题目如下: Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,1],[1,3],[3,1],[3,3],[2,2]] Output

Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,2],[2,1],[1,0],[0,1]]

Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,2],[2,1],[1,0],[0,1]]

Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,2],[2,1],[1,0],[0,1]]

题目如下: Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from these points, with sides not necessarily parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,2],[2,1],[1,0],[

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 线段长+线段中心+字典 日期 题目地址:https://leetcode.com/problems/minimum-area-rectangle-ii/ 题目描述 Given a set of points in the xy-plane, determine the minimum area of any rectangle formed from

Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,1],[1,3],[3,1],[3,3],[2,2]] Output: 4 Ex

Given a set of points in the xy-plane, determine the minimum area of a rectangle formed from these points, with sides parallel to the x and y axes. If there isn't any rectangle, return 0. Example 1: Input: [[1,1],[1,3],[3,1],[3,3],[2,2]] Output: 4 Ex

作者: 负雪明烛 id: fuxuemingzhu 个人博客: http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 确定对角线,找另外两点(4sum) 字典保存出现的x值,y值的点 日期 题目地址:https://leetcode.com/problems/minimum-area-rectangle/description/ 题目描述 Given a set of points in the xy-plane, determine the minimum are

题目标签:HashMap 题目给了我们一组 xy 上的点坐标,让我们找出 能组成矩形里最小面积的那个. 首先遍历所有的点,把x 坐标当作key 存入map, 把重复的y坐标 组成set,当作value 存入map. 然后遍历所有的点,找出 对角的两个点, 再去map里确认是否存在剩下的两个对角点,计算面积,一直保留最小的那个面积. Java Solution: Runtime: 214 ms, faster than 78.80% Memory Usage: 61.5 MB, less than

给定在 xy 平面上的一组点,确定由这些点组成的任何矩形的最小面积,其中矩形的边不一定平行于 x 轴和 y 轴. 如果没有任何矩形,就返回 0. 示例 1: 输入:[[1,2],[2,1],[1,0],[0,1]] 输出:2.00000 解释:最小面积的矩形出现在 [1,2],[2,1],[1,0],[0,1] 处,面积为 2. 示例 2: 输入:[[0,1],[2,1],[1,1],[1,0],[2,0]] 输出:1.00000 解释:最小面积的矩形出现在 [1,0],[1,1],[2,1],

2020-02-10 21:02:13 问题描述: 问题求解: 本题由于可以暴力求解,所以不是特别难,主要是用来熟悉计算几何的一些知识点的. public double minAreaFreeRect(int[][] points) { double res = 2e9; Map<Integer, Set<Integer>> record = new HashMap<>(); for (int[] p : points) { int x = p[0]; int y =

Minimum Value Rectangle time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output You have nn sticks of the given lengths. Your task is to choose exactly four of them in such a way that they can for

Find the total area covered by two rectilinear rectangles in a 2D plane. Each rectangle is defined by its bottom left corner and top right corner as shown in the figure. Assume that the total area is never beyond the maximum possible value of int. Cr

java虚拟机在执行java程序的过程中会把它所管理的内存划分为若干个区域,这些区域都有各自的用途,以及创建和销毁的时间,有的区域随着虚拟机进程的启动而存在,有些区域则依赖着用户的线程的启动和结束而建立和销毁.根据java 虚拟机规范的规定,java虚拟机所管理的内存将会包括以下几个运行时数据区域. 如下图所示: java虚拟机运行时数据区可以分为: 程序计数器  堆  方法区  运行时常量池  本地方法栈  java虚拟机栈.其中程序计数器.java栈.本地方法栈,是根据线程的启动而创建的:而

之前做的时候没想出来...现在来数学推导一波. 题意:从n个木棒中选出4个拼成一个矩形,使得 (周长)2/面积 最小. 解:设矩形宽a长b.我们要最小化下面这个式子: 去掉常数,不妨设b = a + len,则化为下式: 取倒数,也就是最大化下式: 显然对于每个确定的a,len越小越好.所以直接取相邻的即可. 好像还有一种推导方法...... 原式可直接化为a/b + b/a,这就是个耐克函数双钩函数,让a/b趋近1也就是a=b,所以a和b越接近越好.

https://www.luogu.org/problemnew/show/CF1027C #include<cstdio> #include<string> #include<cstdlib> #include<cmath> #include<iostream> #include<cstring> #include<set> #include<queue> #include<algorithm>

传送门 题目大意 有n个木棍,让你选4根使得组成的矩形的周长的平方除以面积最小. 分析 这个题看起来就是一个需要证明的贪心,下面我们来证明一下: 所以我们只需要枚举一边所有的a的可能值,然后b就是比a大的边中最小的,复杂度O(n).总复杂度O(nlogn). 代码 #include<iostream> #include<cstdio> #include<cstring> #include<string> #include<algorithm> #

题意:给定n根木棍,不允许拼接或折断,选择四根组成矩形,求所有合法矩形中周长平方与面积比最小的一个,输出拼成这个矩形的四根木棍 n<=1e6 思路:猜结论:答案必定从相邻的4根中产生 证明见https://www.luogu.org/blog/78371/so1ution-cf1027c #include<cstdio> #include<cstring> #include<string> #include<cmath> #include<ios

函数中的代码是部分代码,详细代码在最后 1 cv2.boundingRect 作用:矩形边框(boundingRect),用于计算图像一系列点的外部矩形边界. cv2.boundingRect(array) -> retval 参数: array - 灰度图像(gray-scale image)或 2D点集( 2D point set ) 返回值:元组 元组(x, y, w, h ) 矩形左上点坐标,w, h 是矩阵的宽.高,例如 (161, 153, 531, 446) 代码示例: conto