Distant Galaxy

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 111    Accepted Submission(s): 53

Problem Description
You
are observing a distant galaxy using a telescope above the Astronomy
Tower, and you think that a rectangle drawn in that galaxy whose edges
are parallel to coordinate axes and contain maximum star systems on its
edges has a great deal to do with the mysteries of universe. However you
do not have the laptop with you, thus you have written the coordinates
of all star systems down on a piece of paper and decide to work out the
result later. Can you finish this task?

 
Input
There
are multiple test cases in the input file. Each test case starts with
one integer N , (1<=N<=100) , the number of star systems on the
telescope. N lines follow, each line consists of two integers: the X
and Y coordinates of the K -th planet system. The absolute value of any
coordinate is no more than 109 , and you can assume that the planets
are arbitrarily distributed in the universe.

N = 0 indicates the end of input file and should not be processed by your program.

 
Output
For each test case, output the maximum value you have found on a single line in the format as indicated in the sample output.
 
Sample Input
10
2 3
9 2
7 4
3 4
5 7
1 5
10 4
10 6
11 4
4 6
0
Sample Output
Case 1: 7
 #include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
struct T
{
int x;
int y;
bool operator <(const T& rhs) const{
return x<rhs.x;
}
}poin[];
// 具体可见大白书,51页
//y[]表示存储所有的y的值,on[]表示左边界,on2表示右边界
int y[],lef[],on[],on2[];
int t, m;
int solve()
{
sort(poin,poin+t);
sort(y,y+t);
m = unique(y,y+t) - y;//去除相同的元素,m表示去除相同的元素后,还有几个
if(m<=) return t; //如果只有两种y值表示所有点都在这两条线上,直接返回
int ans = ;
for(int a = ; a<m; a++)//枚举所有可能的上边界和下边界
for(int b = a+; b<m; b++)
{
int ymin =y[a];int ymax = y[b];
int k = ;
for(int i = ; i<t;i++)
{
if(i== || poin[i].x!=poin[i-].x) //判断前后的x值是否有相同的点
{
k++;
on[k] = on2[k] = ;
lef[k] = k== ? : lef[k-]+ on2[k-] - on[k-];//此时的lef不包括右边界在它上面的点
}
if(poin[i].y>ymin && poin[i].y<ymax) on[k]++;
if(poin[i].y>=ymin && poin[i].y<=ymax) on2[k]++;//on2 - on 表示在线上的点
}
if(k<=) return t;//这个地方表示少于两种的x值,只好返回本身了
int m = ;
for(int j = ;j<=k; j++)
{
ans = max(ans,lef[j]+on2[j]+m);// 大白书51页有详细介绍
m = max(m,on[j] - lef[j]); //
}
}
return ans;
}
int main()
{
int num = ;
while(scanf("%d",&t) && t)
{
for(int i = ;i<t; i++)
{
scanf("%d %d",&poin[i].x,&poin[i].y);
y[i] = poin[i].y;
}
printf("Case %d: %d\n",num,solve());
num++;
}
return ;
}

hdu Distant Galaxy(遥远的银河)的更多相关文章

  1. HDU 5073 Galaxy (2014 Anshan D简单数学)

    HDU 5073 Galaxy (2014 Anshan D简单数学) 题目链接http://acm.hdu.edu.cn/showproblem.php?pid=5073 Description G ...

  2. LA3695 Distant Galaxy

    Distant Galaxy https://vjudge.net/problem/UVALive-3695 You are observing a distant galaxy using a te ...

  3. 2014 Asia AnShan Regional Contest --- HDU 5073 Galaxy

    Galaxy Problem's Link:   http://acm.hdu.edu.cn/showproblem.php?pid=5073 Mean: 在一条数轴上,有n颗卫星,现在你可以改变k颗 ...

  4. hdu 5073 Galaxy(2014acm鞍山亚洲分部 C)

    主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=5073 Galaxy Time Limit: 2000/1000 MS (Java/Others)   ...

  5. hdu 5073 Galaxy(2014acm鞍山亚洲分部 D)

    主题链接:http://acm.hdu.edu.cn/showproblem.php? pid=5073 Galaxy Time Limit: 2000/1000 MS (Java/Others)   ...

  6. HDU 5073 Galaxy(2014鞍山赛区现场赛D题)

    题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=5073 解题报告:在一条直线上有n颗星星,一开始这n颗星星绕着重心转,现在我们可以把其中的任意k颗星星移 ...

  7. ACM学习历程—HDU 5073 Galaxy(数学)

    Description Good news for us: to release the financial pressure, the government started selling gala ...

  8. hdu 5073 Galaxy(2014 鞍山现场赛)

    Galaxy                                                                   Time Limit: 2000/1000 MS (J ...

  9. HDU 5073 Galaxy(Anshan 2014)(数学推导,贪婪)

    Galaxy Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 262144/262144 K (Java/Others) Total S ...

随机推荐

  1. toFixed()方法

    1.定义 toFixed() 方法可把 Number 四舍五入为指定小数位数的数字. 2.示例 Show the number 13.37 with one decimal: <script t ...

  2. Javascript中的对象(二)

    Javascript是一种基于原型的对象语言,而不是我们比较熟悉的,像C#语言基于类的面向对象的语言.在前一篇文章中,我们已经介绍了Javascript中对象定义的创建.接下来我们来介绍一下Javas ...

  3. Wishbone接口Altera JTAG UART

    某些时候,我们在使用Altera FPGA的时候,尤其是涉及SoC系统的时候,通常需要一个串口与PC交互.使用Altera的USB-Blaster免去了外接一个串口.我们可以使用下面所述的IP核通过U ...

  4. dcmtk dcmmkdir 使用帮助

    例子: dcmmkdir --recurse  --input-directory C:\TEMP DICOM --output-file C:\TEMP\DCMDIR --recurse 递归 -- ...

  5. PySpider 框架爬虫错误 HTTP 599: SSL certificate problem: unable to get local issuer certificate解决方案

    首先pyspider all启动pyspider的所有服务,然后访问http://localhost:5000创建一个爬虫任务:taobaomm,点开任务链接编辑http://localhost:50 ...

  6. linux经常使用文字处理命令总结

    linux grep命令 作用 Linux系统中grep命令是一种强大的文本搜索工具,它能使用正則表達式搜索文本,并把匹 配的行打印出来.grep全称是Global Regular Expressio ...

  7. TestNG 五 运行TestNG

    一.并行运行于超时 可以通过在suite标签中使用 parallel 属性来让测试方法运行在不同的线程中.这个属性可以带有如下这样的值: <suite name="My suite&q ...

  8. Python代码可视化

    把以下代码贴到这个网站: http://www.pythontutor.com l1 = [3, [66, 55, 44], (7, 8, 9)] l2 = list(l1) l1.append(10 ...

  9. 倍福TwinCAT(贝福Beckhoff)常见问题(FAQ)-有时候项目会无法编译,重新生成就自动卡死或者自动退出怎么办

    删除所有中文注释,有中文注释则不一定能编译成功.     更多教学视频和资料下载,欢迎关注以下信息: 我的优酷空间: http://i.youku.com/acetaohai123   我的在线论坛: ...

  10. unicode 编码在线转换工具

    字符串 unideo的16进制值