題目鏈接

BZOJ3451

題解

考慮每個點產生的貢獻，即為該點在點分樹中的深度期望值
由於期望的線性，最後的答案就是每個點貢獻之和

對於點對\((i,j)\)，考慮\(j\)成為\(i\)祖先的概率，記為\(P(i,j)\)
那麽
\[ans = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n} P(i,j)\]
由於是隨機選點，\(i\)到\(j\)路徑上所有點第一個被選中的除非是\(j\)，否則\(j\)就不是\(i\)的祖先
由於是隨機的，所以\(P(i,j) = \frac{1}{dis(i,j)}\)
綜上
\[ans = \sum\limits_{i = 1}^{n}\sum\limits_{j = 1}^{n} \frac{1}{dis(i,j)}\]

為了方便計算，我們可以枚舉\(dis\)，計算有多少個長度為\(dis\)的點對
直接枚舉 + 點分是\(O(n^2logn)\)的，我們考慮能不能一起算
當然可以，兩個子樹之間的貢獻合並實際上就是一個生成函數乘積

我們對於一棵分治樹，先求出整棵樹各個深度數量數列形成的生成函數，平方一次
由於會包含回到同一個子樹的情況，在向子樹求一遍減去即可

這樣就優化成了\(O(nlog^2n)\)

#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;
}

BZOJ3451 Tyvj1953 Normal 【期望 + 點分治 + NTT】