BZOJ3456 城市规划【分治NTT】

题目链接

题解

据说这题是多项式求逆

我太弱不会QAQ，只能\(O(nlog^2n)\)分治\(NTT\)

设\(f[i]\)表示\(i\)个节点的简单无向连通图的数量

考虑转移，直接求不好求，我们知道\(n\)个点无向图的数量是\(2^{{n \choose 2}}\)的，考虑用总数减去不连通的

既然图不连通，那么和\(1\)号点联通的点数一定小于\(n\)，我们枚举和\(1\)号点所在联通块大小，就可以得到式子：

\[f[n] = 2^{{n \choose 2}} - \sum\limits_{i = 1}^{n - 1}{n - 1 \choose i - 1}2^{{n - i \choose 2}}f[i]
\]

展开组合数变形得：

\[f[n] = 2^{{n \choose 2}} - (n - 1)!\sum\limits_{i = 1}^{n - 1}\frac{f[i] * i}{i!} * \frac{2^{n - i \choose 2}}{(n - i)!}
\]

分治NTT即可

#include<algorithm>

#include<iostream>

#include<cstring>

#include<cstdio>

#include<cmath>

#define REP(i,n) for (int i = 1; i <= (n); i++)

#define cls(s) memset(s,0,sizeof(s))

#define LL long long int

#define res register

using namespace std;

const int maxn = 500000,maxm = 100005,INF = 1000000000,P = 1004535809;

const int G = 3;

int N,f[maxn],fac[maxn],fv[maxn],inv[maxn],C2[maxn];

int A[maxn],B[maxn],R[maxn],w[2][maxn],L,n,m;

inline int qpow(int a,LL b){

	int ans = 1;

	for (; b; b >>= 1,a = 1ll * a * a % P)

		if (b & 1) ans = 1ll * ans * a % P;

	return ans;

}

inline void NTT(int* a,int f){

	for (res int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);

	for (res int i = 1; i < n; i <<= 1){

		int gn = w[f][i];

		for (res int j = 0; j < n; j += (i << 1)){

			int g = 1,x,y;

			for (res int k = 0; k < i; k++,g = 1ll * g * gn % P){

				x = a[j + k]; y = 1ll * g * a[j + k + i] % P;

				a[j + k] = (x + y) % P; a[j + k + i] = (x - y + P) % P;

			}

		}

	}

	if (f == 1) return;

	int nv = qpow(n,P - 2);

	for (res int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;

}

void solve(int l,int r){

	if (l == r){

		f[l] = ((C2[l] - 1ll * fac[l - 1] * f[l] % P) % P + P) % P;

		return;

	}

	int mid = l + r >> 1,t;

	solve(l,mid);

	m = (mid - l) + (r - l); L = 0;

	for (n = 1; n <= m; n <<= 1) L++;

	for (int i = 0; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));

	for (int i = 0; i < n; i++) A[i] = B[i] = 0;

	t = mid - l + 1;

	for (int i = 0; i < t; i++)

		A[i] = 1ll * f[l + i] * fv[l + i - 1] % P;

	t = r - l;

	B[0] = 0; for (int i = 1; i <= t; i++)

		B[i] = 1ll * C2[i] * fv[i] % P;

	NTT(A,1); NTT(B,1);

	for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * B[i] % P;

	NTT(A,0);

	for (int i = mid + 1; i <= r; i++){

		f[i] = (f[i] + A[i - l]) % P;

	}

	solve(mid + 1,r);

}

int main(){

	scanf("%d",&N);

	fac[0] = fac[1] = inv[0] = inv[1] = fv[0] = fv[1] = 1;

	for (res int i = 2; i <= N; i++){

		fac[i] = 1ll * fac[i - 1] * i % P;

		inv[i] = 1ll * (P - P / i) * inv[P % i] % P;

		fv[i] = 1ll * fv[i - 1] * inv[i] % P;

	}

	for (res int i = 1; i <= N; i++){

		C2[i] = qpow(2,1ll * i * (i - 1) / 2);

	}

	for (res int i = 1; i < maxn; i <<= 1){

		w[1][i] = qpow(G,(P - 1) / (i << 1));

		w[0][i] = qpow(G,(-(P - 1) / (i << 1) % (P - 1) + (P - 1)) % (P - 1));

	}

	solve(1,N);

	printf("%d\n",f[N]);

	return 0;

}

巴特西

BZOJ3456 城市规划【分治NTT】

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巴特西

BZOJ3456 城市规划 【分治NTT】

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BZOJ3456 城市规划【分治NTT】