【BZOJ3456】—城市规划（生成函数+多项式求逆）

传送门

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考虑$f_i$表示$i$个点连通图个数，$g_i$表示$i$个点的图的个数
则$g_i=2^{i\choose 2}$

那$g_n$可以通过枚举1号点所在连通块大小表示

$g_n=\sum_{i=1}^{n}f_i*g_{n-i}*{n-1\choose i-1}$
$\frac{g_n}{(n-1)!}=\sum_{i=1}^{n}f_i*\frac{1}{(i-1)!}*g_{n-i}*\frac{1}{(n-i)!}$

设$A(x)=\sum_{i=1}^{n}\frac{g_i}{(i-1)!}x^{i}$

$B(x)=\sum_{i=1}^{n}\frac{f_i}{(i-1)!}x^i$

$C(x)=\sum_{i=0}^n\frac{g_i}{i!}x^i$

则$A(x)=B(x)C(x)$
$B(x)=\frac{A(x)}{C(x)}$

$\frac{f_n}{(n-1)!}=B(x)[x^i]$

多项式求逆就完了

#include<bits/stdc++.h>
using namespace std;
const int RLEN=1<<20|1;
inline char gc(){
	static char ibuf[RLEN],*ib,*ob;
	(ob==ib)&&(ob=(ib=ibuf)+fread(ibuf,1,RLEN,stdin));
	return (ob==ib)?EOF:*ib++;
}
#define gc getchar
inline int read(){
	char ch=gc();
	int res=0,f=1;
	while(!isdigit(ch))f^=ch=='-',ch=gc();
	while(isdigit(ch))res=(res+(res<<2)<<1)+(ch^48),ch=gc();
	return f?res:-res;
}
#define ll long long
#define re register
#define pii pair<int,int>
#define fi first
#define se second
#define pb push_back
const int mod=1004535809,g=3;
inline int add(int a,int b){return a+b>=mod?a+b-mod:a+b;}
inline void Add(int &a,int b){a=add(a,b);}
inline int dec(int a,int b){return a>=b?a-b:a-b+mod;}
inline void Dec(int &a,int b){a=dec(a,b);}
inline int mul(int a,int b){return 1ll*a*b>=mod?1ll*a*b%mod:a*b;}
inline void Mul(int &a,int b){a=mul(a,b);}
inline int ksm(int a,int b,int res=1){for(;b;b>>=1,a=mul(a,a))(b&1)?(res=mul(res,a)):0;return res;}
const int N=130005;
int rev[N<<2];
#define poly vector<int> 
inline void ntt(poly &f,int lim,int kd){
	for(int i=0;i<lim;i++)if(i>rev[i])swap(f[i],f[rev[i]]);
	for(int mid=1;mid<lim;mid<<=1){
		int now=ksm(g,(mod-1)/(mid<<1));
		for(int i=0;i<lim;i+=mid*2){
			int w=1;
			for(int j=0;j<mid;j++,Mul(w,now)){
				int a0=f[i+j],a1=mul(f[i+j+mid],w);
				f[i+j]=add(a0,a1),f[i+j+mid]=dec(a0,a1);
			}
		}
	}
	if(kd==-1&&(reverse(f.begin()+1,f.begin()+lim),1))
		for(int i=0,inv=ksm(lim,mod-2);i<lim;i++)Mul(f[i],inv);
}
inline void init(int lim){
	for(int i=0;i<lim;i++)rev[i]=(rev[i>>1]>>1)|((i&1)*(lim>>1));
}
inline poly mul(poly a,poly b){
	int deg=a.size()+b.size()-1,lim=1;
	if(deg<=128){
		poly c(deg,0);
		for(int i=0;i<a.size();i++)
		for(int j=0;j<b.size();j++)
			Add(c[i+j],mul(a[i],b[j]));
		return c;
	}
	while(lim<deg)lim<<=1;init(lim);
	a.resize(lim),b.resize(lim);
	ntt(a,lim,1),ntt(b,lim,1);
	for(int i=0;i<lim;i++)Mul(a[i],b[i]);
	ntt(a,lim,-1),a.resize(deg);
	return a;
}
inline poly Inv(poly b,int deg){
	poly a(1,ksm(b[0],mod-2)),c;
	for(int lim=4;lim<(deg<<2);lim<<=1){
		init(lim),c=b,a.resize(lim);
		c.resize(lim>>1),c.resize(lim);
		ntt(a,lim,1),ntt(c,lim,1);
		for(int i=0;i<lim;i++)a[i]=mul(a[i],dec(2,mul(a[i],c[i])));
		ntt(a,lim,-1),a.resize(lim>>1);
	}a.resize(deg);return a;
}
poly f,G,p;
int fac[N],ifac[N],n;
inline int C(int x){
	return (1ll*x*(x-1)/2)%(mod-1);
}
int main(){
	n=read(),ifac[0]=fac[0]=1;
	for(int i=1;i<=n;i++)fac[i]=mul(fac[i-1],i);
	ifac[n]=ksm(fac[n],mod-2);
	for(int i=n-1;i;i--)ifac[i]=mul(ifac[i+1],i+1);
	G.resize(n+1),p.resize(n+1);
	for(int i=0;i<=n;i++)G[i]=mul(ifac[i],ksm(2,C(i)));
	for(int i=1;i<=n;i++)p[i]=mul(ifac[i-1],ksm(2,C(i)));
	G=Inv(G,n+1);
	f=mul(p,G);
	cout<<mul(fac[n-1],f[n]);
}