[题解] BZOJ 3456 洛谷 P4841 [集训队作业2013]城市规划 多项式，分治FFT

题目

令\(f_i\)表示n个点的答案。考虑容斥，用所有连边方案减去有多个连通块的方案。枚举1号点所在的连通块大小：

\(f_i=2^{i(i-1)/2}-\sum_{j>0}^{i-1}f_j \binom{i-1}{j-1}2^{(i-j)(i-j-1)/2}\)

\(\binom{i-1}{j-1}\)表示1号点必须在选出的连通块中，剩下的i-1个点中再选出j-1个。\(2^{(i-j)(i-j-1)/2}\)是剩下的点随意连边，但不跟选出的连通块连边的方案数。

\[\begin{align}
f_i&=2^{i(i-1)/2}-\sum_{j>0}^{i-1}f_j \binom{i-1}{j-1}2^{(i-j)(i-j-1)/2}\\
f_i&=2^{i(i-1)/2}-\sum_{j>0}^{i-1}\frac{f_j (i-1)! 2^{(i-j)(i-j-1)/2}}{(j-1)!(i-j)!}\\
\frac{f_i}{(i-1)!}&=\frac{2^{i(i-1)/2}}{(i-1)!}-\sum_{j>0}^{i-1}\frac{f_j}{(j-1)!}\cdot\frac{2^{(i-j)(i-j-1)/2}}{(i-j)!}\\
\end{align}
\]

后半部分是一个卷积的形式，分治FFT就行了。分治的时候可以先忽略前面的\(\frac{2^{i(i-1)/2}}{(i-1)!}-\)，递归到叶子的时候再把\(\frac{f_i}{(i-1)!}\)用这个东西减。最后输出之前把\(f_n\)乘上\((n-1)!\)。时间复杂度\(O(nlog^2n)\)。

点击查看代码

#include <bits/stdc++.h>

#define rep(i,n) for(int i=0;i<n;++i)
#define repn(i,n) for(int i=1;i<=n;++i)
#define LL long long
#define pii pair <LL,LL>
#define fi first
#define se second
#define mpr make_pair
#define pb push_back

using namespace std;

const LL MOD=1004535809;

LL qpow(LL x,LL a)
{
	LL res=x,ret=1;
	while(a>0)
	{
		if((a&1)==1) ret=ret*res%MOD;
		a>>=1;
		res=res*res%MOD;
	}
	return ret;
}

namespace poly
{
  vector <LL> rev;
  void ntt(vector <LL> &a,LL G)
  {
    LL nn=a.size(),gn,g,x,y;vector <LL> tmp=a;
    rep(i,nn) a[i]=tmp[rev[i]];
    for(int len=1;len<nn;len<<=1)
    {
      gn=qpow(G,(MOD-1)/(len<<1));
      for(int i=0;i<nn;i+=(len<<1))
      {
        g=1;
        for(int j=i;j<i+len;++j,(g*=gn)%=MOD)
        {
          x=a[j];y=a[j+len]*g%MOD;
          a[j]=(x+y)%MOD;a[j+len]=(x-y+MOD)%MOD;
        }
      }
    }
  }
  vector <LL> convolution(vector <LL> a,vector <LL> b,LL G)
  {
    LL nn=1,bt=0,sv=a.size()+b.size()-1;while(nn<a.size()+b.size()-1) nn<<=1LL,++bt;
    while(a.size()<nn) a.pb(0);while(b.size()<nn) b.pb(0);
    rev.clear();
    rep(i,nn)
    {
      rev.pb(0);
      rev[i]=(rev[i>>1]>>1)|((i&1)<<(bt-1));
    }
    ntt(a,G);ntt(b,G);
    rep(i,nn) (a[i]*=b[i])%=MOD;
    ntt(a,qpow(G,MOD-2));
    while(a.size()>sv) a.pop_back();
    LL inv=qpow(nn,MOD-2);
    rep(i,a.size()) (a[i]*=inv)%=MOD;
    return a;
  }
  vector <LL> inverse(vector <LL> a,LL G)
  {
    if(a.size()==1) return vector <LL>{qpow(a[0],MOD-2)};
    vector <LL> aa=a;while(aa.size()>(a.size()+1)>>1) aa.pop_back();
    vector <LL> bb=inverse(aa,G);
    LL nn=1,bt=0,sv=a.size();while(nn<a.size()*2) nn<<=1LL,++bt;
    while(a.size()<nn) a.pb(0);while(bb.size()<nn) bb.pb(0);
    rev.clear();
    rep(i,nn)
    {
      rev.pb(0);
      rev[i]=(rev[i>>1]>>1)|((i&1)<<(bt-1));
    }
    ntt(a,G);ntt(bb,G);
    rep(i,nn) a[i]=(2LL-a[i]*bb[i]%MOD+MOD)*bb[i]%MOD;
    ntt(a,qpow(G,MOD-2));
    while(a.size()>sv) a.pop_back();
    LL inv=qpow(nn,MOD-2);
    rep(i,a.size()) (a[i]*=inv)%=MOD;
    return a;
  }
  vector <LL> sqrt1(vector <LL> a,LL G)//常数项为1
  {
    if(a.size()==1) return vector <LL>{1};
    vector <LL> aa=a;while(aa.size()>(a.size()+1)>>1) aa.pop_back();
    vector <LL> bb=sqrt1(aa,G);while(bb.size()<a.size()) bb.pb(0);
    vector <LL> bbb=inverse(bb,G);
    LL nn=1,bt=0,sv=a.size();while(nn<a.size()*2) nn<<=1LL,++bt;
    while(a.size()<nn) a.pb(0);while(bb.size()<nn) bb.pb(0);while(bbb.size()<nn) bbb.pb(0);
    rev.clear();
    rep(i,nn)
    {
      rev.pb(0);
      rev[i]=(rev[i>>1]>>1)|((i&1)<<(bt-1));
    }
    LL mul=qpow(2,MOD-2);
    ntt(a,G);ntt(bb,G);ntt(bbb,G);
    rep(i,nn) a[i]=mul*(bb[i]+bbb[i]*a[i]%MOD)%MOD;
    ntt(a,qpow(G,MOD-2));
    while(a.size()>sv) a.pop_back();
    LL inv=qpow(nn,MOD-2);
    rep(i,a.size()) (a[i]*=inv)%=MOD;
    return a;
  }
}

LL n,f[130010],fac[130010],inv[130010];

void solve(LL lb,LL ub)
{
  if(lb==ub)
  {
    if(lb==1) f[lb]=1;
    else f[lb]=(qpow(2,lb*(lb-1)/2LL)*inv[lb-1]%MOD-f[lb]+MOD)%MOD;
    return;
  }
  LL mid=(lb+ub)/2;
  solve(lb,mid);
  vector <LL> A,B,C;
  for(int i=lb;i<=mid;++i) A.pb(f[i]);
  rep(i,ub-lb+1) B.pb(qpow(2,(LL)i*(LL)(i-1)/2LL)*inv[i]%MOD);
  C=poly::convolution(A,B,3);
  for(int i=mid+1;i<=ub;++i) (f[i]+=C[i-lb])%=MOD;
  solve(mid+1,ub);
}

int main()
{
  fac[0]=1;repn(i,130005) fac[i]=fac[i-1]*(LL)i%MOD;
  rep(i,130004) inv[i]=qpow(fac[i],MOD-2);
  cin>>n;
  solve(1,n);
  cout<<f[n]*fac[n-1]%MOD<<endl;
	return 0;
}