题目链接:http://www.lydsy.com/JudgeOnline/problem.php?id=1338

1338: Pku1981 Circle and Points单位圆覆盖

Time Limit: 3 Sec  Memory Limit: 162 MB
Submit: 190  Solved: 79
[Submit][Status][Discuss]

Description

You are given N points in the xy-plane. You have a circle of radius one and move it on the xy-plane, so as to enclose as many of the points as possible. Find how many points can be simultaneously enclosed at the maximum. A point is considered enclosed by a circle when it is inside or on the circle. Fig 1. Circle and Points 平面上N个点,用一个半径R的圆去覆盖,最多能覆盖多少个点?

Input

The input consists of a series of data sets, followed by a single line only containing a single character '0', which indicates the end of the input. Each data set begins with a line containing an integer N, which indicates the number of points in the data set. It is followed by N lines describing the coordinates of the points. Each of the N lines has two decimal fractions X and Y, describing the x- and y-coordinates of a point, respectively. They are given with five digits after the decimal point. You may assume 1 <= N <= 300, 0.0 <= X <= 10.0, and 0.0 <= Y <= 10.0. No two points are closer than 0.0001. No two points in a data set are approximately at a distance of 2.0. More precisely, for any two points in a data set, the distance d between the two never satisfies 1.9999 <= d <= 2.0001. Finally, no three points in a data set are simultaneously very close to a single circle of radius one. More precisely, let P1, P2, and P3 be any three points in a data set, and d1, d2, and d3 the distances from an arbitrarily selected point in the xy-plane to each of them respectively. Then it never simultaneously holds that 0.9999 <= di <= 1.0001 (i = 1, 2, 3).

Output

For each data set, print a single line containing the maximum number of points in the data set that can be simultaneously enclosed by a circle of radius one. No other characters including leading and trailing spaces should be printed.

Sample Input

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
0

Sample Output

2
5
5
11

HINT

单位圆覆盖。

n^3算法:考虑覆盖最多的圆,一定有2个点在圆上,所以n^2枚举,o(n)计算覆盖多少点即可。

n^2logn算法:考虑以每个点为圆心做单位圆 ,当一段弧被另一圆覆盖时,表示在这个弧上的点做圆,可覆盖两个点。所以枚举一个点做圆心,再1~n枚举计算交弧的级角区间,sort一下,最大覆盖次数即为答案。

n^2logn代码:

 #include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#define inf 2e9
#define maxn 305
#define pi acos(-1)
using namespace std;
int n,top,ans;
const double eps=1e-;
struct fuck{double x,y;}p[maxn];
struct fuckpp{double ang;int x;}a[maxn];
double dis(fuck x,fuck y){return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));}
double xl(fuck a,fuck b){
double ki=atan(fabs((b.y-a.y)/(b.x-a.x)));
if(b.y-a.y>){
if(b.x-a.x<)
ki=pi-ki;
}
else{
if(b.x<a.x) ki+=pi;
else ki=*pi-ki;
}
return ki;
}
bool comp(fuckpp x,fuckpp y){return x.ang<y.ang;}
int main(){
while(){
scanf("%d",&n);if(n==)break;
for(int i=;i<=n;i++)scanf("%lf %lf",&p[i].x,&p[i].y);
ans=;
for(int i=;i<=n;i++){
top=;
for(int j=;j<=n;j++){
if(i==j)continue;
double k=dis(p[i],p[j]);
if(k>2.0)continue;
double an=acos(k/2.0),ng=xl(p[i],p[j]);
a[++top].ang=ng-an;a[top].x=;
a[++top].ang=ng+an;a[top].x=-;
}
sort(a+,a+top+,comp);
int num=;
for(int i=;i<=top;i++){
num+=a[i].x;ans=max(ans,num);
}
}
printf("%d\n",ans);
}
return ;
}

bzoj1338: Pku1981 Circle and Points单位圆覆盖的更多相关文章

  1. poj1981 Circle and Points 单位圆覆盖问题

    转载请注明出处: http://www.cnblogs.com/fraud/          ——by fraud Circle and Points Time Limit: 5000MS   Me ...

  2. POJ-1981 Circle and Points 单位圆覆盖

    题目链接:http://poj.org/problem?id=1981 容易想到直接枚举两个点,然后确定一个圆来枚举,算法复杂度O(n^3). 这题还有O(n^2*lg n)的算法.将每个点扩展为单位 ...

  3. poj1981Circle and Points(单位圆覆盖最多的点)

    链接 O(n^3)的做法: 枚举任意两点为弦的圆,然后再枚举其它点是否在圆内. 用到了两个函数 atan2反正切函数,据说可以很好的避免一些特殊情况 #include <iostream> ...

  4. poj 1981 Circle and Points

    Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 8131   Accepted: 2899 ...

  5. poj1981 Circle and Points

    地址:http://poj.org/problem?id=1981 题目: Circle and Points Time Limit: 5000MS   Memory Limit: 30000K To ...

  6. poj 1981(单位圆覆盖最多点问题模板)

    Circle and Points Time Limit: 5000MS   Memory Limit: 30000K Total Submissions: 7327   Accepted: 2651 ...

  7. 【POJ 1981 】Circle and Points

    当两个点距离小于直径时,由它们为弦确定的一个单位圆(虽然有两个圆,但是想一想知道只算一个就可以)来计算覆盖多少点. #include <cstdio> #include <cmath ...

  8. hdu 1077(单位圆覆盖问题)

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

  9. Codeforces 1036E Covered Points (线段覆盖的整点数)【计算几何】

    <题目链接> <转载于 >>>  > 题目大意: 在二维平面上给出n条不共线的线段(线段端点是整数),问这些线段总共覆盖到了多少个整数点. 解题分析: 用GC ...

随机推荐

  1. 【LeeetCode】4. Median of Two Sorted Arrays

    There are two sorted arrays nums1 and nums2 of size m and n respectively. Find the median of the two ...

  2. W: http://archive.ubuntukylin.com:10006/ubuntukylin/dists/xenial/InRelease: Signature by key 6CE35A4EBAB676094476BE7CD259B7555E1D3C58 uses weak digest algorithm (SHA1)

    新装ubuntukylin 16.04,sudo apt-get update 时遇到如下问题:   Hit:1 http://archive.ubuntukylin.com:10006/ubuntu ...

  3. IIS8中使用OpenSSL来创建CA并且签发SSL证书

    前言 [转载]http://alvinhu.com/blog/2013/06/12/creating-a-certificate-authority-and-signing-the-ssl-certi ...

  4. layoutSubviews在什么情况下会被调用

    layoutSubviews在以下情况下会被调用: 1.init初始化不会触发layoutSubviews. 2.addSubview会触发layoutSubviews. 3.设置view的Frame ...

  5. HDU 4403 A very hard Aoshu problem

    暴力$dfs$. 先看数据范围,字符串最长只有$15$,也就是说枚举每个字符后面是否放置“$+$”号的复杂度为${2^{15}}$. 每次枚举到一种情况,看哪些位置能放“$=$”号,每个位置都试一下, ...

  6. 跑github上的Symfony项目遇到的问题

    Loading composer repositories with package information Installing dependencies (including require-de ...

  7. dplyr 数据操作 常用函数(1)

    上面介绍完dplyr中,几个主要的操作函数后,我们再进一步了解dplyr中那些函数可能我们会经常要用到. 这里主要根据dplyr包作者的书籍目录来把它列出来. 1.add_rownames 添加行名称 ...

  8. Java NIO Path接口和Files类配合操作文件

    Java NIO Path接口和Files类配合操作文件 @author ixenos Path接口 1.Path表示的是一个目录名序列,其后还可以跟着一个文件名,路径中第一个部件是根部件时就是绝对路 ...

  9. linux ll命令参数的详解

    用法:ls [选项]... [文件]... 列出 FILE 的信息(默认为当前目录). 如果不指定-cftuvSUX 或--sort 选项,则根据字母大小排序. 长选项必须使用的参数对于短选项时也是必 ...

  10. H264的coded_block_pattern编码块模式

    1 词汇约定 CodedBlockPatternLuma:一个宏块的亮度分量的coded_block_pattern CodedBlockPatternChroma:一个宏块的色度分量的coded_b ...