[集训队作业2013]城市规划
有标号简单无向连通图计数。
sol1
设 f(n) 表示 n 个点的简单无向连通图的个数, g(n) 表示 n 个点的简单无向图的个数。
考虑求出 g(n), 枚举 1 所在连通块的大小, 有:
\[\begin{align}
g(n) &= \sum_{i=1}^{n}\binom{n-1}{i-1}f(i)g(n-i)\\
&= \sum_{i=0}^{n-1}\binom{n-1}{i}f(i+1)g(n-1-i)\\
\end{align}
\]
多项式求逆即可。
具体地, \(\hat {\lhd G(z)} = \hat G(z)\cdot\hat{\lhd F(z)}\), \(\hat{\lhd F(z)} = \hat G(z)^{-1}\cdot \hat{\lhd G(z)}\)。
sol2
不一定连通的图就是连通图的无序拼接。
\[\begin{align}
G(z) &= \sum_{n} \frac{F(z)^n}{n!}z^n\\
&= e^{F(z)}
\end{align}
\]
所以 \(F(z) = \ln G(z)\)
#include<bits/stdc++.h>
typedef long long LL;
using namespace std;
const int N = 8e5 + 23, mo = 1004535809;
LL ksm(LL a, LL b) {
LL res = 1ll;
for(; b; b>>=1, a=a*a%mo)
if(b & 1) res = res * a % mo;
return res;
}
const LL g = 3ll, ig = ksm(g, mo - 2);
void Dervt(int n, LL *a, LL *b) {
for(int i = 1; i < n; ++i) b[i - 1] = a[i] * (LL)i % mo;
b[n - 1] = 0ll;
}
void Integ(int n, LL *a, LL *b) {
for(int i = 1; i < n; ++i) b[i] = a[i - 1] * ksm(i, mo - 2) % mo;
b[0] = 0ll;
}
int rv[N];
LL t[N];
void ntt_init(int n) {
for(int i = 1; i < n; ++i) rv[i] = (rv[i>>1]>>1) | (i&1 ? n>>1 : 0);
}
void ntt(LL *a, int n, int type) {
for(int i = 1; i < n; ++i) if(i < rv[i]) swap(a[i], a[rv[i]]);
for(int m = 2; m <= n; m = m<<1) {
LL w = ksm(type == 1 ? g : ig, (mo - 1) / m);
for(int i = 0; i < n; i = i + m) {
LL tmp = 1ll;
for(int j = 0; j < (m>>1); ++j) {
LL p = a[i+j], q = a[i+j+(m>>1)] * tmp % mo;
a[i+j] = (p+q) % mo, a[i+j+(m>>1)] = (p-q+mo) % mo;
tmp = tmp * w % mo;
}
}
}
if(type == -1) {
LL Inv = ksm(n, mo - 2);
for(int i = 0; i < n; ++i) a[i] = a[i] * Inv % mo;
}
}
void poly_inv(int deg, LL *a, LL *b) {
if(deg == 1) { b[0] = ksm(a[0], mo - 2); return; }
poly_inv((deg + 1) >> 1, a, b);
int len = 1; while(len < (deg << 1)) len = len << 1;
ntt_init(len);
for(int i = 0; i < deg; ++i) t[i] = a[i];
for(int i = deg; i < len; ++i) t[i] = b[i] = 0ll;
ntt(b, len, 1), ntt(t, len, 1);
for(int i = 0; i < len; ++i) b[i] = b[i] * (2ll - t[i] * b[i] % mo) % mo;
ntt(b, len, -1);
for(int i = deg; i < len; ++i) b[i] = 0ll;
}
LL b[N], c[N];
void poly_ln(int deg, LL *a) {
poly_inv(deg, a, b), Dervt(deg, a, c);
int len = 1; while(len < (deg << 1)) len = len << 1;
ntt_init(len);
for(int i = deg; i < len; ++i) b[i] = c[i] = 0ll;
ntt(b, len, 1), ntt(c, len, 1);
for(int i = 0; i < len; ++i) b[i] = b[i] * c[i] % mo;
ntt(b, len, -1);
Integ(deg, b, a);
}
LL fac[N], ifac[N];
int n;
LL G[N], F[N];
int main()
{
scanf("%d", &n);
if(n == 0) {
puts("0");
return 0;
}
fac[0] = 1ll;
for(int i = 1; i <= n; ++i) fac[i] = fac[i-1] * (LL)i % mo;
ifac[n] = ksm(fac[n], mo - 2);
for(int i = n; i > 0; --i) ifac[i-1] = ifac[i] * (LL)i % mo;
// for(int i = 0; i <= n; ++i) {
// cout << fac[i] << ' ' << fac[i] * ifac[i] % mo << '\n';
// }
++n;
for(int i = 0; i < n; ++i) G[i] = ksm(2, (LL)i * (LL)(i-1ll) / 2ll) * ifac[i] % mo;
poly_ln(n, G);
--n;
cout << G[n] * fac[n] % mo;
return 0;
}
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