hduoj 1077 Catching Fish 求单位圆最多覆盖点个数
Catching Fish
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1217 Accepted Submission(s): 466
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 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.
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
5
5
11
hduoj 1077 Catching Fish 求单位圆最多覆盖点个数的更多相关文章
- 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 ...
- (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.
- 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 ...
- Catching Fish[HDU1077]
Catching Fish Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)To ...
- 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 ...
- [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 ...
- [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> # ...
- 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 ...
- 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
- [Webpack 2] Intro to the Production Webpack Course
There are several lessons that will build on top of this project. It is a fairly standard, small web ...
- 2014-08-05 pig
Pig的数据类型能够分为两种:一种是scalar类型,包含单一的value,一种是complex类型,包含有其他的类型. 对于scalar类型: int,long,float,double,chara ...
- 使用Java辅助类(CountDownLatch、CyclicBarrier、Semaphore)并发编程
在java 1.5中,提供了一些非常有用的辅助类来帮助我们进行并发编程,比如CountDownLatch,CyclicBarrier和Semaphore,今天我们就来学习一下这三个辅助类的用法 一.C ...
- Nginx vs Apache--reference
May 14th, 2014 - By Walker Rowe https://anturis.com/blog/nginx-vs-apache/ What is the Nginx web and ...
- UDP打洞和心跳包设计
一.设备终端class DeviceClient { int deviceID; int IP; int port; char connectID[16]; time_t lastTime; stru ...
- 所有Mac用户都需要知道的9个实用终端命令行<转>
转自 http://www.macx.cn/thread-2075903-1-1.html 通常情况下,只有高端用户才会经常用到终端应用.这并不意味着命令行非常难学,有的时候命令行可以轻松.快速的解决 ...
- (转)apache的keepalive和keepalivetimeout(apache优化)
KeepAlive指的是保持连接活跃,类似于Mysql的永久连接. 如果将KeepAlive设置为On,那么来自同一客户端的请求就不需要再一次连接,避免每次请求都要新建一个连接而加重服务器的负担. ...
- 学习java随笔第十篇:java线程
线程生命周期 线程的生命周期:新建状态.准备状态.运行状态.等待/阻塞状态.死亡状态 示意图: 定义.创建及运行线程 线程: package threadrun; //定义一个实现Runnable接口 ...
- [日历] C#修改CNDate日历帮助类 (转载)
点击下载 CNDate.rar 主要功能如下 .传回公历y年m月的总天数 .根据日期值获得周一的日期 .获取农历 #region 私有方法 private static long[] lunarInf ...
- C++的MFC,与C#的.NET
转载:http://blog.sina.com.cn/s/blog_7f5bde5c0101hk5n.html 以下摘自各问答网站.博客论坛: [1]MFC早已过时,现在C++多数是用来编写底层方法而 ...