题解 [集训队作业2013]城市规划

题解

设 (f_n) 表示 (n) 个点的无向联通图个数，(g_n) 表示 (n) 个点的无向图个数，显然 (g_n=2^{ binom{n}{2}})。

那么可得

[g_n=sum_{i=1}^n binom{n-1}{i-1}f_ig_{n-i} ]

意思就是枚举 (1) 所在的最大联通块的大小，统计方案数。

将 (g_n=2^{ binom{n}{2}}) 带入原式，并进行化简

[2^{ binom{n}{2}}=sum_{i=1}^n binom{n-1}{i-1}f_i2^{ binom{n-i}{2}}\ 2^{ binom{n}{2}}=sum_{i=1}^nfrac{(n-1)!}{(i-1)!(n-i)!}f_i2^{ binom{n-i}{2}}\ frac{2^{ binom{n}{2}}}{(n-1)!}=sum_{i=1}^nfrac{f_i2^{ binom{n-i}{2}}}{(i-1)!(n-i)!} ]

发现这是个卷积形式，设

[F(x)=sum_{i=1}^{+infty}frac{f_i}{(i-1)!}x^i\ G(x)=sum_{i=0}^{+infty}frac{2^{ binom{i}{2}}}{i!}x^i\ H(x)=sum_{i=1}^{+infty}frac{2^{ binom{i}{2}}}{(i-1)!}x^i ]

则

[H=F*G\ F=H*G^{-1} ]

先对 (G) 求逆，再与 (H) 卷积即可，复杂度 (mathcal{O m (nlogn)})。

Code

#include<bits/stdc++.h>
#define Re register
#define ri Re signed
#define pd(i) ++i
#define bq(i) --i
namespace IO{
    char buf[1<<21],*p1=buf,*p2=buf;
    #define gc() p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?(-1):*p1++
    struct nanfeng_stream{
        template<typename T>inline nanfeng_stream &operator>>(T &x) {
            Re bool f=false;x=0;Re char ch=gc();
            while(!isdigit(ch)) f|=ch=='-',ch=gc();
            while(isdigit(ch)) x=(x<<1)+(x<<3)+(ch^48),ch=gc();
            return x=f?-x:x,*this;
        }
    }cin;
}
using IO::cin;
namespace nanfeng{
    #define FI FILE *IN
    #define FO FILE *OUT
    template<typename T>inline T cmax(T x,T y) {return x>y?x:y;}
    template<typename T>inline T cmin(T x,T y) {return x>y?y:x;}
    using ll=long long;
    static const int N=1<<19,MOD=1004535809;
    ll h[N],g[N],f[N],gt[N],gc[N],frac[N],inv[N],w1[N],w2[N];
    int R[N],n,len,st;
    auto fpow=[](int x,int y) {
        ll res=1;
        while(y) {
            if (y&1) res=res*x%MOD;
            x=1ll*x*x%MOD;
            y>>=1;
        }
        return res;
    };
    auto MD=[](ll x) {return x>=MOD?x-MOD:x;};
    auto NTT1=[](ll *a) {
        for (ri i(0);i<st;pd(i)) if (R[i]>i) std::swap(a[R[i]],a[i]);
        for (ri t(st>>1),d(1);d<st;t>>=1,d<<=1) 
            for (ri i(0);i<st;i+=d<<1) 
                for (ri j(0);j<d;pd(j)) {
                    const ll tmp=w1[t*j]*a[i+j+d]%MOD;
                    a[i+j+d]=(a[i+j]-tmp+MOD)%MOD;
                    a[i+j]=MD(a[i+j]+tmp);
                }
    };
    auto NTT2=[](ll *a) {
        for (ri i(0);i<st;pd(i)) if (R[i]>i) std::swap(a[R[i]],a[i]);
        for (ri t(st>>1),d(1);d<st;t>>=1,d<<=1) 
            for (ri i(0);i<st;i+=d<<1) 
                for (ri j(0);j<d;pd(j)) {
                    const ll tmp=w2[t*j]*a[i+j+d]%MOD;
                    a[i+j+d]=(a[i+j]-tmp+MOD)%MOD;
                    a[i+j]=MD(a[i+j]+tmp);
                }
        ll Inv=fpow(st,MOD-2);
        for (ri i(0);i<st;pd(i)) a[i]=a[i]*Inv%MOD;
    };
    void calc(int deg) {
        if (deg==1) return (void)(gt[0]=fpow(g[0],MOD-2));
        calc(deg+1>>1);
        len=0,st=1;
        while(st<=deg<<1) st<<=1,++len;
        for (ri i(0);i<st;pd(i)) R[i]=(R[i>>1]>>1)|((i&1)<<(len-1));
        w1[1]=fpow(3,(MOD-1)/st),w2[1]=fpow(w1[1],MOD-2);
        for (ri i(2);i<st;pd(i)) w1[i]=w1[i-1]*w1[1]%MOD,w2[i]=w2[i-1]*w2[1]%MOD;
        memcpy(gc,g,sizeof(ll)*deg);
        memset(gc+deg,0,sizeof(ll)*(st-deg));
        NTT1(gt),NTT1(gc);
        for (ri i(0);i<st;pd(i)) gt[i]=(2ll-gc[i]*gt[i]%MOD+MOD)%MOD*gt[i]%MOD;
        NTT2(gt);
        memset(gt+deg,0,sizeof(ll)*(st-deg));
    }
    inline int main() {
        // FI=freopen("nanfeng.in","r",stdin);
        // FO=freopen("nanfeng.out","w",stdout);
        cin >> n;
        ++n;
        frac[0]=inv[0]=1;
        for (ri i(1);i<=n;pd(i)) frac[i]=frac[i-1]*i%MOD;
        inv[n]=fpow(frac[n],MOD-2);
        for (ri i(n-1);i;bq(i)) inv[i]=inv[i+1]*(i+1)%MOD;
        g[0]=1;
        for (ri i(1);i<=n;pd(i)) {
            ll tmp=fpow(2,(1ll*i*(i-1)>>1)%(MOD-1));
            h[i]=tmp*inv[i-1]%MOD,g[i]=tmp*inv[i]%MOD;
        }
        w1[0]=w2[0]=1;
        calc(n);
        st=1,len=0;
        while(st<=n<<1) st<<=1,++len;
        memset(gt+n,0,sizeof(ll)*(st-n+1));
        for (ri i(0);i<st;pd(i)) R[i]=(R[i>>1]>>1)|((i&1)<<(len-1));
        w1[1]=fpow(3,(MOD-1)/st),w2[1]=fpow(w1[1],MOD-2);
        for (ri i(2);i<st;pd(i)) w1[i]=w1[i-1]*w1[1]%MOD,w2[i]=w2[i-1]*w2[1]%MOD;
        NTT1(gt),NTT1(h);
        for (ri i(0);i<st;pd(i)) h[i]=h[i]*gt[i]%MOD;
        NTT2(h);
        printf("%lld
",h[n-1]*frac[n-2]%MOD);
        return 0;
    }
}
int main() {return nanfeng::main();}