聪明的猴子(Clever Monkeys) 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 long long INF = (1LL << 60);

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int m;
    if (!(cin >> m)) return 0;
    vector<long long> jmp(m);
    for (int i = 0; i < m; ++i) cin >> jmp[i];
    int n;
    cin >> n;
    vector<long long> x(n), y(n);
    for (int i = 0; i < n; ++i) cin >> x[i] >> y[i];
    vector<long long> dist(n, INF);
    vector<char> vis(n, 0);
    dist[0] = 0;
    long long maxEdgeSq = 0;
    for (int step = 0; step < n; ++step) {
        int u = -1;
        for (int i = 0; i < n; ++i) {
            if (!vis[i] && (u == -1 || dist[i] < dist[u])) u = i;
        }
        vis[u] = 1;
        maxEdgeSq = max(maxEdgeSq, dist[u]);
        for (int v = 0; v < n; ++v) {
            if (!vis[v]) {
                long long dx = x[u] - x[v], dy = y[u] - y[v];
                long long w = dx * dx + dy * dy;
                if (w < dist[v]) dist[v] = w;
            }
        }
    }
    int ans = 0;
    for (int i = 0; i < m; ++i) {
        if (jmp[i] * jmp[i] >= maxEdgeSq) ++ans;
    }
    cout << ans << "\n";
    return 0;
}