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. POJ 3126 Prime Path(素数路径)

    POJ 3126 Prime Path(素数路径) Time Limit: 1000MS    Memory Limit: 65536K Description - 题目描述 The minister ...

  2. java static方法不能被重写@Override

    重写方法的目的是为了多态,或者说:重写是实现多态的前提,即重写是发生在继承中且是针对非static方法的. 语法上子类允许出现和父类只有方法体不一样其他都一模一样的static方法,但是在父类引用指向 ...

  3. scss切页面

    html <div class="data-list"> <div class="data-list-item"> <div cl ...

  4. 设计模式(五)Builder Pattern建造者模式

    在我们日常生活中,如构建一个飞船,一个手机,一栋建筑,都会有非常复杂的组装,这时候应该用到建造者模式 以建造一个飞船为例 案例:造小页飞船 1.飞船各部分元件 package com.littlepa ...

  5. Eclipse调试DEBUG时快速查看某个变量的值的快捷键、快速跳转到某行的快捷键

    Eclipse调试DEBUG时快速查看某个变量的值的快捷键 Ctrl + Shift + i

  6. sort-选择排序

    void sort_select(vector<int> &v) { for(int i=0;i<v.size()-1;i++) { int min=v[i]; int in ...

  7. Axure 设置条件的操作

    登录的三种场景: 1.用户名为空,只输入密码时,执行三个动作:跳出提示内容(用户名为空).光标定位在用户名的文本框中.清空密码的文本框: 2.密码为空,只输入用户名,执行两个操作:跳出提示内容(密码为 ...

  8. Go语言学习之6 反射详解

    1.反射: 定义: 反射就是程序能够在运行时检查变量和值,求出它们的类型.                   可以在运行时动态获取变量的相关信息                   Import ( ...

  9. compile FFMPEG under windows

    这个文章应该算是中文版最好的了.但是还有一些需要修正. 特别是winsock2的处理 win10 msys2 vs2015 ffmpeg3.2.2 编译 这个英文版的才是最好的 Building ff ...

  10. ultragrid

    foreach (UltraGridColumn aCol in this.ultraGrid1.DisplayLayout.Bands[0].Columns){this.ultraGrid1.Dis ...