OpenCASCADE圆与平面求交 eryar@163.com 在 解析几何求交之圆与二次曲面中分析了OpenCASCADE提供的类IntAna_IntConicQuad可以用来计算圆与二次曲面之间的交点,这个算法是将平面Plane作为二次曲面的一个特例来处理,最后主要是对三角函数方程进行求解. 当直接使用圆和平面作为参数时,IntAna_IntConicQuad重载了函数Perform来对圆和平面进行求交计算,这时的算法与前面解三角函数不同,代码如下: void IntAna_IntConic

传统解法 题目来自 leetcode 335. Self Crossing. 题意非常简单,有一个点,一开始位于 (0, 0) 位置,然后有规律地往上,左,下,右方向移动一定的距离,判断是否会相交(self crossing). 一个很容易想到的方案就是求出所有线段,然后用 O(n^2) 的时间复杂度两两判断线段是否相交,而线段相交的判断,可以列个二元一次方程求解(交点).这个解法非常容易想到,但是实际操作起来比较复杂,接下去介绍利用向量的解法. 向量解法 简单回顾下向量,具体的自行谷歌.向量就

一定要注意位运算的优先级!!!我被这个卡了好久 判断线段相交模板题. 叉积,点积,规范相交,非规范相交的简单模板 用了“链表”优化之后还是$O(n^2)$的暴力,可是为什么能过$10^5$的数据? #include<cmath> #include<cstdio> #include<cstring> #include<algorithm> #define N 100005 using namespace std; struct Point { double x

IRelationalOperator 接口: 1. Provides access to members that determine if a certain spatial relationship exists between two geometries. Members     Description Contains Indicates if this geometry contains the other geometry.前者是否包含后者! Crosses Indicates

Pick-up sticks Time Limit: 3000MS   Memory Limit: 65536K Total Submissions: 7699   Accepted: 2843 Description Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find

先说一下题目大意:给定一些线段,这些线段顺序编号,这时候如果两条线段相交,则把他们加入到一个集合中,问给定一个线段序号,求在此集合中有多少条线段. 这个题的难度在于怎么判断线段相交,判断玩相交之后就是怎么找个他们之间的联系,这时候就要用到并查集了. 步骤: 1.判断两条线段相交 2. 用并查集实现查找线段个数和添加到集合中 关于这个判断线段相交的问题.我搞了一晚上加上一下午,刚开始自己想了一种数学上的相交,就是先求出两条线段所在的线性方程,然后求出他们的交点,最后在判断这个交点在不在这两个线段之

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>

1.组件配置 首先,要下载.NET for Postgresql的驱动,npgsql,EF6,以及EntityFramework6.Npgsql,版本号 3.1.1.0. 由于是mvc项目,所以,把相应的配置文件写在web.config里面,如下: <configSections> <!-- For more information on Entity Framework configuration, visit http://go.microsoft.com/fwlink/?LinkI

POJ_1066_Treasure Hunt_判断线段相交 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-art technology they are able to determine that the lower floor of the

POJ_2653_Pick-up sticks_判断线段相交 Description Stan has n sticks of various length. He throws them one at a time on the floor in a random way. After finishing throwing, Stan tries to find the top sticks, that is these sticks such that there is no stick o

POJ_1556_The Doors_判断线段相交+最短路 Description You are to find the length of the shortest path through a chamber containing obstructing walls. The chamber will always have sides at x = 0, x = 10, y = 0, and y = 10. The initial and final points of the path

POJ_3304_Segments_线段判断是否相交 Description Given n segments in the two dimensional space, write a program, which determines if there exists a line such that after projecting these segments on it, all projected segments have at least one point in common.

start )             );         )) )) );         XYZ xyz12 = lCurve1.Curve.get_EndPoint();         XYZ xyz21 = lCurve2.Curve.get_EndPoint();         XYZ xyz22 = lCurve2.Curve.get_EndPoint();         );         XYZ xyz2 = );         XYZ xyz3 = );      

Treasure Hunt Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7857   Accepted: 3247 Description Archeologists from the Antiquities and Curios Museum (ACM) have flown to Egypt to examine the great pyramid of Key-Ops. Using state-of-the-ar

POJ2653 判断线段相交的方法 先判断直线是否相交 再判断点是否在线段上 复杂度是常数的 题目保证最后答案小于1000 故从后往前尝试用后面的线段 "压"前面的线段 排除不可能的答案 就可以轻松AC了. #include<iostream> #include<stdio.h> #include<stdlib.h> #include<string.h> #include<math.h> #include<algorit

题目链接:https://vjudge.net/problem/POJ-3449 题意:给出若干几何体,判断每个几何体与其它几何体的相交情况,并依次输出. 思路: 首先要知道的是根据正方形对角线的两个点怎么求其它两个点,比如已知(x0,y0),(x2,y2),那么: x1+x3=x0+x2, x1-x3=y2-y0, y1+y3=y0+y2, y1-y3=x0-x2 之后就暴力枚举了,枚举所有几何体的所有边,用线段相交判断几何体相交.这题的输入输出很恶心. AC代码: #include<cstd

2019 杭电多校 1 1013 题目链接:HDU 6590 比赛链接:2019 Multi-University Training Contest 1 Problem Description After returning with honour from ICPC(International Cat Programming Contest) World Finals, Tom decides to say goodbye to ICPC and start a new period of l

题目链接:POJ 3805 Problem Description Numbers of black and white points are placed on a plane. Let's imagine that a straight line of infinite length is drawn on the plane. When the line does not meet any of the points, the line divides these points into t

题意不难理解,给出多个多边形,输出多边形间的相交情况(嵌套不算相交),思路也很容易想到.枚举每一个图形再枚举每一条边 恶心在输入输出,不过还好有sscanf(),不懂可以查看cplusplus网站 根据正方形对角的两顶点求另外两个顶点公式: x2 = (x1+x3-y3+y1)/2; y2 = (x3-x1+y1+y3)/2; x4= (x1+x3+y3-y1)/2; y4 = (-x3+x1+y1+y3)/2; 还有很多细节要处理 #include <iostream> #include &

POJ 2826 An Easy Problem?! -- 思路来自kuangbin博客 下面三种情况比较特殊,特别是第三种 G++怎么交都是WA,同样的代码C++A了 #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <cmath> using namespace std; const double eps = 1e-8;

题目链接:https://www.51nod.com/onlineJudge/questionCode.html#!problemId=1264 题意:中文题诶- 思路:对于直线a1a2, b1b2,我们可以通过跨立实验判断其是否相交(通过计算斜率也可以啦,不过好像有点麻烦): 其公式为:s=((向量)a1a2叉乘a1b1)*(a1a2)叉乘a1b2),若 s <=0 ,则有点b1, b2分别在线段a1a2的两边或者在同一条直线上: 我们可以很直观的知道如果a1, b2分别位于b1b2两边并且b