来源 poj 2318

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.

用叉积的方法判断点在向量的左右边就可以了

int direction(point p1,point p2,point p3)//p1是向量起点,p2是终点,p3是判断点,>0则在左边<0在右侧

{

return (p1.x-p3.x)(p2.y-p3.y)-(p1.y-p3.y)(p2.x-p3.x);

}

#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include <iomanip>
#include<cmath>
#include<float.h>
#include<string.h>
#include<algorithm>
#define sf scanf
#define pf printf
#define mm(x,b) memset((x),(b),sizeof(x))
#include<vector>
#include<queue>
#include<stack>
#include<map>
#define rep(i,a,n) for (int i=a;i<n;i++)
#define per(i,a,n) for (int i=a;i>=n;i--)
typedef long long ll;
typedef long double ld;
typedef double db;
const ll mod=1e9+100;
const db e=exp(1);
const db eps=1e-8;
using namespace std;
const double pi=acos(-1.0);
const int inf=0xfffffff;
struct point
{
int x,y;
point(int a=0,int b=0)
{
x=a;y=b;
}
}t;
struct xl
{
point p1,p2;//p1是下面的点
int sum;
xl(int x=0,int y=0,int a=0,int b=0)
{
p1=point(x,y);
p2=point(a,b);sum=0;
}
}a[5005];
//bool cmp(xl a,xl b)
//{
// return a.p1.x<b.p1.x;
//}
int direction(point p1,point p2,point p3)//p1是向量起点,p2是终点,p3是判断点,》0则在左边<0在右侧
{
return (p1.x-p3.x)*(p2.y-p3.y)-(p1.y-p3.y)*(p2.x-p3.x);
}
int main()
{
int n,m,X1,X2,Y1,Y2,x,y;
int jud=1;
while(1)
{
cin>>n;
if(!n)
return 0;
if(!jud)
pf("\n");
cin>>m>>X1>>Y1>>X2>>Y2;
rep(i,0,n)
{
cin>>x>>y;
a[i]=xl(y,Y2,x,Y1);
}
a[n]=xl();
// sort(a,a+n,cmp);
while(m--)
{
int temp=0;
cin>>t.x>>t.y;
rep(i,0,n)
{
if(direction(a[i].p1,a[i].p2,t)>0)
{
a[i].sum++;temp=1;
break;
}
}
if(!temp)
a[n].sum++;
}
rep(i,0,n+1)
{
pf("%d: %d\n",i,a[i].sum);
}
jud=0;
}
}

E - TOYS的更多相关文章

  1. 【POJ】2318 TOYS(计算几何基础+暴力)

    http://poj.org/problem?id=2318 第一次完全是$O(n^2)$的暴力为什么被卡了-QAQ(一定是常数太大了...) 后来排序了下点然后单调搞了搞..(然而还是可以随便造出让 ...

  2. poj 2318 TOYS (二分+叉积)

    http://poj.org/problem?id=2318 TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 101 ...

  3. POJ 2318 TOYS (计算几何,叉积判断)

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

  4. POJ 2318 TOYS && POJ 2398 Toy Storage(几何)

    2318 TOYS 2398 Toy Storage 题意 : 给你n块板的坐标,m个玩具的具体坐标,2318中板是有序的,而2398无序需要自己排序,2318要求输出的是每个区间内的玩具数,而231 ...

  5. poj 2318 TOYS

    TOYS 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. 思路:这道题很水,只是要知道会使用叉乘来表示点在线的上面还是下面: 当a ...

  6. 【POJ】2318 TOYS ——计算几何+二分

    TOYS Time Limit: 2000MS   Memory Limit: 65536K Total Submissions: 10281   Accepted: 4924 Description ...

  7. Codeforces Round #346 (Div. 2) C Tanya and Toys

    C. Tanya and Toys 题目链接http://codeforces.com/contest/659/problem/C Description In Berland recently a ...

  8. 2016NEFU集训第n+3场 G - Tanya and Toys

    Description In Berland recently a new collection of toys went on sale. This collection consists of 1 ...

  9. 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 ...

  10. POJ2318 TOYS[叉积 二分]

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

随机推荐

  1. 一篇文章让你学透Linux系统中的more命令

    Linux 下有很多实用工具可以让你在终端界面查看文本文件.其中一个就是 more. more 跟我之前另一篇文章里写到的工具 —— less 很相似.它们之间的主要不同点在于 more 只允许你向前 ...

  2. ionic tab显示到顶部去了

    添加配置调整显示位置 .config(function($stateProvider, $urlRouterProvider,$ionicConfigProvider) { $ionicConfigP ...

  3. 使用 vmstat, mpstat 和 sar 查看系统运行参数

    vmstat 字段含义 Procs r: The number of processes waiting for run time. 等待执行的进程, 数字越大意味着CPU越忙, 果该数字经常大于逻辑 ...

  4. C++11中的右值引用及move语义编程

    C++0x中加入了右值引用,和move函数.右值引用出现之前我们只能用const引用来关联临时对象(右值)(造孽的VS可以用非const引用关联临时对象,请忽略VS),所以我们不能修临时对象的内容,右 ...

  5. WCF中记录SOAP消息日志

    Windows Communication Foundation (WCF) 默认启用消息日志,WCF使用了在System.Diagnostics名称空间中定义的跟踪机制.在这个跟踪机制中,通过在配置 ...

  6. C#-MVC开发微信应用(8)--菜单管理的实现

    之前讲解了微信后台管理页面的操作来管理菜单,下面我们在简单的来看一下,代码是如何实现的. 我们要实现获取微信的菜单.创建菜单.删除菜单等操作. 01.首先定义菜单操作的接口: /// <summ ...

  7. c链表之oc AutoReleasePool

    直接贴 原文吧: http://blog.sunnyxx.com/2014/10/15/behind-autorelease/

  8. vue中单选框,利用不存在的值标示选中状态

    1.效果预览 2.index.html <!DOCTYPE html> <html lang="en"> <head> <meta cha ...

  9. docker打包centos增加中文支持

    docker打包centos增加中文支持 前言 使用的某个包的返回值,在本机测试时返回结果是中文,结果打包到docker后返回结果变英文了:猜测是系统语言的问题,进入docker测试了一下,发现果然是 ...

  10. spring batch中用到的表

    1,批量表的前缀:{prefix}来自类AbstractJdbcBatchMetadataDao中的变量DEFAULT_TABLE_PREFIX 2,{prefix}job_execution:存放j ...