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
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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-9;
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>0){
        if(b.x-a.x<0)
        ki=pi-ki;
    }
    else{
        if(b.x<a.x) ki+=pi;
        else ki=2*pi-ki;
    }
    return ki;
}
bool comp(fuckpp x,fuckpp y){return x.ang<y.ang;}
int main(){
    while(1){
        scanf("%d",&n);if(n==0)break;
        for(int i=1;i<=n;i++)scanf("%lf %lf",&p[i].x,&p[i].y);
        ans=1;
        for(int i=1;i<=n;i++){
            top=0;
            for(int j=1;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=1;
                a[++top].ang=ng+an;a[top].x=-1;
            }
            sort(a+1,a+top+1,comp);
            int num=1;
            for(int i=1;i<=top;i++){
                num+=a[i].x;ans=max(ans,num);
            }
        }
        printf("%d\n",ans);
    }
    return 0;
}