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…

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): 13549    Accepted Submission(s): 6645 Problem Description Many geometry(几何)problems were designed in the ACM/…

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…

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…

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,…

pid=1086">You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 6997    Accepted Submission(s): 3385 Problem Description Many geometry(几何)problems were designe…

http://acm.hdu.edu.cn/showproblem.php?pid=1086 You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 8861    Accepted Submission(s): 4317 Problem Description Many…

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…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 6837 Accepted Submission(s): 3303 Problem Description Many geometry(几何)problems were designed in the ACM/ICPC. A…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 9596    Accepted Submission(s): 4725 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): 199    Accepted Submission(s): 132   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): 12959    Accepted Submission(s): 6373 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): 6340    Accepted Submission(s): 3064 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)                                         Problem Description Many geometry(几何)problems wer…

Geometry Problem HDU - 6242 Alice is interesting in computation geometry problem recently. She found a interesting problem and solved it easily. Now she will give this problem to you : You are given NN distinct points (Xi,Yi)(Xi,Yi) on the two-dimens…

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&…

http://acm.hdu.edu.cn/showproblem.php?pid=1086 分析:简单计算几何题,相交判断直接用模板即可. 思路:将第k条直线与前面k-1条直线进行相交判断,因为题目中不排除多条直线相交于同一个点的重复情况. 代码: #include <iostream> #include <stdio.h> #include <string.h> #include <algorithm> #include <queue> #i…

ACM-ICPC Live Archive 又是搞了一个晚上啊!!! 总算是得到一个教训,误差总是会有的,不过需要用方法排除误差.想这题才几分钟,敲这题才半个钟,debug就用了一个晚上了!TAT 有一定几何基础的很容易想到这里的碰撞一定是其中一个顶点撞到另一个三角形的边上,于是就可以暴力枚举每一个顶点的最先碰撞的时间.注意的是,求直线相交以后,判交点是否在线段内,对于1e7的数据,千万不要判点在线上!TAT 因为已经知道是交点了,只要判断是不是在两点之间就好了. 代码如下:(把onseg改少了…

链接:http://acm.hdu.edu.cn/showproblem.php?pid=1086 You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 7167    Accepted Submission(s): 3480 Problem Description Ma…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 10259    Accepted Submission(s): 5074 Problem Description Many geometry(几何)problems were designed in the ACM/…

传送门:You can Solve a Geometry Problem too 题意:给n条线段,判断相交的点数. 分析:判断线段相交模板题,快速排斥实验原理就是每条线段代表的向量和该线段的一个端点与 另一条线段的两个端点构成的两个向量求叉积,如果线段相交那么另一条线段两个端点必定在该线段的两边,则该线段代表的向量必定会顺时针转一遍逆时针转一遍,叉积必定会小于等于0,同样对另一条线段这样判断一次即可. #include <algorithm> #include <cstdio>…

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)Total Submission(s): 11918    Accepted Submission(s): 5908 Problem Description Many geometry(几何)problems were designed in the ACM/…

Alice is interesting in computation geometry problem recently. She found a interesting problem and solved it easily. Now she will give this problem to you : You are given NN distinct points (Xi,Yi)(Xi,Yi) on the two-dimensional plane. Your task is to…

pid=5323" target="_blank" style="">链接 Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 511    Accepted Submission(s): 114 Problem Descriptio…

Solve this interesting problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 1479    Accepted Submission(s): 423 Problem Description Have you learned something about segment tree? If not, don’…

Geometry Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others)Total Submission(s): 1722    Accepted Submission(s): 304Special Judge Problem Description Alice is interesting in computation geometry problem recen…