uva 11800 Determine the Shape】的更多相关文章

vjudge上题目链接:Determine the Shape 第二道独自 A 出的计算几何水题,题意就是给你四个点,让你判断它是哪种四边形:正方形.矩形.菱形.平行四边形.梯形 or 普通四边形. 按照各个四边形的特征层层判断就行,只是这题四个点的相对位置不确定(点与点之间是相邻还是相对并不确定),所以需要枚举各种情况,幸好 4 个点一共只有 3 种不同的情况:abcd.abdc.acbd 画个图就能一目了然,接下来就是编码了,无奈因为一些细节问题我还是调试了还一会 o(╯□╰)o #incl…
http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2900 题目大意: 给定平面上的4个点,没有3点共线,你的任务是判断着4个点能组成什么样的四边形.(正方形,矩形,菱形,平行四边形,菱形,梯形,普通四边形) 思路: 从后面开始判断,梯形一组平行边即可. 平行四边形要两组,而菱形对角线垂直,矩形有一个角为直角,正方形既是矩形又是菱形. 至于判断…
题意: 给四个点,判断四边形的形状.可能是正方形,矩形,菱形,平行四边形,梯形或普通四边形. 解法: 开始还在纠结怎么将四个点按序排好,如果直接处理的话,有点麻烦,原来凸包就可搞,直接求个凸包,然后点就自动按逆时针排好了,然后就判断就可以了,判断依据题目下面有,主要是用到点积和叉积,判断垂直用点积,判断平行用叉积. 代码: #include <iostream> #include <cstdio> #include <cstring> #include <cstd…
题目传送门 题意:给了四个点,判断能构成什么图形,有优先规则 分析:正方形和矩形按照点积为0和长度判断,菱形和平行四边形按向量相等和长度判断,梯形按照叉积为0判平行.因为四个点是任意给出的,首先要进行凸包排序,可能会有三点共线的情况. /************************************************ * Author :Running_Time * Created Time :2015/10/22 星期四 13:27:33 * File Name :UVA_11…
原文:ArcGIS Engine开发的ArcGIS 版本管理的功能 转自:http://blog.csdn.net/linghe301/article/details/7965901 这是以前的ArcGIS Engine开发成果,主要是Geodatabase方面的,模仿ArcGIS版本的流程系统环境: VS2010.ArcGIS Engine10.DevExpress721(第三方控件,比较常用易于下载) ---------------------------------------------…
From: https://alexanderetz.com/2015/07/25/understanding-bayes-updating-priors-via-the-likelihood/ Reading note. In a previous post I outlined the basic idea behind likelihoods and likelihood ratios. Likelihoods are relatively straightforward to under…
Mr. Pippo's Pizza 题目连接: https://www.hackerrank.com/contests/ieeextreme-challenges/challenges/mr-pippos-pizza Description Mr. Pippo wants to start a new pizza shop. Everything about his pizzas is unique - the recipe is unique, the taste is unique, and…
BACKGROUND The present invention relates to video processing systems. Advances in imaging technology have led to high resolution cameras for personal use as well as professional use. Personal uses include digital cameras and camcorders that can captu…
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