【LG4841】城市规划

【LG4841】城市规划

题面

洛谷

题解

记\(t_i\)表示\(i\)个点的无向图个数，显然\(t_i=2^{C_i^2}\)。

设\(f_i\)表示\(i\)个点的无向连通图个数，容斥一下，枚举\(1\)号点所在连通块的大小，再让剩下的点随便构成联通图，

则有：

\[f_i=t_i-\sum_{j=1}^{i-1}f_j*C_{i-1}^{j-1}*t_{i-j}
\]

展开组合数：

\[f_i=t_i-\sum_{j=1}^{i-1}f_j*t_{i-j}*\frac {(i-1)!}{(i-j)!(j-1)!}\\
\Leftrightarrow \frac {f_i}{(i-1)!}=\frac {t_i}{(i-1)!}-\sum_{j=1}^{i-1}\frac {f_j}{(j-1)!}*\frac {t_{i-j}}{(i-j)!}
\]

设\(g_i=\frac {f_i}{(i-1)!},h_i=\frac {t_i}{i!}\)，那么

\[g_i=i*h_i-\sum_{j=1}^{i-1}g_i*h_i
\]

分治+\(NTT\)即可。

代码

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
using namespace std;
const int Mod = 1004535809;
int fpow(int x, int y) {
    int res = 1;
    while (y) {
        if (y & 1) res = 1ll * res * x % Mod;
        x = 1ll * x * x % Mod;
        y >>= 1;
    }
    return res;
}
const int G = 3, iG = fpow(G, Mod - 2);
const int MAX_N = 4e5 + 5;

int Limit, rev[MAX_N];
void NTT(int *p, int op) {
    for (int i = 0; i < Limit; i++) if (i < rev[i]) swap(p[i], p[rev[i]]);
    for (int i = 1; i < Limit; i <<= 1) {
        int rot = fpow(op == 1 ? G : iG, (Mod - 1) / (i << 1));
        for (int j = 0; j < Limit; j += (i << 1)) {
            int w = 1;
            for (int k = 0; k < i; k++, w = 1ll * w * rot % Mod) {
                int x = p[j + k], y = 1ll * w * p[i + j + k] % Mod;
                p[j + k] = (x + y) % Mod, p[i + j + k] = (x - y + Mod) % Mod;
            }
        }
    }
    if (op == -1) {
        int inv = fpow(Limit, Mod - 2);
        for (int i = 0; i < Limit; i++) p[i] = 1ll * p[i] * inv % Mod;
    }
}

int g[MAX_N], h[MAX_N];
void Div(int l, int r) {
    if (l == r) return ;
    static int A[MAX_N], B[MAX_N], C[MAX_N];
    int mid = (l + r) >> 1;
    Div(l, mid);
    int p = 0;
    for (Limit = 1; Limit <= (r - l) * 2; Limit <<= 1, ++p) ;
    for (int i = 0; i < Limit; i++) rev[i] = (rev[i >> 1] >> 1) | ((i & 1) << (p - 1));
    for (int i = 0; i <= mid - l; i++) A[i] = g[i + l];
    for (int i = 0; i < r - l; i++) B[i] = h[i + 1];
    for (int i = mid - l + 1; i < Limit; i++) A[i] = 0;
    for (int i = r - l; i < Limit; i++) B[i] = 0;
    NTT(A, 1); NTT(B, 1);
    for (int i = 0; i < Limit; i++) C[i] = 1ll * A[i] * B[i] % Mod;
    NTT(C, -1);
    for (int i = mid + 1; i <= r; i++) g[i] = (g[i] - C[i - 1 - l] + Mod) % Mod;
    Div(mid + 1, r);
}
int N, fac[MAX_N], ifc[MAX_N];
int C(int n, int m) {
    if (m > n) return 0;
    else return 1ll * fac[n] * ifc[m] % Mod * ifc[n - m] % Mod;
}
int main () {
#ifndef ONLINE_JUDGE
    freopen("cpp.in", "r", stdin);
#endif
    cin >> N;
    fac[0] = 1; for (int i = 1; i <= N; i++) fac[i] = 1ll * fac[i - 1] * i % Mod;
    ifc[N] = fpow(fac[N], Mod - 2); for (int i = N - 1; ~i; i--) ifc[i] = 1ll * ifc[i + 1] * (i + 1) % Mod;
    for (int i = 1; i <= N; i++) h[i] = 1ll * fpow(2, 1ll * i * (i - 1) / 2 % (Mod - 1)) * ifc[i] % Mod;
    for (int i = 1; i <= N; i++) g[i] = 1ll * ifc[i - 1] * fpow(2, 1ll * i * (i - 1) / 2 % (Mod - 1)) % Mod;
    Div(1, N);
    printf("%d\n", (int)(1ll * g[N] * fac[N - 1] % Mod));
    return 0;
}