TOYS

http://poj.org/problem?id=2318

Time Limit: 2000MS   Memory Limit: 65536K
Total Submissions: 19301   Accepted: 9106

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 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

The input file contains one or more problems. The first line of a problem consists of six integers, n m x1 y1 x2 y2. The number of cardboard partitions is n (0 < n <= 5000) and the number of toys is m (0 < m <= 5000). The coordinates of the upper-left corner and the lower-right corner of the box are (x1,y1) and (x2,y2), respectively. The following n lines contain two integers per line, Ui Li, indicating that the ends of the i-th cardboard partition is at the coordinates (Ui,y1) and (Li,y2). You may assume that the cardboard partitions do not intersect each other and that they are specified in sorted order from left to right. The next m lines contain two integers per line, Xj Yj specifying where the j-th toy has landed in the box. The order of the toy locations is random. You may assume that no toy will land exactly on a cardboard partition or outside the boundary of the box. The input is terminated by a line consisting of a single 0.

Output

The output for each problem will be one line for each separate bin in the toy box. For each bin, print its bin number, followed by a colon and one space, followed by the number of toys thrown into that bin. Bins are numbered from 0 (the leftmost bin) to n (the rightmost bin). Separate the output of different problems by a single blank line.

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

As the example illustrates, toys that fall on the boundary of the box are "in" the box.
 
基础几何题,懒得写二分,直接暴力过了= =
 #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

  1. POJ2318 TOYS[叉积 二分]

    TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 14433   Accepted: 6998 Description ...

  2. 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 ...

  3. 【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 ...

  4. 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 ...

  5. 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 ...

  6. POJ 2318 TOYS (计算几何,叉积判断)

    TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 8661   Accepted: 4114 Description ...

  7. 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 ...

  8. 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 ...

  9. 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 ...

  10. 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

  1. h3c端口汇聚的几种情况-S5500-S5048

    S5500和S5048E聚合的几种情况,分享下:1.S5048E用手工的话,S5500用静态方式2.S5048E用静态的话,S5500用动态方式3.S5048E和S5048E聚合的话,用相同方式即可. ...

  2. 阿里云启用IPV6

    ping过别人的IPv6网址之后,可以确定,局域网是不支持IPv6的.所以要使用隧道技术建立两台机器之间的IPv6连接 1.发现测试用服务器上没有IPv6地址.所以测试服务器的内核应该是没有IPv6模 ...

  3. [UE4]多播代理

    1. 第一种 DECLARE_DYNAMIC_MULTICAST_DELEGATE_TwoParams(FReceiveDelegateEvent, FString, Value1, FString, ...

  4. iframe引入百度地图显示企业位置

    步骤一:打开下面这个地址:http://api.map.baidu.com/lbsapi/creatmap/index.html     步骤二:定位中心点     在打开的页面左侧,输入企业的详细地 ...

  5. html5 如何实现客户端验证上传文件的大小

    在HTML 5中,现在可以在客户端进行文件上传时的校验了,比如用户选择文件后,可以 马上校验文件的大小和属性等.本文章向码农介绍html5 如何实现客户端验证上传文件的大小,感兴趣的码农可以参考一下. ...

  6. 不设目标也能通关「马里奥」的AI算法,全靠好奇心学习

    在强化学习中,设计密集.定义良好的外部奖励是很困难的,并且通常不可扩展.通常增加内部奖励可以作为对此限制的补偿,OpenAI.CMU 在本研究中更近一步,提出了完全靠内部奖励即好奇心来训练智能体的方法 ...

  7. Linux火焰图-centos

    centos7.5mini安装 yum install -y yum-utils perf debuginfo-install -y perf #debuginfo-install下载了305MB的文 ...

  8. ORM( ORM查询13种方法3. 单表的双下划线的使用 4. 外键的方法 5. 多对多的方法 ,聚合,分组,F查询,Q查询,事务 )

    必知必会13条 <1> all(): 查询所有结果 <2> get(**kwargs): 返回与所给筛选条件相匹配的对象,返回结果有且只有一个,如果符合筛选条件的对象超过一个或 ...

  9. python基础补充内容

    知识内容: 1.三元运算表达式 2.python代码编写规范 3.模块导入与使用 4.python文件名 5.python脚本的"__name__"属性 6.python之禅 一. ...

  10. PHP大小写是否敏感问题的汇总

      一.大小写敏感1. 变量名区分大小写view sourceprint?     <?php    $abc = 'abcd';    echo $abc; //输出 'abcd'     e ...