TOYS(叉积)
TOYS
http://poj.org/problem?id=2318
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 19301 | Accepted: 9106 |
Description
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 is rebellious and obeys his parents by simply throwing his toys into the box. All the toys get mixed up, and it is impossible for John to find his favorite toys.
John's parents came up with the following idea. They put cardboard partitions into the box. Even if John keeps throwing his toys into the box, at least toys that get thrown into different bins stay separated. The following diagram shows a top view of an example toy box.
For this problem, you are asked to determine how many toys fall into each partition as John throws them into the toy box.
Input
Output
Sample Input
5 6 0 10 60 0
3 1
4 3
6 8
10 10
15 30
1 5
2 1
2 8
5 5
40 10
7 9
4 10 0 10 100 0
20 20
40 40
60 60
80 80
5 10
15 10
25 10
35 10
45 10
55 10
65 10
75 10
85 10
95 10
0
Sample Output
0: 2
1: 1
2: 1
3: 1
4: 0
5: 1 0: 2
1: 2
2: 2
3: 2
4: 2
Hint
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<queue>
#include<vector>
using namespace std; struct Point{
double x,y;
}; struct Line{
Point a,b;
}line[]; double Cross(Point A,Point B,Point C){
return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);
} int ans[]; int main(){
int n,m;
double x1,y1,x2,y2,x,y;
int co=;
while(cin>>n){
if(!n) break;
memset(ans,,sizeof(ans));
cin>>m>>x1>>y1>>x2>>y2;
Point a,b;
if(co) cout<<endl;
for(int i=;i<n;i++){
cin>>x>>y;
a.x=x,a.y=y1;
b.x=y,b.y=y2;
line[i].a=a,line[i].b=b;
}
int j;
for(int i=;i<m;i++){
cin>>x>>y;
a.x=x,a.y=y;
for(j=;j<n;j++){
if(Cross(line[j].b,line[j].a,a)>){
break;
}
}
ans[j]++;
}
for(int i=;i<=n;i++){
cout<<i<<": "<<ans[i]<<endl;
}
co++;
}
}
TOYS(叉积)的更多相关文章
- POJ2318 TOYS[叉积 二分]
TOYS Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 14433 Accepted: 6998 Description ...
- POJ 2318 TOYS (叉积+二分)
题目: Description Calculate the number of toys that land in each bin of a partitioned toy box. Mom and ...
- 【POJ 2318】TOYS 叉积
用叉积判断左右 快速读入写错了卡了3小时hhh #include<cmath> #include<cstdio> #include<cstring> #includ ...
- POJ 2318 TOYS 叉积
题目大意:给出一个长方形盒子的左上点,右下点坐标.给出n个隔板的坐标,和m个玩具的坐标,求每个区间内有多少个玩具. 题目思路:利用叉积判断玩具在隔板的左方或右方,并用二分优化查找过程. #includ ...
- 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 prob ...
- POJ 2318 TOYS (计算几何,叉积判断)
TOYS Time Limit: 2000MS Memory Limit: 65536K Total Submissions: 8661 Accepted: 4114 Description ...
- POJ2318:TOYS(叉积判断点和线段的关系+二分)&&POJ2398Toy Storage
题目:http://poj.org/problem?id=2318 题意: 给定一个如上的长方形箱子,中间有n条线段,将其分为n+1个区域,给定m个玩具的坐标,统计每个区域中的玩具个数.(其中这些线段 ...
- POJ-2318 TOYS,暴力+叉积判断!
TOYS 2页的提交记录终于搞明白了. 题意:一个盒子由n块挡板分成n+1块区 ...
- POJ 2318 TOYS(叉积+二分)
题目传送门:POJ 2318 TOYS Description Calculate the number of toys that land in each bin of a partitioned ...
- POJ 2318 TOYS【叉积+二分】
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 ...
随机推荐
- Mac网络连接问题
场景:同一个网络,其他电脑和手机可以访问远程网络端口为443的网站,如博客园,唯独我的电脑不能访问 解决方法:查看路由器的子网掩码和DNS地址,将IP设置为手动,输入ip.子网掩码和DNS即可
- jquery粘贴操作
今天忘记记录一个点了,关于input字体默认浅色,聚焦变深的问题. 图一,默认浅色 图二,聚焦出现下拉框“最近搜索”记录,点击“程序员” 图三,input值变为“程序员”,颜色没有变深(复制粘贴也不变 ...
- 搭建GlusterFS文件系统
(1)环境准备 创建两个虚拟机配置如下 把仅主机第二张网卡配置如下: GlusterFS1 GlusterFS2 上传文件到opt目录下 文件内容如下 (2)GlusterFS安装配置 1.安装Glu ...
- JedisCluster中应用的Apache Commons Pool对象池技术
对象池技术在服务器开发上应用广泛.在各种对象池的实现中,尤其以数据库的连接池最为明显,可以说是每个服务器必须实现的部分. apache common pool 官方文档可以参考:https://c ...
- Storm集成Kafka的Trident实现
原本打算将storm直接与flume直连,发现相应组件支持比较弱,topology任务对应的supervisor也不一定在哪个节点上,只能采用统一的分布式消息服务Kafka. 原本打算将结构设 ...
- php 编程笔记分享 - 非常实用
php opendir()列出目录下所有文件的两个实例 php opendir()函数讲解及遍历目录实例 php move_uploaded_file()上传文件实例及遇到问题的解决方法 php使用m ...
- Noip往年题目整理
Noip往年题目整理 张炳琪 一.历年题目 按时间倒序排序 年份 T1知识点 T2知识点 T3知识点 得分 总体 2016day1 模拟 Lca,树上差分 期望dp 144 挺难的一套题目,偏思维难度 ...
- Spring 学习之AOP
1. 走进面前切面编程 编程范式: 面向过程编程,c语言: 面向对象编程:c++,java,c#; 函数式编程: 事件驱动编程: 面向切面编程: AOP是一种编程范式,不是编程语言:解决特定问题,不能 ...
- c++官方文档-命名空间
#include<stdio.h> #include<iostream> #include<queue> #include<map> #include& ...
- MVC,MVP 和 MVVM 模式如何选择?
转摘:http://www.linuxidc.com/Linux/2015-10/124622.htm 前言 做客户端开发.前端开发对MVC.MVP.MVVM这些名词不了解也应该大致听过,都是为了解决 ...