POJ2318TOYS(叉积判断点与直线位置)

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POJ2318TOYS(叉积判断点与直线位置)

题目链接题意:一个矩形被分成了n + 1块,然后给出m个点,求每个点会落在哪一块中,输出每块的点的个数就是判断点与直线的位置,点在直线的逆时针方向叉积 < 0,点在直线的顺时针方向叉积 > 0 // 可以选择二分查找 #include <iostream> #include <algorithm> #include <cstdio> #include <cstring> using namespace std; typedef long l…

POJ 2318 叉积判断点与直线位置

TOYS Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his…

POJ2318 TOYS(叉积判断点与直线的关系+二分)

Calculate the number of toys that land in each bin of a partitioned toy box. Mom and dad have a problem - their child John never puts his toys away when he is finished playing with them. They gave John a rectangular box to put his toys in, but John i…

poj2318（叉积判断点在直线左右+二分）

题目链接:https://vjudge.net/problem/POJ-2318 题意:有n条线将矩形分成n+1块,m个点落在矩形内,求每一块点的个数. 思路: 最近开始肝计算几何,之前的几何题基本处于挂机状态,但听别人说几何题不会太难,所以打算把几何给过了. 先引入叉积的一个重要性质,O为原点: OP^OQ>0 : P在Q的顺时针方向. OP^OQ<0 : P在Q的逆时针方向. OP^OQ=0 : O,P,Q共线. 那么我们就可以利用该性质判断一个点P在直线AB的左侧当且仅当:PA^PB&l…

POJ 2398 - Toy Storage 点与直线位置关系

Toy Storage Time Limit: 1000MS Memory Limit: 65536K Total Submissions: 5439 Accepted: 3234 Description Mom and dad have a problem: their child, Reza, never puts his toys away when he is finished playing with them. They gave Reza a rectangular box…

判断两条直线的位置关系 POJ 1269 Intersecting Lines

两条直线可能有三种关系:1.共线 2.平行(不包括共线) 3.相交. 那给定两条直线怎么判断他们的位置关系呢.还是用到向量的叉积例题:POJ 1269 题意:这道题是给定四个点p1, p2, p3, p4,直线L1,L2分别穿过前两个和后两个点.来判断直线L1和L2的关系这三种关系一个一个来看: 1. 共线. 如果两条直线共线的话,那么另外一条直线上的点一定在这一条直线上.所以p3在p1p2上,所以用get_direction(p1, p2, p3)来判断p3相对于p1p2的关…

POJ 1269 Intersecting Lines (判断直线位置关系)

题目链接:POJ 1269 Problem 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 a line becau…