城市规划——生成函数

题面

　　Bzoj3456

解析

　　设$g_i$为$i$个点的无向图个数，即$g_i=2^{frac{i*(i-1)}{2}}$，$f_i$为无向连通图个数，枚举点1所在连通块，有下式：$$g_i=sum_{j=1}^i inom{i-1}{j-1}f_i g_{i-j} \ g_i=sum_{j=1}^i frac{(i-1)!}{(j-1)!(i-j)!}f_ig_{i-j}\ frac{g_i}{(i-1)!}=sum_{j=1}^{i}frac{f_j}{(j-1)!}*frac{g_{i-j}}{(i-j)!}$$

　　设$A(x)=sum_{i=1}^{infty}frac{g_i}{(i-1)!}x^i$, $F(x)=sum_{i=1}^{infty}frac{f_i}{(i-1)!}x^i$, $B(x)=sum_{i=0}^{infty}frac{g_i}{i!}x^i$，则：$$A(x)=F(x)B(x)\ F(x)=frac{A(x)}{B(x)}$$

　　多项式求逆即可

　　复杂度$O(N log N)$

　代码：

#include<cstdio>
#include<iostream>
#include<algorithm>
using namespace std;
typedef long long ll;
const int maxn = 260005, mod = 1004535809, g = 3;

int add(int x, int y)
{
    return x + y < mod? x + y: x + y - mod;
}

int rdc(int x, int y)
{
    return x - y < 0? x - y + mod: x - y;
}

ll qpow(ll x, ll y)
{
    y %= (mod - 1);
    ll ret = 1;
    while(y)
    {
        if(y&1)
            ret = ret * x % mod;
        x = x * x % mod;
        y >>= 1;
    }
    return ret;
}

int n, lim, bit, rev[maxn<<1];
ll ginv, fac[maxn], fnv[maxn], c[maxn<<1], G[maxn<<1], gg[maxn<<1], f[maxn<<1];

void init()
{
    ginv = qpow(g, mod - 2);
    fac[0] = 1;
    for(int i = 1; i <= n; ++i)
        fac[i] = i * fac[i-1] % mod;
    fnv[n] = qpow(fac[n], mod - 2);
    G[n] = fnv[n] * qpow(2, 1LL * n * (n - 1) >> 1) % mod;
    for(int i = n - 1; i >= 0; --i)
    {
        fnv[i] = fnv[i+1] * (i + 1) % mod;
        f[i+1] = fnv[i] * qpow(2, 1LL * i * (i + 1) >> 1) % mod;
        G[i] = fnv[i] * qpow(2, 1LL * i * (i - 1) >> 1) % mod;
    }
}

void NTT_init(int x)
{
    lim = 1;
    bit = 0;
    while(lim <= x)
    {
        lim <<= 1;
        ++ bit;
    }
    for(int i = 1; i < lim; ++i)
        rev[i] = (rev[i>>1] >> 1) | ((i & 1) << (bit - 1));
}

void NTT(ll *x, int y)
{
    for(int i = 1; i < lim; ++i)
        if(i < rev[i])
            swap(x[i], x[rev[i]]);
    ll wn, w, u, v;
    for(int i = 1; i < lim; i <<= 1)
    {
        wn = qpow((y == 1)? g: ginv, (mod - 1) / (i << 1));
        for(int j = 0; j < lim; j += (i << 1))
        {
            w = 1;
            for(int k = 0; k < i; ++k)
            {
                u = x[j+k];
                v = x[j+k+i] * w % mod;
                x[j+k] = add(u, v);
                x[j+k+i] = rdc(u, v);
                w = w * wn % mod;
            }
        }
    }
    if(y == -1)
    {
        ll linv = qpow(lim, mod - 2);
        for(int i = 0; i < lim; ++i)
            x[i] = x[i] * linv % mod;
    }
}

void get_inv(ll *x, ll *y, int len)
{
    if(len == 1)
    {
        x[0] = qpow(y[0], mod - 2);
        return ;
    }
    get_inv(x, y, (len + 1) >> 1);
    for(int i = 0; i < len; ++i)
        c[i] = y[i];
    NTT_init(len << 1);
    NTT(x, 1);
    NTT(c, 1);
    for(int i = 0; i < lim; ++i)
    {
        x[i] = rdc(add(x[i], x[i]), (c[i] * x[i] % mod) * x[i] % mod);
        c[i] = 0;
    }
    NTT(x, -1);
    for(int i = len; i < lim; ++i)
        x[i] = 0;
}

int main()
{
    scanf("%d", &n);
    init();
    get_inv(gg, G, n + 1);
    NTT_init((n + 1) << 1);
    NTT(gg, 1);
    NTT(f, 1);
    for(int i = 0; i < lim; ++i)
        f[i] = f[i] * gg[i] % mod;
    NTT(f, -1);
    printf("%lld
", f[n] * fac[n-1] % mod);
    return 0;
}