题目链接
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 = sumlimits_{i = 1}^{n}sumlimits_{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 = sumlimits_{i = 1}^{n}sumlimits_{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
",i,ans[i]);
for (int i = 1; i <= n; i++) Ans += 1.0 / i * ans[i];
printf("%.4lf
",Ans);
return 0;
}