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CF 804D Expected diameter of a tree

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题目描述

https://www.luogu.com.cn/problem/CF804D

简要题意：给定一个森林和 $q$ 个询问，每次询问给定一个无序二元组 $(x,y)$，问将 $x$ 和 $y$ 所在的树随机连接起来后的直径的期望

$n,q\le 10^5$

Solution

首先我们对于每棵树求出树上从所有点出发的最长路径，这个有经典的 $O(n)$ 做法

那么答案即为 $\frac{1}{nm}\sum_{i=1}^n\sum_{j=1}^m max\lbrace f_{1,i}+f_{2,j},d_1,d_2\rbrace$

注意到询问满足对二元组的自然根号

所以我们枚举较小的树上的点，然后二分即可

时间复杂度 $O(n\sqrt n\log n)$

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#include <iostream>
#include <cstdio>
#include <vector>
#include <algorithm>
#include <map>
#include <iomanip>
#define maxn 100010
#define ll long long
using namespace std;

int n, m, q;

struct Edge {
    int to, next;
} e[maxn * 2]; int c1, head[maxn];
inline void add_edge(int u, int v) {
    e[c1].to = v; e[c1].next = head[u]; head[u] = c1++;
}

int fa[maxn];
void init_fa(int n) { for (int i = 1; i <= n; ++i) fa[i] = i; }

int find(int x) { return fa[x] == x ? x : fa[x] = find(fa[x]); }

inline void merge(int x, int y) {
    int fx = find(x), fy = find(y);
    if (fx == fy) return ;
    fa[fx] = fy;
}

int f[maxn], g[maxn], fr[maxn];
void dfs(int u, int fa) {
    f[u] = g[u] = -1;
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = 1; if (v == fa) continue; 
        dfs(v, u);
        if (f[u] < f[v] + w) g[u] = f[u], f[u] = f[v] + w, fr[u] = v; 
        else if (g[u] < f[v] + w) g[u] = f[v] + w; 
    }
    if (f[u] < 0) f[u] = 0, fr[u] = u;
    else if (g[u] < 0) g[u] = 0;
}

void Dfs(int u, int fa) {
    for (int i = head[u]; ~i; i = e[i].next) {
        int v = e[i].to, w = 1, t; if (v == fa) continue;
        if (fr[u] == v) t = g[u] + w;
        else t = f[u] + w;
        if (f[v] < t) g[v] = f[v], f[v] = t, fr[v] = u;
        else if (g[v] < t) g[v] = t; 
        Dfs(v, u); 
    } 
}

vector<int> A[maxn]; 
vector<int> suf[maxn];
map<pair<int, int>, double> mp; 
double query(int x, int y) {
    if (mp.find({ x, y }) != mp.end()) return mp[{ x, y }];
    double ans = 0; int mx = max(A[x].back(), A[y].back());
    ll s = 0, t, sum = 1ll * A[x].size() * A[y].size(); 
    for (auto u : A[x]) {
        auto it = lower_bound(A[y].begin(), A[y].end(), mx - u - 1);
        if (it == A[y].end()) continue;
        t = A[y].end() - it; 
        s += t; ans += (u + 1) * t + suf[y][it - A[y].begin()]; 
    }
    ans += (sum - s) * mx;
    return mp[{ x, y }] = ans / sum;
} 

int main() { fill(head, head + maxn, -1); 
    ios::sync_with_stdio(false);
    cin.tie(nullptr); cout.tie(nullptr);

    cin >> n >> m >> q; init_fa(n); 
    for (int i = 1; i <= m; ++i) {
        int x, y; cin >> x >> y;
        add_edge(x, y); add_edge(y, x);
        merge(x, y); 
    }
    for (int i = 1; i <= n; ++i) if (find(i) == i) dfs(i, 0);
    for (int i = 1; i <= n; ++i) if (find(i) == i) Dfs(i, 0); 
    for (int i = 1; i <= n; ++i) A[find(i)].push_back(f[i]);
    for (int i = 1; i <= n; ++i)
        if (find(i) == i) {
            sort(A[i].begin(), A[i].end()); 
            int sz = A[i].size();
            suf[i].resize(sz); suf[i][sz - 1] = A[i][sz - 1]; 
            for (int j = sz - 2; ~j; --j) 
                suf[i][j] = suf[i][j + 1] + A[i][j];
        }
    for (int i = 1; i <= q; ++i) {
        int x, y; cin >> x >> y;
        x = find(x); y = find(y);
        if (x == y) { cout << "-1\n"; continue; }
        if (A[x].size() > A[y].size() || A[x].size() == A[y].size() && x > y) swap(x, y);
        cout << fixed << setprecision(7) << query(x, y) << "\n";
    } 
    return 0;
}