zoukankan      html  css  js  c++  java
  • TOYS(叉积)

    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= =
     1 #include<iostream>
     2 #include<cstdio>
     3 #include<cstring>
     4 #include<cmath>
     5 #include<string>
     6 #include<algorithm>
     7 #include<queue>
     8 #include<vector>
     9 using namespace std;
    10 
    11 struct Point{
    12     double x,y;
    13 };
    14 
    15 struct Line{
    16     Point a,b;
    17 }line[5005];
    18 
    19 double Cross(Point A,Point B,Point C){
    20     return (B.x-A.x)*(C.y-A.y)-(B.y-A.y)*(C.x-A.x);
    21 }
    22 
    23 int ans[5005];
    24 
    25 int main(){
    26     int n,m;
    27     double x1,y1,x2,y2,x,y;
    28     int co=0;
    29     while(cin>>n){
    30         if(!n) break;
    31         memset(ans,0,sizeof(ans));
    32         cin>>m>>x1>>y1>>x2>>y2;
    33         Point a,b;
    34         if(co) cout<<endl;
    35         for(int i=0;i<n;i++){
    36             cin>>x>>y;
    37             a.x=x,a.y=y1;
    38             b.x=y,b.y=y2;
    39             line[i].a=a,line[i].b=b;
    40         }
    41         int j;
    42         for(int i=0;i<m;i++){
    43             cin>>x>>y;
    44             a.x=x,a.y=y;
    45             for(j=0;j<n;j++){
    46                 if(Cross(line[j].b,line[j].a,a)>0){
    47                     break;
    48                 }
    49             }
    50             ans[j]++;
    51         }
    52         for(int i=0;i<=n;i++){
    53             cout<<i<<": "<<ans[i]<<endl;
    54         }
    55         co++;
    56     }
    57 }
    View Code
  • 相关阅读:
    redux核心知识
    react性能优化要点
    react-router4的使用备注
    selenium启动Chrome配置参数问题
    Navicat15最新版本破解和破解报错总结
    Silence主题美化-部署
    vscode打开文件,中文显示乱码(已解决)
    Windows下Charles从下载安装到证书设置和浏览器抓包
    python下的selenium和chrome driver的安装
    Python 直接赋值、浅拷贝和深度拷贝解析
  • 原文地址:https://www.cnblogs.com/Fighting-sh/p/9804185.html
Copyright © 2011-2022 走看看