You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 13549    Accepted Submission(s): 6645 Problem Description Many geometry(几何)problems were designed in the ACM/…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10204    Accepted Submission(s): 5042 Problem Description Many geometry(几何)problems were designed in the ACM/…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6425    Accepted Submission(s): 3099 Problem Description Many geometry(几何)problems were designed in the ACM/I…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6932    Accepted Submission(s): 3350 Problem Description Many geometry(几何)problems were designed in the ACM/I…

题目:给出一些线段,判断有几个交点. 问题:如何判断两条线段是否相交? 向量叉乘(行列式计算):向量a(x1,y1),向量b(x2,y2): 首先我们要明白一个定理:向量a×向量b(×为向量叉乘),若结果小于0,表示向量b在向量a的顺时针方向:若结果大于0,表示向量b在向量a的逆时针方向:若等于0,表示向量a与向量b平行.(顺逆时针是指两向量平移至起点相连,从某个方向旋转到另一个向量小于180度).如下图: 在上图中,OA×OB = 2 > 0, OB在OA的逆时针方向:OA×OC = -2 <…

链接:传送门 题意:给出 n 个线段找到交点个数 思路:数据量小,直接暴力判断所有线段是否相交 /************************************************************************* > File Name: hdu1086.cpp > Author: WArobot > Blog: http://www.cnblogs.com/WArobot/ > Created Time: 2017年05月07日 星期日 23时34…

称号: You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 145 Accepted Submission(s): 100   Problem Description Many geometry(几何)problems were designed in the ACM/IC…

Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. And now, I also prepare a geometry problem for this final exam. According to the experience of many ACMers, geometry problems are always much trouble, but this problem is ve…

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=1086 判断两条线段是否有交点,我用的是跨立实验法: 两条线段分别是A1到B1,A2到B2,很显然,如果这两条线段有交点,那么可以肯定的是: A1-B1,A2-B1这两个向量分别在B2-B1的两边,判断是不是在两边可以用向量的叉积来判断,这里就不说了,同理B1-A1,B2-A1在A2-A1的两边,当同时满足这两个条件时,说明这两条线段是有交点的. #include<cstdio> #include&…

数学题,证明AB和CD.只需证明C.D在AB直线两侧,并且A.B在CD直线两侧.公式为:(ABxAC)*(ABxAD)<= 0 and(CDxCA)*(CDxCB)<= 0 #include <stdio.h> #define MAXNUM 105 typedef struct { double x1, y1; double x2, y2; } line_st; line_st lines[MAXNUM]; int cal(int i, int j) { double ab_x,…