zoukankan      html  css  js  c++  java
  • bzoj1338: Pku1981 Circle and Points单位圆覆盖

    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:

     1 #include<cstdio>
     2 #include<cstring>
     3 #include<iostream>
     4 #include<algorithm>
     5 #include<cmath>
     6 #define inf 2e9
     7 #define maxn 305
     8 #define pi acos(-1)
     9 using namespace std;
    10 int n,top,ans;
    11 const double eps=1e-9;
    12 struct fuck{double x,y;}p[maxn];
    13 struct fuckpp{double ang;int x;}a[maxn];
    14 double dis(fuck x,fuck y){return sqrt((x.x-y.x)*(x.x-y.x)+(x.y-y.y)*(x.y-y.y));}
    15 double xl(fuck a,fuck b){
    16     double ki=atan(fabs((b.y-a.y)/(b.x-a.x)));
    17     if(b.y-a.y>0){
    18         if(b.x-a.x<0)
    19         ki=pi-ki;
    20     }
    21     else{
    22         if(b.x<a.x) ki+=pi;
    23         else ki=2*pi-ki;
    24     }
    25     return ki;
    26 }
    27 bool comp(fuckpp x,fuckpp y){return x.ang<y.ang;}
    28 int main(){
    29     while(1){
    30         scanf("%d",&n);if(n==0)break;
    31         for(int i=1;i<=n;i++)scanf("%lf %lf",&p[i].x,&p[i].y);
    32         ans=1;
    33         for(int i=1;i<=n;i++){
    34             top=0;
    35             for(int j=1;j<=n;j++){
    36                 if(i==j)continue;
    37                 double k=dis(p[i],p[j]);
    38                 if(k>2.0)continue;
    39                 double an=acos(k/2.0),ng=xl(p[i],p[j]);
    40                 a[++top].ang=ng-an;a[top].x=1;
    41                 a[++top].ang=ng+an;a[top].x=-1;
    42             }
    43             sort(a+1,a+top+1,comp);
    44             int num=1;
    45             for(int i=1;i<=top;i++){
    46                 num+=a[i].x;ans=max(ans,num);
    47             }
    48         }
    49         printf("%d
    ",ans);
    50     }
    51     return 0;
    52 }
    View Code
  • 相关阅读:
    0108 创建表约束
    Mybatis 将数据库中查出的记录,一对多返回,即分组,然后返回每个组的所有数据
    SQL主表、从表
    MySQL中添加、删除字段,使用SQL语句操作
    git 将远程工作分支合并到本地dev分支
    MySQL inner join 和 left join 的区别
    Mysql union 和 order by 同时使用需要注意的问题
    The used SELECT statements have a different number of columns
    Every derived table must have its own alias(MySQL报错:每个派生表都必须有自己的别名)
    MySQL 日期格式化及字符串、date、毫秒互相转化
  • 原文地址:https://www.cnblogs.com/longshengblog/p/5736783.html
Copyright © 2011-2022 走看看