Problem:
Problem D: Circle and Points

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

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

Output for the Sample Input

2
5
5
11

Solution in C++ (judge verdict: Accepted)

#include <iostream>
#include <cstdio>
#include <vector>
#include <map>
#include <complex>
#include <cmath>
#include <algorithm>
using namespace std;

// * XY座標
#define X real()
#define Y imag()

// * 点の表現
typedef complex<double> P;

// * 許容する誤差ε
const double EPS = 1e-10;
// * Infinity
const double INF = 1e12;
// * 円周率
const double PI = acos(-1.0);
// * 
const P INF_P(INF,INF);

// * complex<double> の順序を定義する
namespace std {
	bool operator < (const P& a, const P& b) {
		return (real(a) != real(b))? real(a) < real(b) : imag(a) < imag(b);
	}
}

// * 点座標のデバッグ出力
void print(P p) {
	cout << "point" << p << ";" << endl;
}

// * 2乗する
double sq(double x) {
	return x * x;
}

// * 2点間の距離
double dist(P p1, P p2) {
	return abs(p1 - p2);
}

// * 2つのスカラーが等しいかどうか
bool equal(double a, double b) {
	return ( fabs( a - b ) < EPS );
}
// * 2つのベクトルが等しいかどうか
bool eq_v(P a, P b) {
	return ( equal( a.X, b.X ) && equal( a.Y, b.Y ) );
}

// * 内積 (dot product) : a・b = |a||b|cosΘ
double dot(P a, P b) {
	return real( conj(a) * b );
}

// * 外積 (cross product) : a×b = |a||b|sinΘ
double cross(P a, P b) {
	return imag( conj(a) * b ) ;
}

// ベクトル a の単位法線ベクトルを求める
vector<P> normal_unit(P a) {
	vector<P> vp;
	vp.push_back( (a * P(0,  1)) / abs(a) );
	vp.push_back( (a * P(0, -1)) / abs(a) ) ;
	return vp;
}

// 原点を軸に点 p を角度 a (ラジアン)だけ回転させた点を返す
P rot(P p, double a){
	double x = p.X * cos(a) - p.Y * sin(a);
	double y = p.X * sin(a) + p.Y * cos(a);
	return P(x,y);
}

int main(){
	int n;
	while( cin >> n , n ){
		vector<P> v;
		int ans=1;
		for(int i=0 ; i < n ; i++ ){
			double x, y;
			scanf("%lf %lf", &x, &y);
			v.push_back( P(x,y) );
		}
		for(int i=0 ; i < v.size() ; i++ ){
			for(int j=i+1 ; j < v.size() ; j++ ){
				if( i == j ) continue;
				P p1 = v[i];
				P p2 = v[j];
				double d = dist(p1,p2);
				if( d >= 2.0 ) continue;
				
				double angle = atan2( p2.Y-p1.Y , p2.X-p1.X );
				P mid = rot( p2-p1 , -angle ) / 2.0;
				double h = sqrt(1.0 - sq(d/2.0));
				P p3 = P( mid.X , h );
				P p4 = P( mid.X , -h );
				int cnt1=0, cnt2=0;
				for(int k=0 ; k < v.size() ; k++ ){
					P p = rot( v[k]-p1 , -angle );
					if( dist(p3,p) <= 1.0 ) cnt1++;
					if( dist(p4,p) <= 1.0 ) cnt2++;
				}
				ans = max( ans , max(cnt1,cnt2) );
			}
		}
		printf("%d\n", ans);
	}
}