最大深度和选点(STA-Station) 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;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> head(n + 1, -1), to(2 * n), nxt(2 * n);
    int ec = 0;
    auto addEdge = [&](int u, int v) {
        to[ec] = v;
        nxt[ec] = head[u];
        head[u] = ec++;
    };
    for (int i = 1; i < n; ++i) {
        int u, v;
        cin >> u >> v;
        addEdge(u, v);
        addEdge(v, u);
    }
    vector<int> par(n + 1, 0), sz(n + 1, 1), dep(n + 1, 0), order;
    order.reserve(n);
    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 e = head[u]; e != -1; e = nxt[e]) {
            int v = to[e];
            if (!vis[v]) {
                vis[v] = true;
                par[v] = u;
                dep[v] = dep[u] + 1;
                order.push_back(v);
            }
        }
    }
    vector<long long> f(n + 1, 0);
    for (int i = 1; i <= n; ++i) f[1] += dep[i];
    for (int i = n - 1; i >= 0; --i) {
        int u = order[i];
        if (par[u]) sz[par[u]] += sz[u];
    }
    for (int i = 1; i < n; ++i) {
        int u = order[i];
        int p = par[u];
        f[u] = f[p] + n - 2LL * sz[u];
    }
    int bestNode = 1;
    for (int i = 2; i <= n; ++i) {
        if (f[i] > f[bestNode]) bestNode = i;
    }
    cout << bestNode << "\n";
    return 0;
}