求平面上第K近的点

Problem description：

Given $N$ points on the plane. Find the $Kth$ closest point to the origin $(0,0)$. (Here, the distance between two points on a plane is the Euclidean distance.)

Input：

Line 1: $N$ $K$

Others: point coordinate

Output：

Line 1: $Kth$ closest point to the origin $(0,0)$

Sample Input 1：

2 1
1 3
-2 2

Sample Output 1：

-2 2

Solution1：

#include <iostream>
#include <cstdio>
#include <cmath>
#include<algorithm>
using namespace std;

struct point{
    
    
	long long int x;
	long long int y;
	unsigned long long int distance;
};

bool cmp(point a, point b){
    
    
	return a.distance < b.distance;
}

int main() 
{
    
    
	int n, k;
	scanf("%d%d",&n, &k);
	const int N=n, K=k;
	
	point points[N];
	
	int i;
	for(i=0; i<N; i++){
    
    
		scanf("%lld%lld",&points[i].x, &points[i].y);
		points[i].distance = ((points[i].x)*(points[i].x)) + ((points[i].y)*(points[i].y));
	}
	
	sort(points, points+N, cmp);
	
	cout << points[K-1].x << ' ' << points[K-1].y << endl;
		
	return 0;
}

注意1:

1. 算法首先求出所有点到$(0,0)$的距离，然后sort排序直接求出距离$(0,0)$第K近的点。
2. 算法复杂度和空间复杂度较高。
3. 可通过创建结构体存储点的坐标。
4. sort函数有三个参数，第三个参数可以制定排序规则。