Check if the two rectangles intersect

最新推荐文章于 2024-08-28 11:31:52 发布
最新推荐文章于 2024-08-28 11:31:52 发布
阅读量108 收藏
点赞数
CC 4.0 BY-SA版权
分类专栏： algorithm
版权声明：本文为博主原创文章，遵循 CC 4.0 BY-SA 版权协议，转载请附上原文出处链接和本声明。
本文链接：https://blog.youkuaiyun.com/iteye_15479/article/details/82088653
本文探讨了如何通过底边左坐标和顶边右坐标来判断两个矩形是否相交，涉及几何运算与条件判断。

Given the bottom left and top right coordinates of the two rectangles, check whether the two rectangles intersect or not.
Java:实现一个四叉树（整数坐标）算法（附完整源码）
希望我的博客，能帮上你解决学习中工作中所遇到的问题
08-29 486
Java:实现一个四叉树（整数坐标）算法（附完整源码）
[代码大模型]MHPP: Exploring the Capabilities and Limitations of Language Models Beyond Basic Code Generat
强化学习曾小健
06-23 1122
近年来，大型语言模型（LLMs）的进步在代码生成方面大大提高了性能，尤其是在功能级别。例如，GPT-4在HumanEval上的通过率为88.4%。然而，这引发了质疑现有基准在全面评估功能级别代码生成能力方面的充分性。我们的研究分析了两个常见的基准，HumanEval和MBPP，并发现，由于质量、难度和粒度等方面的限制，这些基准可能无法充分评估LLMs的代码生成能力。
HDU2056 Rectangles【水题】【相交面积】
程序员充电站（itcharge）
01-03 1221
题目大意：给你两个矩形对角线两端的坐标，输出这两个矩形的相交面积 思路：假设两个矩形相交，则相交的矩形面积横坐标为四个横坐标中间的 两个横坐标，纵坐标为四个纵坐标中间的两个纵坐标，然后计算面积。若 两个矩形不相交，则相交面积为0.00。 那么怎么判断是否相交呢。思路很简单，分别计算出矩形1和矩形2最小和 最大和横、纵坐标，若矩形1的最小横坐标>=矩形2的最大横坐标 或者 矩形1的最大横坐标<=矩形2的最小横坐标 则两个矩形不可能相交，同理， 纵坐标也是如此。
B. Two Tables
he_qingjun的博客
08-03 360
题目：https://codeforces.com/contest/1555/problem/B 又是简单题，对我来说太难了，我是菜狗。。。下面是官方解，做了个学习笔记 You have an axis-aligned rectangle room with width W and height H, so the lower left corner is in point (0,0) and the upper right corner is in (W,H). There is a rectangula
hdu 3885 Find the Difference
喵呜的Blog
07-27 1489
Find the DifferenceTime Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 89    Accepted Su
What are the lesser known but useful data structures?
纸上得来终觉浅
08-31 2209
(adsbygoogle = window.adsbygoogle || []).push({}); What are the lesser known but useful data structures? There are some data structures around that are really useful but are unknown to most programmers
华为OD机试算法精选题解汇总：提前准备，稳步通关的关键
专注于全栈开发领域
08-28 1340
通过提前准备和针对性的练习，华为OD机试并非不可战胜的难关。我们为你准备的《华为OD机试算法精选题解汇总》将成为你备考过程中的有力工具，帮助你更好地理解题目，掌握解题技巧，从容应对即将到来的挑战。华为的机试不仅是一次技术能力的考验，更是对个人应变能力与综合素质的全面衡量。希望每位应试者都能通过充分的准备，在华为的招聘过程中脱颖而出，迈向职业生涯的新高峰。
《NX二次开发经典案例》专栏目录（上篇）
王牌飞行员_里海的博客
05-24 2177
本专栏订阅后是永久阅读的。欢迎一起学习NX二次开发案例，逐步积累宝贵的经验，早日成为行业专家。
2D Rotated Rectangle Collision
weixin_30352191的博客
10-10 158
Introduction While working on a project for school, I found it necessary to perform a collision check between sprites that had been translated and rotated. I wanted to use bounding boxes because a pe...
intersect 矩形
最新发布
02-28
Returns True if there exists an intersection between two rectangles, otherwise returns False. """ # Unpack coordinate values from parameters into more readable variables names ax_min, ay_min, ax...
cv2.rotatedrectangleintersecti
07-28
on function is used to calculate the intersection area between two rotated rectangles. The function is called `cv2.rotatedRectangleIntersection(rect1, rect2)`. It takes two parameters, `rect1` and `...
算法：去除字符串中的重复字母
渴死的火烈鸟
11-06 1013
问题：去除字符串中的重复字符，不能使用额外空间（1或者2个变量除外）。 很简单的练手题,但是发现很难写正确。 [code="java"] public class DuplicateChar { //return length of the final string public static int removeDuplicateChar(char []str){ ...
怎么不开根号算平方根？
渴死的火烈鸟
12-24 610
写一个函数，不用开根号算平方根。 这是网上一个题目，一开始一筹莫展，看了答案恍然大悟，就是用二分法去逼近。 [code="java"] const double error = 0.000000001f; double findSqrt(double t){ double high = t; double low = 0; whil...
nvidia面试题
渴死的火烈鸟
01-04 535
八道题目，第一题有六个小题，一道小题0.5分，剩下的七道题目一题1分。全英文试卷，要求英文作答。 1. (1)what is an abstract class (2)myclass a;myclass b;b=a; myclass a;myclass b=a;区别 (3)stl有哪三类？ (4)容器有哪几种？ (5)virtual destruction作用？ (6)...
计算数组元素的乘积
渴死的火烈鸟
11-03 446
Given an array a[i], your task is to generate an array b[i], b[i] is the product of the numbers in the array except the i th number.   My solution:   DP problem?   Define two functions: f(i)=a...
sort n integers of range 0~n^2-1 in O(n) time
渴死的火烈鸟
06-24 434
This is one problem introduced in "Introduction to Algorithm". We can use radix-sort to solve it.   package alg; import java.util.ArrayList; import java.util.Iterator; /* * This example sh...
8-Queen Problem
渴死的火烈鸟
09-03 333
Problem: Write an algorithm to print all ways of arranging eight queens on a chess board so that none of them share the same row, column or diagonal. My Code: ____________________________________...
[leetcode] Palindrome Partition
渴死的火烈鸟
04-12 308
http://leetcode.com/onlinejudge#question_131 Question: Given a string s, partition s such that every substring of the partition is a palindrome. Return all possible palindrome partitioning of...
How to generate permutations
渴死的火烈鸟
09-07 195
The following algorithm generates the next permutation lexicographically after a given permutation. It changes the given permutation in-place. Find the largest index  k   such that  ...
We want to design a divide and conquer algorithm for computing the union of a collection of rectangles. The input rectangles are aligned with the axes and they are all stabbed by the y-axis. Each rectangle is represented by the coordinates of its top-left and bottom-right corners, and the union is representation by a sequence of interior-disjoint rectangles listed from top to bottom. We require that no two consecutive rectangles in the representation can be merged into a single rectangle.
05-30
if len(rectangles) == 0: return [] if len(rectangles) == 1: return rectangles mid = len(rectangles) // 2 left, right = rectangles[:mid], rectangles[mid:] left = merge_rectangles(left) right = ...
评论
添加红包

请填写红包祝福语或标题

个

红包个数最小为10个

元

红包金额最低5元

当前余额3.43元 前往充值 >
需支付：10.00元
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付元
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明：

1.余额是钱包充值的虚拟货币，按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载，可以购买VIP、付费专栏及课程。

余额充值