Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7674    Accepted Submission(s): 3252 Problem Description There are many secret openings in the floor which are covered by a big…

题目链接:hdu 1115 计算几何求多边形的重心,弄清算法后就是裸题了,这儿有篇博客写得很不错的: 计算几何-多边形的重心 代码如下: #include<cstdio> #include<cstring> #include<algorithm> #include<cmath> using namespace std; ; struct point { double x,y; point() {} point(double x, double y): x(x…

Lifting the Stone Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 4819    Accepted Submission(s): 2006 Problem Description There are many secret openings in the floor which are covered by a big…

Shape of HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4972    Accepted Submission(s): 2250 Problem Description 话说上回讲到海东集团推选老总的事情,最终的结果是XHD以微弱优势当选,从此以后,“徐队”的称呼逐渐被“徐总”所取代,海东集团(HDU)也算是名副其实了.…

最大三角形 Time Limit: 5000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 4015    Accepted Submission(s): 1433 Problem Description 老师在计算几何这门课上给Eddy布置了一道题目,题目是这样的:给定二维的平面上n个不同的点,要求在这些点里寻找三个点,使他们构成的三角形拥有的面积最大.Eddy对这道…

Shape of HDU Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 7797    Accepted Submission(s): 3526 Problem Description 话说上回讲到海东集团推选老总的事情,最终的结果是XHD以微弱优势当选,从此以后,"徐队"的称呼逐渐被"徐总"所取代,海东集…

Problem How Many Triangles (HDU 5784) 题目大意 给定平面上的n个点(n<2000),询问可以组成多少个锐角三角形. 解题分析 直接统计锐角三角形较困难,考虑问题的反面,统计直角三角形.钝角三角形.平角三角形(暂时这么叫吧QAQ). 首先枚举三角形的一个端点A,对其他点进行象限为第一关键字,极角为第二关键字排序. 然后使用三个指针,进行O(n)的扫描. 具体做法为用 i 指针指向三角形的第二个端点B.我们可以假想通过平移和旋转,把A点置于平面直角坐标系的原点,…

昨天用vim练了一道大水题,今天特地找了道稍难一点的题.不过也不是很难,简单的计算几何而已.练习用vim编码,用gdb调试,结果居然1A了,没调试...囧... 做法很简单,无非就是两种情况:①三个巫师构成一个钝角(极限情况是直角)三角形,那么所画的圆应该是钝角所对边为直径的圆:②三个巫师构成一个锐角三角形,那么所画的圆应该是三角形的外接圆. 就这样纸,上了点模板,代码如下: /* * Author : ben */ #include <cstdio> #include <cstdlib…

题目链接[http://acm.hdu.edu.cn/showproblem.php?pid=6206] 题意: 给出不共线的三个点,和一个点(x,y),然后判断(x,y)在不在这三个点组成的圆外. 题解: 咋一看很简单,实际也很简单,但是坐标都很大,会爆long double,怎么办?只有用java了. 公式: a = ((y2 - y1) * (y3 * y3 - y1 * y1 + x3 * x3 - x1 * x1) - (y3 - y1) * (y2 * y2 - y1 * y1 +…

openGL Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 170    Accepted Submission(s): 77 Problem Description Jiaoshou selected a course about “openGL” this semester. He was quite interested in m…