Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 9996   Accepted: 2632 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9) …
题目链接 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12040   Accepted: 3125 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point:…
题目链接:http://poj.org/problem?id=1410 Intersection Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 12822   Accepted: 3347 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle.…
题目: Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9) end point: (11,2) rectangle: left-top: (1,5) right-bottom: (7,1)  Figure 1: Line segment doe…
Intersection 大意:给你一条线段,给你一个矩形,问是否相交. 相交:线段全然在矩形内部算相交:线段与矩形随意一条边不规范相交算相交. 思路:知道详细的相交规则之后题事实上是不难的,可是还有个坑点就是题目里明明说给的是矩形左上角跟右下角的点,但实际上不是,须要又一次推断一下...真坑. struct Point { double x, y; } A, B, C, D; struct Line { Point a, b; } L; int n; double xmult(Point p1…
题目链接:POJ 1410 Description You are to write a program that has to decide whether a given line segment intersects a given rectangle. An example: line: start point: (4,9) end point: (11,2) rectangle: left-top: (1,5) right-bottom: (7,1) Figure 1: Line se…
LINK 题意:询问是否存在直线,使得所有线段在其上的投影拥有公共点 思路:如果投影拥有公共区域,那么从投影的公共区域作垂线,显然能够与所有线段相交,那么题目转换为询问是否存在直线与所有线段相交.判断相交先求叉积再用跨立实验.枚举每个线段的起始结束点作为直线起点终点遍历即可. /** @Date : 2017-07-12 14:35:44 * @FileName: POJ 3304 基础线段交判断.cpp * @Platform: Windows * @Author : Lweleth (Sou…
题意: 给一条线段,和一个矩形,问线段是否与矩形相交或在矩形内. 解法: 判断是否在矩形内,如果不在,判断与四条边是否相交即可.这题让我发现自己的线段相交函数有错误的地方,原来我写的线段相交函数就是单纯做了两次跨立实验,在下图这种情况是错误的: 这样的话线段与右边界的两次跨立实验(叉积<=0)都会通过,但是并不相交. 所以要加快速排斥. 还有就是这题题目说给出的不一定是左上角,右下角依次的顺序.所以干脆重新自己定义左上角,右下角. 代码: #include <iostream> #inc…
题目传送门 题意:一个矩形和一条线段,问是否有相交 分析:考虑各种情况.坑点:给出的矩形的两个端点是无序的,还有线段完全在矩形内也算相交 /************************************************ * Author :Running_Time * Created Time :2015/10/27 星期二 13:17:49 * File Name :POJ_1410.cpp ******************************************…
题目链接 题意 判断线段和矩形是否有交点(矩形的范围是四条边及内部). 思路 判断线段和矩形的四条边有无交点 && 线段是否在矩形内. 注意第二个条件. Code #include <cstdio> #include <cmath> #include <iostream> #include <cstring> #define inf 0x3f3f3f3f #define eps 1e-6 #define maxn 110 using name…