Catching Fish

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1217    Accepted Submission(s): 466

Problem Description
Ignatius
likes catching fish very much. He has a fishnet whose shape is a circle
of radius one. Now he is about to use his fishnet to catch fish. All
the fish are in the lake, and we assume all the fish will not move when
Ignatius catching them. Now Ignatius wants to know how many fish he can
catch by using his fishnet once. We assume that the fish can be regard
as a point. So now the problem is how many points can be enclosed by a
circle of radius one.

Note: If a fish is just on the border of the fishnet, it is also caught by Ignatius.

 
Input
The
input contains several test cases. The first line of the input is a
single integer T which is the number of test cases. T test cases follow.
Each
test case starts with a positive integer N(1<=N<=300) which
indicate the number of fish in the lake. Then N lines follow. Each line
contains two floating-point number X and Y (0.0<=X,Y<=10.0). You
may assume no two fish will at the same point, and no two fish are
closer than 0.0001, no two fish in a test case are approximately at a
distance of 2.0. In other words, if the distance between the fish and
the centre of the fishnet is smaller 1.0001, we say the fish is also
caught.
 
Output
For each test case, you should output the maximum number of fish Ignatius can catch by using his fishnet once.
 
Sample Input
4
3
6.47634 7.69628
5.16828 4.79915
6.69533 6.20378
6
7.15296 4.08328
6.50827 2.69466
5.91219 3.86661
5.29853 4.16097
6.10838 3.46039
6.34060 2.41599
8
7.90650 4.01746
4.10998 4.18354
4.67289 4.01887
6.33885 4.28388
4.98106 3.82728
5.12379 5.16473
7.84664 4.67693
4.02776 3.87990
20
6.65128 5.47490
6.42743 6.26189
6.35864 4.61611
6.59020 4.54228
4.43967 5.70059
4.38226 5.70536
5.50755 6.18163
7.41971 6.13668
6.71936 3.04496
5.61832 4.23857
5.99424 4.29328
5.60961 4.32998
6.82242 5.79683
5.44693 3.82724
6.70906 3.65736
7.89087 5.68000
6.23300 4.59530
5.92401 4.92329
6.24168 3.81389
6.22671 3.62210
 
Sample Output
2
5
5
11
 
Author
Ignatius.L
 
Recommend
We have carefully selected several similar problems for you:  1056 1079 1071 1099 1066
 
 
枚举两点,计算已知过两点和半径的圆心。
O(n^3)

hduoj 1077 Catching Fish 求单位圆最多覆盖点个数的更多相关文章

  1. HDU 1077 Catching Fish(用单位圆尽可能围住多的点)

    传送门:http://acm.hdu.edu.cn/showproblem.php?pid=1077 Catching Fish Time Limit: 10000/5000 MS (Java/Oth ...

  2. (水题)HDU - 1077 - Catching Fish - 计算几何

    http://acm.hdu.edu.cn/showproblem.php?pid=1077 很明显这样的圆,必定有两个点在边界上.n平方枚举圆,再n立方暴力判断.由于没有给T,所以不知道行不行.

  3. hdu4106 区间k覆盖问题(连续m个数,最多选k个数) 最小费用最大流 建图巧妙

    /** 题目:hdu4106 区间k覆盖问题(连续m个数,最多选k个数) 最小费用最大流 建图巧妙 链接:http://acm.hdu.edu.cn/showproblem.php?pid=4106 ...

  4. Catching Fish[HDU1077]

    Catching Fish Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)To ...

  5. TZOJ 2392 Bounding box(正n边形三点求最小矩形覆盖面积)

    描述 The Archeologists of the Current Millenium (ACM) now and then discover ancient artifacts located ...

  6. [ACM_暴力] 最多交换k个数的顺序,求a[i]的最大连续和

    /* http://codeforces.com/contest/426/problem/C 最多交换k个数的顺序,求a[i]的最大连续和 爆解 思路:Lets backtrack interval ...

  7. [hdu5251]矩形面积 旋转卡壳求最小矩形覆盖

    旋转卡壳求最小矩形覆盖的模板题. 因为最小矩形必定与凸包的一条边平行,则枚举凸包的边,通过旋转卡壳的思想去找到其他3个点,构成矩形,求出最小面积即可. #include<cstdio> # ...

  8. 容斥原理应用(求1~r中有多少个数与n互素)

    问题:求1~r中有多少个数与n互素. 对于这个问题由容斥原理,我们有3种写法,其实效率差不多.分别是:dfs,队列数组,位运算. 先说说位运算吧: 用二进制1,0来表示第几个素因子是否被用到,如m=3 ...

  9. Acdream1084 寒假安排 求n!中v因子个数

    题目链接:pid=1084">点击打开链接 寒假安排 Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 128000/64000 ...

随机推荐

  1. SVN安装手册

    转:http://www.cnblogs.com/newsea/archive/2012/04/28/2474818.html 常用工具2012-04-04 21:24        by       ...

  2. Linux和Windows互传文件命令(转)

    sftp>lls 显示当前目录内容 sftp>lcd g:\ 进入本地g盘 sftp>put phpMyAdmin.zip 将本地g盘下的phpMyAdmin.zip上传到/home ...

  3. [置顶] String StringBuffer StringBuilder的区别剖析

    这是一道很常见的面试题目,至少我遇到过String/StringBuffer/StringBuilder的区别:String是不可变的对象(final)类型,每一次对String对象的更改均是生成一个 ...

  4. &&与&

    if((2>1)&&(4>3))System.out.printf("两边都是true"); else   System.out.println(&qu ...

  5. Properties文件,Data,Calendar类的使用

    package cn.hncu.day9; import java.io.FileInputStream;import java.io.FileNotFoundException;import jav ...

  6. xshell连接本地Linux虚拟机!

    终端输入ifconfig获取本地虚拟机的IP地址; 安装openssh-server sudo apt-get install openssh-server 查看server是否启动: ps -ef ...

  7. Eclipse - 添加 PyDev 插件

    1. 安装PyDev插件 启用Eclipse.在Help菜单中,选择Install New Software···, 然后点击Add按钮.在Location中输入:http://pydev.org/u ...

  8. Linux删除文件Argument list too long问题的解决方案

    方法一:使用find find . -name 文件 | xargs rm -f 但文件数量过多,find命令也会出现问题: -bash: /bin/find: Argument list too l ...

  9. Bash从路径中获取文件名

    #!/bin/bash basename /etc/hosts

  10. linux下实现redis共享session的tomcat集群

    为了实现主域名与子域名的下不同的产品间一次登录,到处访问的效果,因此采用rediss实现tomcat的集群效果.基于redis能够异步讲缓存内容固化到磁盘上,从而当服务器意外重启后,仍然能够让sess ...