n - i)!}
\]

发现是一个卷积的形式

令

\[F(x) = \sum\limits_{n = 1}^{+\infty} \frac{f_n}{(n - 1)!} x^n
\]

\[G(x) = \sum\limits_{n = 0}^{+\infty} \frac{2^{{n \choose 2}}}{n!} x^n
\]

\[H(x) = \sum\limits_{n = 1}^{+\infty} \frac{2^{{n \choose 2}}}{(n - 1)!} x^n
\]

则有

\[H(x) = F(x)G(x)
\]

那么

\[F(x) = H(x)G^{-1}(x)
\]

多项式求逆后再求一次卷积就可求得答案

复杂度\(O(nlogn)\)，快了不少

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 300005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
const int G = 3,P = 1004535809;
int f[maxn],g[maxn],h[maxn],gv[maxn],N;
int fac[maxn],inv[maxn],fv[maxn];
int c[maxn],R[maxn];
inline int qpow(int a,LL b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
void NTT(int* a,int n,int f){
	for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
	for (int i = 1; i < n; i <<= 1){
		int gn = qpow(G,(P - 1) / (i << 1));
		for (int j = 0; j < n; j += (i << 1)){
			int g = 1,x,y;
			for (int k = 0; k < i; k++,g = 1ll * g * gn % P){
				x = a[j + k]; y = 1ll * g * a[j + k + i] % P;
				a[j + k] = (x + y) % P; a[j + k + i] = ((x - y) % P + P) % P;
			}
		}
	}
	if (f == 1) return;
	int nv = qpow(n,P - 2); reverse(a + 1,a + n);
	for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
void work(int deg,int* a,int* b){
	if (deg == 1){b[0] = qpow(a[0],P - 2); return;}
	work((deg + 1) >> 1,a,b);
	int L = 0,n = 1;
	while (n < (deg << 1)) n <<= 1,L++;
	for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	for (int i = 0; i < deg; i++) c[i] = a[i];
	for (int i = deg; i < n; i++) c[i] = 0;
	NTT(c,n,1); NTT(b,n,1);
	for (int i = 0; i < n; i++)
		b[i] = 1ll * ((2ll - 1ll * c[i] * b[i] % P) % P + P) % P * b[i] % P;
	NTT(b,n,-1);
	for (int i = deg; i < n; i++) b[i] = 0;
}
void init(){
	fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;
	for (int i = 2; i <= N; i++){
		fac[i] = 1ll * fac[i - 1] * i % P;
		inv[i] = 1ll * (P - P / i) * inv[P % i] % P;
		fv[i] = 1ll * fv[i - 1] * inv[i] % P;
	}
	g[0] = 1;
	for (int i = 1; i <= N; i++){
		h[i] = 1ll * qpow(2,1ll * i * (i - 1) / 2) * fv[i - 1] % P;
		g[i] = 1ll * qpow(2,1ll * i * (i - 1) / 2) * fv[i] % P;
	}
}
int main(){
	N = read();
	init();
	work(N + 1,g,gv);
	int L = 0,n = 1;
	while (n <= (N << 1)) n <<= 1,L++;
	for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	NTT(h,n,1); NTT(gv,n,1);
	for (int i = 0; i < n; i++)
		f[i] = 1ll * h[i] * gv[i] % P;
	NTT(f,n,-1);
	int ans = 1ll * f[N] * fac[N - 1] % P;
	printf("%d\n",ans);
	return 0;
}