大量的工作沟通(Work Communication) 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 <queue>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    const int LOG = 18;
    vector<vector<int>> up(LOG, vector<int>(n, 0));
    vector<vector<int>> adj(n);
    for (int i = 1; i < n; ++i) {
        int f;
        cin >> f;
        up[0][i] = f;
        adj[f].push_back(i);
    }
    vector<int> dep(n, 0), maxAnc(n, 0);
    queue<int> q;
    dep[0] = 1;
    maxAnc[0] = 0;
    q.push(0);
    while (!q.empty()) {
        int u = q.front();
        q.pop();
        for (int v : adj[u]) {
            dep[v] = dep[u] + 1;
            maxAnc[v] = max(v, maxAnc[u]);
            q.push(v);
        }
    }
    for (int k = 1; k < LOG; ++k) {
        for (int i = 0; i < n; ++i) {
            up[k][i] = up[k - 1][up[k - 1][i]];
        }
    }
    auto lca = [&](int u, int v) {
        if (dep[u] < dep[v]) swap(u, v);
        int d = dep[u] - dep[v];
        for (int k = 0; k < LOG; ++k) {
            if ((d >> k) & 1) u = up[k][u];
        }
        if (u == v) return u;
        for (int k = LOG - 1; k >= 0; --k) {
            if (up[k][u] != up[k][v]) {
                u = up[k][u];
                v = up[k][v];
            }
        }
        return up[0][u];
    };
    int Q;
    cin >> Q;
    for (int i = 0; i < Q; ++i) {
        int m, cur;
        cin >> m >> cur;
        for (int j = 1; j < m; ++j) {
            int x;
            cin >> x;
            cur = lca(cur, x);
        }
        cout << maxAnc[cur] << "\n";
    }
    return 0;
}