BZOJ3451 Tyvj1953 Normal 【期望 + 点分治 + NTT】

题目链接

BZOJ3451

题解

考虑每个点产生的贡献，即为该点在点分树中的深度期望值

由于期望的线性，最后的答案就是每个点贡献之和

对于点对\((i,j)\)，考虑\(j\)成为\(i\)祖先的概率，记为\(P(i,j)\)

那么

\[ans = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n} P(i,j)
\]

由于是随机选点，\(i\)到\(j\)路径上所有点第一个被选中的除非是\(j\)，否则\(j\)就不是\(i\)的祖先

由于是随机的，所以\(P(i,j) = \frac{1}{dis(i,j)}\)

综上

\[ans = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n} \frac{1}{dis(i,j)}
\]

为了方便计算，我们可以枚举\(dis\)，计算有多少个长度为\(dis\)的点对

直接枚举 + 点分是\(O(n^2logn)\)的，我们考虑能不能一起算

当然可以，两个子树之间的贡献合并实际上就是一个生成函数乘积

我们对于一棵分治树，先求出整棵树各个深度数量数列形成的生成函数，平方一次

由于会包含回到同一个子树的情况，在向子树求一遍减去即可

这样就优化成了\(O(nlog^2n)\)

#include<algorithm>
#include<iostream>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<map>
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define mp(a,b) make_pair<int,int>(a,b)
#define cls(s) memset(s,0,sizeof(s))
#define cp pair<int,int>
#define LL long long int
using namespace std;
const int maxn = 150005,maxm = 100005,INF = 1000000000;
inline int read(){
	int out = 0,flag = 1; char c = getchar();
	while (c < 48 || c > 57){if (c == '-') flag = -1; c = getchar();}
	while (c >= 48 && c <= 57){out = (out << 3) + (out << 1) + c - 48; c = getchar();}
	return out * flag;
}
int n,h[maxn],ne = 1;
struct EDGE{int to,nxt;}ed[maxn];
inline void build(int u,int v){
	ed[++ne] = (EDGE){v,h[u]}; h[u] = ne;
	ed[++ne] = (EDGE){u,h[v]}; h[v] = ne;
}
const int G = 3,P = 998244353;
int R[maxn];
inline int qpow(int a,int b){
	int re = 1;
	for (; b; b >>= 1,a = 1ll * a * a % P)
		if (b & 1) re = 1ll * re * a % P;
	return re;
}
void NTT(int* a,int n,int f){
	for (int i = 0; i < n; i++) if (i < R[i]) swap(a[i],a[R[i]]);
	for (int i = 1; i < n; i <<= 1){
		int gn = qpow(G,(P - 1) / (i << 1));
		for (int j = 0; j < n; j += (i << 1)){
			int g = 1,x,y;
			for (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) % P;
			}
		}
	}
	if (f == 1) return;
	int nv = qpow(n,P - 2); reverse(a + 1,a + n);
	for (int i = 0; i < n; i++) a[i] = 1ll * a[i] * nv % P;
}
LL ans[maxn];
int F[maxn],fa[maxn],siz[maxn],vis[maxn],N,rt;
void getrt(int u){
	siz[u] = 1; F[u] = 0;
	Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
		fa[to] = u; getrt(to);
		siz[u] += siz[to];
		F[u] = max(F[u],siz[to]);
	}
	F[u] = max(F[u],N - siz[u]);
	if (F[u] < F[rt]) rt = u;
}
int dep[maxn],md;
int A[maxn],B[maxn];
void dfs(int u){
	A[dep[u]]++; siz[u] = 1; md = max(md,dep[u]);
	Redge(u) if (!vis[to = ed[k].to] && to != fa[u]){
		fa[to] = u; dep[to] = dep[u] + 1; dfs(to);
		siz[u] += siz[to];
	}
}
void dfs1(int u){
	B[dep[u]]++; md = max(md,dep[u]);
	Redge(u) if (!vis[to = ed[k].to] && to != fa[u])
		dfs1(to);
}
void solve(int u){
	vis[u] = true; siz[u] = N; fa[u] = 0;
	for (int i = 0; i <= N; i++) A[i] = B[i] = 0;
	dep[u] = 0; A[0] = 1; md = 0;
	Redge(u) if (!vis[to = ed[k].to]){
		fa[to] = u; dep[to] = 1; dfs(to);
	}
	int m = (md << 1),L = 0,n = 1;
	while (n <= m) n <<= 1,L++;
	for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
	for (int i = md + 1; i < n; i++) A[i] = 0;
	NTT(A,n,1);
	for (int i = 0; i < n; i++) A[i] = 1ll * A[i] * A[i] % P;
	NTT(A,n,-1);
	for (int i = 0; i < n; i++) ans[i + 1] += 1ll * A[i];
	Redge(u) if (!vis[to = ed[k].to]){
		md = 1; dfs1(to);
		m = (md << 1),L = 0,n = 1;
		while (n <= m) n <<= 1,L++;
		for (int i = 1; i < n; i++) R[i] = (R[i >> 1] >> 1) | ((i & 1) << (L - 1));
		NTT(B,n,1);
		for (int i = 0; i < n; i++) B[i] = 1ll * B[i] * B[i] % P;
		NTT(B,n,-1);
		for (int i = 0; i < n; i++) ans[i + 1] -= 1ll * B[i];
		for (int i = 0; i < n; i++) B[i] = 0;
	}
	Redge(u) if (!vis[to = ed[k].to]){
		N = siz[to]; F[rt = 0] = INF; getrt(to);
		solve(rt);
	}
}
int main(){
	n = read();
	for (int i = 1; i < n; i++) build(read() + 1,read() + 1);
	F[rt = 0] = INF; N = n; getrt(1);
	solve(rt);
	double Ans = 0;
	//REP(i,n) printf("dis %d  cnt %lld\n",i,ans[i]);
	for (int i = 1; i <= n; i++) Ans += 1.0 / i * ans[i];
	printf("%.4lf\n",Ans);
	return 0;
}