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: Wrong Answer)

#include <iostream>
#include <string>
#include <sstream>
#include <cmath>
#include <vector>

using namespace std;

typedef double ld;

struct Point {
    ld x, y;
    Point(ld x, ld y) : x(x), y(y) {}
    string to_s() {
        ostringstream os;
        os << x << ',' << y;
        return os.str();
    }
    Point operator+(const Point& o) const {
        return Point(x+o.x, y+o.y);
    }
    Point operator/(const int d) const {
        return Point(x/d, y/d);
    }
};

struct Circle {
    ld x, y;
    ld r;
    Circle(ld x=0, ld y=0, ld r=1) : x(x), y(y), r(r) {}
    const static ld eps = 1e-8;
    bool isIntersect(const Circle& o) const {
        ld dx = o.x - x;
        ld dy = o.y - y;
        return (r + o.r) * (r + o.r) > (dx*dx + dy*dy);
    }
    pair<Point, Point> intersectionPoint(const Circle& o) const {
        ld dx = o.x - x;
        ld dy = o.y - y;
        ld d = sqrt(dx*dx + dy*dy);

        ld slope = dy / dx;
        ld theta = atan(slope);
        ld theta_p = acos((r*r - o.r*o.r + d*d) / (2*r*d));

        return make_pair(
                Point(x + r*cos(theta - theta_p), y + r*sin(theta - theta_p)),
                Point(x + r*cos(theta + theta_p), y + r*sin(theta + theta_p)));
    }
    bool contains(ld px, ld py) const {
        return r * r + eps > (px - x) * (px - x) + (py - y) * (py - y);
    }
    string to_s() {
        ostringstream os;
        os << x << ',' << y << ':' << r << endl;
        return os.str();
    }
};

int countContainingPoints(const Circle& c, const vector<Circle>& ps) {
    int cnt = 0;
    for (int i = 0; i < ps.size(); i++) 
        if (c.contains(ps[i].x, ps[i].y)) 
            cnt++;
    return cnt;
}

Point midPoint(pair<Point, Point>& pp) {
    return (pp.first + pp.second) / 2;
}

int N;
int main() {
    vector<Circle> cs;
    while (cin >> N, N) {
        cs.clear(); cs.resize(N);
        for (int i = 0; i < N; i++) cin >> cs[i].x >> cs[i].y;

        vector<Point> ips; // Intersection PointS
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                if (i == j) continue;
                if (!cs[i].isIntersect(cs[j])) continue;
                pair<Point, Point> ip = cs[i].intersectionPoint(cs[j]);
                ips.push_back(ip.first);
                ips.push_back(ip.second);
                //ips.push_back(midPoint(ip));
            }
        }

        int maxPoint = 0;
        for (int i = 0; i < ips.size(); i++) {
            maxPoint = max(maxPoint, countContainingPoints(Circle(ips[i].x, ips[i].y, 1), cs));
        }
        cout << maxPoint << endl;
    }
    return 0;
}