计蒜客竞赛基础，最近点对问题

最新推荐文章于 2021-01-27 21:47:52 发布

stan_smith

最新推荐文章于 2021-01-27 21:47:52 发布

阅读量293

点赞数 1

CC 4.0 BY-SA版权

分类专栏：算法文章标签：二分

本文链接：https://blog.youkuaiyun.com/qq_35539135/article/details/54645969

算法专栏收录该内容

12 篇文章

订阅专栏

本文介绍一种高效算法来解决点集中的最近点对问题。通过递归地将点集分为两部分并分别求解，然后寻找跨越两部分的更近距离。此算法使用了分治策略，并结合排序优化跨区间距离比较过程。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
#define INF 100002
using namespace std;

struct point {
	double x;
	double y;
};
point p[100002];
point p1[100002];
point p2[100002];
//point flagp[6];

bool cmpx(point a, point b) {
	if (a.x != b.x) {
		return a.x < b.x;
	}
	else {
		return a.y < b.y;
	}
}

bool cmpy(point a, point b) {
	if (a.y != b.y) {
		return a.y < b.y;
	}
	else {
		return a.x > b.x;
	}
}


double func(int number, int left, int right) {
	double len1, len2, minlen, l;
	int num1, num2, k = 0;
	if (number == 2) {
		return sqrt((p[left].x - p[right].x) * (p[left].x - p[right].x) +

			(p[left].y - p[right].y) * (p[left].y - p[right].y));
	}
	else if (number < 2) {
		return INF;
	}
	else {
		num1 = number / 2;
		num2 = number - num1;
		len1 = func(num1, left, left + num1 - 1);
		len2 = func(num2, left + num1, right);
		minlen = len1 < len2 ? len1 : len2;
		l = (p[(left + right - 1) / 2].x + p[(left + right - 1) / 2 + 1].x) / 2;

		for (int i = left; i <= (left + right - 1) / 2; i++) {
			if (fabs(p[i].x - l) < minlen) {
				p1[k].x = p[i].x;
				p1[k].y = p[i].y;
				k++;
			}
		}
		sort(p1, p1 + k, cmpy);
		int k2 = 0;
		for (int i = (left + right - 1) / 2 + 1; i <= right; i++) {
			if (fabs(l - p[i].x) < minlen) {
				p2[k2].x = p[i].x;
				p2[k2].y = p[i].y;
				k2++;
			}
		}
		sort(p2, p2 + k2, cmpy);
		double flaglen = INF;
		for (int i = 0; i < k; i++) {
			for (int j = 0; j < k2; j++) {
				if (p2[j].y - p1[i].y < minlen) {
					flaglen = sqrt((p2[j].x - p1[i].x) * (p2

						[j].x - p1[i].x) + (p2[j].y - p1[i].y) * (p2[j].y - p1[i].y));
					if (flaglen < minlen) {
						minlen = flaglen;
					}
				}
			}
		}
		return minlen;
	}
}

int main() {
	int num;
	cin >> num;

	for (int i = 0; i < num; i++) {
		cin >> p[i].x >> p[i].y;
	}
	sort(p, p + num, cmpx);
	cout << func(num, 0, num - 1) << endl;

	return 0;
}

这篇文章给的思路很清晰！http://blog.youkuaiyun.com/zyh15/article/details/49050841