Luogu 4841 城市规划

BZOJ 3456 权限题

太菜了推不出式子

我们设$f(n)$表示$n$个点的无向连通图的数量，那么有

$$f(n) = 2^{inom{n}{2}} - sum_{i = 1}^{n - 1}inom{n - 1}{i - 1}f(i)2^{inom{n - i}{2}}$$

思路就是全部减去不合法的，枚举$1$号点所在的联通块的大小，剩下随便生成一张无向图。

拆开组合数，简单变形一下

$$f(n) = 2^{inom{n}{2}} - (n - 1)!sum_{i = 1}^{n - 1}frac{2^{inom{n - i}{2}}}{(n - i)!}frac{f(i)}{(i - 1)!}$$

记$h(i) = frac{f(i)}{(i - 1)!}$，$g(i) = frac{2^{inom{i}{2}}}{i!}$

$$(n - 1)!h(n) = 2^{inom{n}{2}} - (n - 1)!sum_{i = 1}^{n - 1}g(i)h(n - i)$$

发现这个式子可以用分治FFT优化，直接上就做完了。

时间复杂度$O(nlog^2n)$。

Code：

#include <cstdio>
#include <cstring>
using namespace std;
typedef long long ll;

const int N = 3e5 + 5;
const ll P = 1004535809LL;

int n, lim, pos[N];
ll f[N], g[N], fac[N], inv[N], a[N], b[N];

template <typename T>
inline void read(T &X) {
    X = 0; char ch = 0; T op = 1;
    for (; ch > '9'|| ch < '0'; ch = getchar())
        if (ch == '-') op = -1;
    for (; ch >= '0' && ch <= '9'; ch = getchar())
        X = (X << 3) + (X << 1) + ch - 48;
    X *= op;
}
 
inline ll fpow(ll x, ll y) {
    ll res = 1;
    for (; y > 0; y >>= 1) {
        if (y & 1) res = res * x % P;
        x = x * x % P;
    }
    return res;
}

template <typename T>
inline void swap(T &x, T &y) {
    T t = x; x = y; y = t;
}

inline void prework(int len) {
    int l = 0;
    for (lim = 1; lim <= len; lim <<= 1, ++l);
    for (int i = 0; i < lim; i++)
        pos[i] = (pos[i >> 1] >> 1) | ((i & 1) << (l - 1));
} 

inline void ntt(ll *c, int opt) {
    for (int i = 0; i < lim; i++)    
        if (i < pos[i]) swap(c[i], c[pos[i]]);
    for (int i = 1; i < lim; i <<= 1) {
        ll wn = fpow(3, (P - 1) / (i << 1));
        if (opt == -1) wn = fpow(wn, P - 2);
        for (int len = i << 1, j = 0; j < lim; j += len) {
            ll w = 1;
            for (int k = 0; k < i; k++, w = w * wn % P) {
                ll x = c[j + k], y = w * c[j + k + i] % P;
                c[j + k] = (x + y) % P, c[j + k + i] = (x - y + P) % P;
            }
        }
    }
    
    if (opt == -1) {
        ll invP = fpow(lim, P - 2);
        for (int i = 0; i < lim; i++) c[i] = c[i] * invP % P;
    }
}

inline ll getC(int x) {
    if (x < 2) return 0;
    return (1LL * x * (x - 1) / 2);
}

void solve(int l, int r) {
    if (l == r) {
        f[l] = (fpow(2, getC(l) % (P - 1)) - f[l] * fac[l - 1] % P + P) % P;
        f[l] = f[l] * inv[l - 1] % P;
        return;
    }
    
    int mid = ((l + r) >> 1);
    solve(l, mid);
    
    prework(r - l + 1);
    for (int i = 0; i < lim; i++) a[i] = b[i] = 0LL;
    for (int i = l; i <= mid; i++) a[i - l] = f[i];
    for (int i = 1; i <= r - l; i++) b[i - 1] = g[i];
    ntt(a, 1), ntt(b, 1);
    for (int i = 0; i < lim; i++) a[i] = a[i] * b[i] % P;
    ntt(a, -1);
    
    for (int i = mid + 1; i <= r; i++)
        f[i] = (f[i] + a[i - l - 1] % P) % P;
    
    solve(mid + 1, r);
}

int main() {
    read(n);
    
    fac[0] = 1LL;
    for (int i = 1; i <= n; i++) fac[i] = fac[i - 1] * i % P;
    inv[n] = fpow(fac[n], P - 2);
    for (int i = n - 1; i >= 0; i--) inv[i] = inv[i + 1] * (i + 1) % P;
    
    for (int i = 1; i <= n; i++) 
        g[i] = fpow(2LL, getC(i) % (P - 1)) * inv[i] % P;
    
    solve(1, n);
    
    printf("%lld
", f[n] * fac[n - 1] % P);
    return 0;
}