【nyoj-1132-promise me a medal(求线段交点)】的更多相关文章

nyoj-1132-promise me a medal(求线段交点)

题目链接 /* Name:nyoj-1132-promise me a medal Copyright: Author: Date: 2018/4/26 20:26:22 Description: 向量之间的运算,不熟悉就绕蒙逼了 用 ACM国际大学生程序设计竞赛 算法与实现的模板 有个坑: 如果第二条线段的起点是第一条线段的终点,那么不需要计算 1 1 2 1 3 1 3 1 4 输出 yes 1.0 3.0 */ #include <iostream> #include <cstdi…

谈谈"求线段交点"的几种算法(js实现,完整版)

"求线段交点"是一种非常基础的几何计算, 在很多游戏中都会被使用到. 下面我就现学现卖的把最近才学会的一些"求线段交点"的算法总结一下, 希望对大家有所帮助. 本文讲的内容都很初级, 主要是面向和我一样的初学者, 所以请各位算法帝们轻拍啊 嘎嘎 引用 已知线段1(a,b) 和线段2(c,d) ,其中a b c d为端点, 求线段交点p .(平行或共线视作不相交) =============================== 算法一: 求两条线段所在直线的交点, 再…

poj1408(求线段交点)

求出所有线段的交点,然后利用叉乘求四边形面积即可. // // main.cpp // poj1408 // // Created by 陈加寿 on 15/12/31. // Copyright (c) 2015年 chenhuan001. All rights reserved. // #include <iostream> #include <cmath> #include <vector> #include <string.h> #include &…

hdu 2857:Mirror and Light（计算几何，点关于直线的对称点，求两线段交点坐标）

Mirror and Light Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 650    Accepted Submission(s): 316 Problem Description The light travels in a straight line and always goes in the minimal path b…

hdu 2528:Area（计算几何，求线段与直线交点 + 求多边形面积）

Area Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 225    Accepted Submission(s): 77 Problem Description 电子科大清水河校区是电子科大大力兴建的未来主校区,于07年秋正式迎接学生入住,目前有07.08级本科生及部分研究生在此校区学习.生活.清水河校区位于成都高新西区的中部地带,占…

【计算几何初步-线段相交】【HDU1089】线段交点

You can Solve a Geometry Problem too Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 7847    Accepted Submission(s): 3834 Problem Description Many geometry(几何)problems were designed in the ACM/…

js判断向量叉点 并求出交点坐标

 代码如下可以直接运行,判断向量相交并求出交点坐标 <!DOCTYPE html> <html> <head> <meta http-equiv="Content-Type" content="text/html; charset=utf-8"> <title>判断线段相交并求交点</title> <style> .title{text-align:center;} .wrapper…

UVa 11437:Triangle Fun（计算几何综合应用，求直线交点，向量运算，求三角形面积）

Problem ATriangle Fun Input: Standard Input Output: Standard Output In the picture below you can see a triangle ABC. Point D, E and F divides the sides BC, CA and AB into ratio 1:2 respectively. That is CD=2BD, AE=2CE and BF=2AF. A, D; B, E and C, F…

POJ_1269_Intersecting Lines_求直线交点

POJ_1269_Intersecting Lines_求直线交点 Description We all know that a pair of distinct points on a plane defines a line and that a pair of lines on a plane will intersect in one of three ways: 1) no intersection because they are parallel, 2) intersect in…

sgu 129 Inheritance 凸包,线段交点,计算几何 难度:2

129. Inheritance time limit per test: 0.25 sec. memory limit per test: 4096 KB The old King decided to divide the Kingdom into parts among his three sons. Each part is a polygonal area. Taking into account the bad temper of the middle son the King ga…