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
- Spring Cloud构建微服务架构(七)消息总线
先回顾一下,在之前的Spring Cloud Config的介绍中,我们还留了一个悬念:如何实现对配置信息的实时更新.虽然,我们已经能够通过/refresh接口和Git仓库的Web Hook来实现Gi ...
- 学习笔记之Jira
Jira | Issue & Project Tracking Software | Atlassian https://www.atlassian.com/software/jira The ...
- [原]DataGridView 回车不换行代码 AND 编辑时的字符控制
// 让 dataGridView1 在遇到回车时不响应 protected override bool ProcessCmdKey(ref Message msg, Keys keyData) { ...
- DIV+CSS如何让文字垂直居中?(转)
此篇文章转自网络,但是我忘了原文地址,如果有人知道,麻烦告知一声~ 在说到这个问题的时候,也许有人会问CSS中不是有vertical-align属性来设置垂直居中的吗?即使是某些浏览器不支持我只需做少 ...
- nginx 限流配置
上配置 http { include mime.types; default_type application/octet-stream; #log_format main '$remote_addr ...
- python simplejson and json 使用及区别
''' import simplejson as json #几个主要函数:dump.dumps.load.loads,带s跟不带s的区别: 带s的是对 字符串的处理,而不带 s的是对文件对像的处理. ...
- [UE4]C++创建对象的三种方式
#include <iostream> using namespace std; class A { private: int n; public: A(int m):n(m) { } ~ ...
- (jsp/html)网页上嵌入播放器(常用播放器代码整理) http://www.jb51.net/article/37267.htm
网页上嵌入播放器,只要在HTML上添加以上代码就OK了,下面整理了一些常用的播放器代码,总有一款适合你,感兴趣的朋友可以参考下哈,希望对你有所帮助 这个其实很简单,只要在HTML上添加以上代码就O ...
- css实现文本两行或多行文本溢出显示省略号
word-break: break-all; text-overflow: ellipsis; display: -webkit-box; /** 对象作为伸缩盒子模型显示 **/ -webkit-b ...
- ES6进一步整理
内容: 1.变量及赋值 2.函数 3.数组及json 4.字符串 5.面向对象 6.Promise 7.generator 8.模块 1.变量及赋值 (1)ES5变量定义 var: 可以重复定 ...