TOYS(计算几何-入门)】的更多相关文章

题目 ‘^’代表叉乘 ‘•’代表点乘 点积:a•b=ax*bx+ay*by 叉积:a^b=ax*by-bx*ay…
rt,计算几何入门: TOYS 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 toy…
POJ2318 本题需要运用to left test不断判断点处于哪个分区,并统计分区的点个数(保证点不在边界和界外),用来做叉积入门题很合适 //计算几何-叉积入门题 //Time:157Ms Memory:828K #include<iostream> #include<cstring> #include<cstdio> #include<algorithm> using namespace std; #define MAXN 5005 struct P…
TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10281   Accepted: 4924 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 w…
TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 12015   Accepted: 5792 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 w…
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…
跨产品的利用率推断点线段向左或向右,然后你可以2分钟 代码: #include <cstdio> #include <cstring> #include <algorithm> using namespace std; const int N = 5005; int n, m, x1, y1, x2, y2; struct Point { int x, y; Point() {} Point(int x, int y) { this->x = x; this-&g…
题目链接:https://cn.vjudge.net/problem/POJ-2318 题意 在一个矩形内,给出n-1条线段,把矩形分成n快四边形 问某些点在那个四边形内 思路 二分+判断点与位置关系 提交过程 WA*n x1和x2,y1和y2在复制的时候没分清(哭 WA 可能存在二分问题? AC 代码 #define PI 3.1415926 #include <cmath> #include <cstdio> #include <vector> #include &…
今天开始学习计算几何,百度了两篇文章,与君共勉! 计算几何入门题推荐 计算几何基础知识 题意:有一个盒子,被n块木板分成n+1个区域,每个木板从左到右出现,并且不交叉. 有m个玩具(可以看成点)放在这个盒子里,问每个区域分别有多少个玩具. 思路:首先,用叉积判断玩具是否在木板的左边,再用二分找到符合的最右边的木板,该木板答案加一. #include<stdio.h> #include<string.h> struct point{ int x,y; point(){} point(…
嘟嘟嘟 题面:先告诉你一个矩形抽屉的坐标,然后\(n\)个隔板将抽屉分成了\(n + 1\)格(格子从\(0\)到\(n - 1\)标号),接下来随机输入\(m\)个玩具的坐标.问最后每一个格子里有多少个玩具. 仔细想想就是一道计算几何入门题. 对于一个玩具\((x_0, y_0)\),我们只要找到在他的左面且离他最近的隔板\(i\),则这个玩具就在第\(i\)格里. 怎么判断隔板\(AB\)(规定\(A\)点是隔板的上端,\(B\)点是隔板下端)是否在玩具的左边?用叉积即可:只要\(\over…