ALG4：最近点对(the closest pair of points )

问题描述如下：

详情见：最近点对问题描述

具体实现：

• 由于设置本题的 OJ 设置了时间限制，规定了只能用分治的思想实现
• 分成三部分处理：
• 根据中值将待处理的点集分成三部分
• 左边求出最小值 d l e f t d_{left} dleft​
• 右边求出最小值 d r i g h t d_{right} dright​
• 中间的点带求出最小值 d m i d d_{mid} dmid​,(根据鸽巢原理，中间的点带中需要求的点不会超过6个)
• 将左右两边的点集处理方式类似，递归求解子问题

具体的代码实现如下：
(已通过全部的23个测试样例)

#include <iostream>
#include <iomanip>
#include <vector>
#include <algorithm>
#include <math.h>

using namespace std;


class Point
{
public:
	double x;
	double y;
	Point()
	{
		x = 0.00;
		y = 0.00;
	}
};
double dist(Point a, Point b)
{//计算两个点之间的距离
	return sqrt(double((a.x - b.x) * (a.x - b.x) + (a.y - b.y) * (a.y - b.y)));
}
bool cmp1(Point a, Point b)
{//制定外层排序规则
	return a.x < b.x;
}
bool cmp2(Point a, Point b)
{//指定内层排序规则
	return a.y < b.y;
}

double min_dist(Point* point_list, int left_index, int right_index)
{
	//递归出口
	if (right_index - left_index == 1)
		return dist(point_list[left_index], point_list[right_index]);
	//处理两边的情况
	int mid_index = (left_index + right_index) >> 1;
	double min_left = min_dist(point_list, left_index, mid_index);
	double min_right = min_dist(point_list, mid_index, right_index);
	double d = min(min_left, min_right);//找到左右两边的最小距离
	//下面找中间的点的最小值
	vector<Point> mid_list;
	for (int left = mid_index; left >= left_index; left--)
	{//从中间开始向左找
		if (point_list[mid_index].x - point_list[left].x <= d)
			mid_list.push_back(point_list[left]);
		else
			break;
	}
	for (int right = mid_index + 1; right <= right_index; right++)
	{//从中间开始向右找
		if (point_list[right].x - point_list[mid_index].x <= d)
			mid_list.push_back(point_list[right]);
		else
			break;
	}
	sort(mid_list.begin(), mid_list.end(), cmp2);
	//对于中间节点序列进行处理
	double min_mid = 0xffffffffffffffff;
	for (int i = 0; i < mid_list.size() - 1; i++)
		for (int j = i + 1; j < mid_list.size(); j++)
			if (mid_list[j].y - mid_list[i].y <= d)
				min_mid = min(min_mid, dist(mid_list[i], mid_list[j]));
			else
				break;
	return min(min_mid, d);
}






int main()
{
	int n;
	cin >> n;
	Point* point_list = new Point[n];
	for (int i = 0; i < n; i++)
	{
		double x, y;
		cin >> x >> y;
		point_list[i].x = x;
		point_list[i].y = y;
	}
	sort(point_list, point_list + n, cmp1);
	cout << fixed << setprecision(2) << pow(min_dist(point_list, 0, n - 1), 2) << endl;
	return 0;
}

仅供复习时整理参考，用于学习交流

标签：index,closest,point,double,Point,list,ALG4,pair,left
来源： https://blog.csdn.net/m0_57476889/article/details/117197891