最远点对(Farthest Pair) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

const int NEG = -100000000;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    vector<vector<int>> adj(n + 1);
    for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        adj[u].push_back(v);
        adj[v].push_back(u);
    }
    vector<int> par(n + 1, 0), order;
    vector<bool> vis(n + 1, false);
    vis[1] = true;
    order.push_back(1);
    for (int idx = 0; idx < (int)order.size(); ++idx) {
        int u = order[idx];
        for (int v : adj[u]) {
            if (!vis[v]) {
                vis[v] = true;
                par[v] = u;
                order.push_back(v);
            }
        }
    }
    vector<int> d0(n + 1, NEG), d1(n + 1, NEG);
    for (int i = 1; i <= n; ++i) {
        if (a[i] == 0) d0[i] = 0;
        else d1[i] = 0;
    }
    int ans = 0;
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        int p = par[u];
        if (p) {
            ans = max(ans, d0[p] + d1[u] + 1);
            ans = max(ans, d1[p] + d0[u] + 1);
            d0[p] = max(d0[p], d0[u] + 1);
            d1[p] = max(d1[p], d1[u] + 1);
        }
    }
    cout << ans << "\n";
    return 0;
}