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 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.
 
计算几何的入门题。题目的意思是在一个矩形箱中放物品,其中,会用n块板隔成n+1个空间,求每个物体分别在哪个空间内。
那么,由于n比较小,直接模拟就是了,模拟的核心就是判断物品与线段(相当于直线)的位置,到底在左边还是右边.这个怎么判断?用叉积就是了。
但是尴尬的是,TLE了,没办法,多组数据。但是,我们能很快想到,二分,这是有序的呀。
 #include<cmath>
 #include<cstring>
 #include<cstdio>
 #include<algorithm>
 #define LL long long
 using namespace std;
 ];
 ],D[];
 int read(){
     ,f=; char ch=getchar();
     '){if (ch=='-') f=-f; ch=getchar();}
     +ch-',ch=getchar();
     return x*f;
 }
 int cross(int x_0,int y_0,int x_1,int y_1){
     return x_0*y_1-x_1*y_0;
 }
 int main(){
     n=read();
     while (n){
         m=read(),U[]=D[]=read(),uy=read(),U[n+]=D[n+]=read(),dy=read();
         ; i<=n; i++) U[i]=read(),D[i]=read();
         memset(ans,,sizeof ans);
         ; i<=m; i++){
             int x=read(),y=read();
             ,R=n+,mid,pos;
             while (L<=R){
                 mid=(L+R)>>;
                 ) R=mid-,pos=mid; ;
             }
             ans[pos-]++;
         }
         ; i<=n; i++) printf("%d: %d\n",i,ans[i]);
         putchar('\n');
         n=read();
     }
     ;
 } 
 

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. HDU 5976 Detachment(拆分)

    HDU 5976 Detachment(拆分) 00 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)   Problem D ...

  2. 小程序之image图片实现宽度100%,高度自适应

    哇 今天搞了半天  图片一直变形啊啊啊啊 宽度100%   高度给100%   给auto   完全不管用啊啊啊啊 然后最后最终!!!! 首先我们要给我们的图片一个100%的宽度!让它自适应!! .g ...

  3. Pandas 基础(3) - 生成 Dataframe 的几种方式

    这一节想总结一下 生成 Dataframe 的几种方式: CSV Excel python dictionary List of tuples List of dictionary 下面分别一一介绍具 ...

  4. 【二十二】mysqli事务处理与预处理总结

    事务处理 事务基本原理 如果不开启事务,执行一条sql,马上会持久化数据.可见:默认的mysql对sql语句的执行是自动提交的! 如果开启了事务,就是关闭了自动提交的功能,改成了commit执行自动提 ...

  5. Windows android appium python3 环境搭建

    安装nodejs https://www.cnblogs.com/sea-stream/p/10520624.html java 环境变量配置: https://www.cnblogs.com/sea ...

  6. idea关于热部署插件JRebel的使用教程

    1. idea安装JRebel热部署插件 在1处输入jrebel然后在3处点击install安装按钮就可以了,安装好以后如下图: 2. 激活JRebel help -> JRebel -> ...

  7. 力扣(LeetCode)1.两数之和

    给定一个整数数组 nums 和一个目标值 target,请你在该数组中找出和为目标值的那 两个 整数,并返回他们的数组下标. 你可以假设每种输入只会对应一个答案.但是,你不能重复利用这个数组中同样的元 ...

  8. js,java时间处理

    1.JS获取时间格式为“yyyy-MM-dd HH:mm:ss”的字符串 function getTimeStr(){ var myDate = new Date(); var year = myDa ...

  9. C# : 泛型的继承关系实现的一个可以存放不同数据类型的链表

    以下定义的是一个链表结点类型: internal sealed class Node<T> { public T m_data; public Node<T> m_next; ...

  10. Word Ladder(双向BFS)

    2018-10-02 23:46:38 问题描述: 问题求解: 显然是个解空间遍历问题,每次修改其中一位,由于步长是1,所以可以使用BFS进行解空间的遍历.