奶牛无线网络(Moo Network G) 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 <numeric>
#include <algorithm>
using namespace std;

struct Pt {
    long long x, y;
    int id;
    bool operator<(const Pt& o) const {
        if (x != o.x) return x < o.x;
        return y < o.y;
    }
};

struct Edge {
    int u, v;
    long long w;
    bool operator<(const Edge& o) const { return w < o.w; }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<Pt> pts(n);
    for (int i = 0; i < n; ++i) {
        cin >> pts[i].x >> pts[i].y;
        pts[i].id = i;
    }
    sort(pts.begin(), pts.end());
    vector<int> lastAtY(11, -1);
    vector<Edge> edges;
    edges.reserve(n * 11);
    for (int i = 0; i < n; ++i) {
        for (int y = 0; y <= 10; ++y) {
            int j = lastAtY[y];
            if (j != -1) {
                long long dx = pts[i].x - pts[j].x;
                long long dy = pts[i].y - pts[j].y;
                edges.push_back({pts[i].id, pts[j].id, dx * dx + dy * dy});
            }
        }
        lastAtY[pts[i].y] = i;
    }
    sort(edges.begin(), edges.end());
    vector<int> p(n);
    iota(p.begin(), p.end(), 0);
    auto find = [&](int x) {
        while (p[x] != x) x = p[x] = p[p[x]];
        return x;
    };
    long long ans = 0;
    for (const auto& e : edges) {
        int ru = find(e.u), rv = find(e.v);
        if (ru != rv) {
            p[ru] = rv;
            ans += e.w;
        }
    }
    cout << ans << "\n";
    return 0;
}