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2 分钟
P7883 平面最近点对(加强加强版)
备战NOIP 1= !!!
题目
题解
枚举肯定过不了,考虑分治优化
每个区间内的好算,关键是如何合并?
记表示两子区间返回的最小值
在点的附近距离内找点,可能更新答案
按纵坐标枚举即可
锅
代码
#include<bits/stdc++.h>#define PII pair<long long, long long>using namespace std;#define x first#define y secondconst int N = 4e5 + 10;PII p[N];int n;#define ll long longlong long merge(int l, int r){// cout << l << " " << r << "----------\n"; if(l == r) return 0x3f3f3f3f3f3f3f3f; int mid = (l + r) >> 1; ll h1 = merge(l, mid); ll h2 = merge(mid + 1, r); ll h = min(h1, h2); vector <PII> B; for(int i = l;i <= r;i++) { if((p[i].x - p[mid].x) * (p[i].x - p[mid].x) < h) B.push_back({p[i].y, p[i].x}); } sort(B.begin(), B.end());// for(auto i : B)// {// cout << i.x << " " << i.y << endl;// } for(int i = 0;i < B.size();i++) { for(int q = i + 1;q < B.size();q++) { if((B[i].first - B[q].first) * (B[i].first - B[q].first) > h) break; h = min(h, (B[i].first - B[q].first) * (B[i].first - B[q].first) + (B[i].second - B[q].second) * (B[i].second - B[q].second)); } }// cout << l << " " << r << " " << h << endl; return h;}
int main(){ ios::sync_with_stdio(0), cin.tie(0), cout.tie(0); cin >> n; for(int i = 1;i <= n;i++) cin >> p[i].x >> p[i].y;
sort(p + 1, p + 1 + n);
cout << merge(1, n) << endl; return 0;}
P7883 平面最近点对(加强加强版)
https://blog.histcat.top/posts/p7883/