Boundary（计算几何）

圆上的一个点（原点）已经确定了，我们枚举一下另一个定点 P，然后枚举点 A ，此时不共线的三点已经可以确定一个圆了，换句话说，三个点就可以确定下来圆心的位置了，当点 P 和点 O 都确定下来后，枚举点 A 所得到的所有圆心的众数再加一，就是以此圆心做圆后可以经过的点的个数，利用map维护最大值就是答案了，找圆心的任务直接交给模板就好了。

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
#include<map>
using namespace std;
// `计算几何模板`
const double eps = 1e-8;
const double inf = 1e20;
const double pi = acos(-1.0);
const int maxp = 1010;
//`Compares a double to zero`
int sgn(double x){
	if(fabs(x) < eps)return 0;
	if(x < 0)return -1;
	else return 1;
}
//square of a double
inline double sqr(double x){return x*x;}
/*
 * Point
 * Point()               - Empty constructor
 * Point(double _x,double _y)  - constructor
 * input()             - double input
 * output()            - %.2f output
 * operator ==         - compares x and y
 * operator <          - compares first by x, then by y
 * operator -          - return new Point after subtracting curresponging x and y
 * operator ^          - cross product of 2d points
 * operator *          - dot product
 * len()               - gives length from origin
 * len2()              - gives square of length from origin
 * distance(Point p)   - gives distance from p
 * operator + Point b  - returns new Point after adding curresponging x and y
 * operator * double k - returns new Point after multiplieing x and y by k
 * operator / double k - returns new Point after divideing x and y by k
 * rad(Point a,Point b)- returns the angle of Point a and Point b from this Point
 * trunc(double r)     - return Point that if truncated the distance from center to r
 * rotleft()           - returns 90 degree ccw rotated point
 * rotright()          - returns 90 degree cw rotated point
 * rotate(Point p,double angle) - returns Point after rotateing the Point centering at p by angle radian ccw
 */
struct Point{
	double x,y;
	Point(){}
	Point(double _x,double _y){
		x = _x;
		y = _y;
	}
	void input(){
		scanf("%lf%lf",&x,&y);
	}
	void output(){
		printf("%.2f %.2f\n",x,y);
	}
	bool operator == (Point b)const{
		return sgn(x-b.x) == 0 && sgn(y-b.y) == 0;
	}
	bool operator < (Point b)const{
		return sgn(x-b.x)== 0?sgn(y-b.y)<0:x<b.x;
	}
	Point operator -(const Point &b)const{
		return Point(x-b.x,y-b.y);
	}
	//叉积
	double operator ^(const Point &b)const{
		return x*b.y - y*b.x;
	}
	//点积
	double operator *(const Point &b)const{
		return x*b.x + y*b.y;
	}
	//返回长度
	double len(){
		return hypot(x,y);//库函数
	}
	//返回长度的平方
	double len2(){
		return x*x + y*y;
	}
	//返回两点的距离
	double distance(Point p){
		return hypot(x-p.x,y-p.y);
	}
	Point operator +(const Point &b)const{
		return Point(x+b.x,y+b.y);
	}
	Point operator *(const double &k)const{
		return Point(x*k,y*k);
	}
	Point operator /(const double &k)const{
		return Point(x/k,y/k);
	}
	//`计算pa  和  pb 的夹角`
	//`就是求这个点看a,b 所成的夹角`
	//`测试 LightOJ1203`
	double rad(Point a,Point b){
		Point p = *this;
		return fabs(atan2( fabs((a-p)^(b-p)),(a-p)*(b-p) ));
	}
	//`化为长度为r的向量`
	Point trunc(double r){
		double l = len();
		if(!sgn(l))return *this;
		r /= l;
		return Point(x*r,y*r);
	}
	//`逆时针旋转90度`
	Point rotleft(){
		return Point(-y,x);
	}
	//`顺时针旋转90度`
	Point rotright(){
		return Point(y,-x);
	}
	//`绕着p点逆时针旋转angle`
	Point rotate(Point p,double angle){
		Point v = (*this) - p;
		double c = cos(angle), s = sin(angle);
		return Point(p.x + v.x*c - v.y*s,p.y + v.x*s + v.y*c);
	}
}p[2010];
/*
 * Stores two points
 * Line()                         - Empty constructor
 * Line(Point _s,Point _e)        - Line through _s and _e
 * operator ==                    - checks if two points are same
 * Line(Point p,double angle)     - one end p , another end at angle degree
 * Line(double a,double b,double c) - Line of equation ax + by + c = 0
 * input()                        - inputs s and e
 * adjust()                       - orders in such a way that s < e
 * length()                       - distance of se
 * angle()                        - return 0 <= angle < pi
 * relation(Point p)              - 3 if point is on line
 *                                  1 if point on the left of line
 *                                  2 if point on the right of line
 * pointonseg(double p)           - return true if point on segment
 * parallel(Line v)               - return true if they are parallel
 * segcrossseg(Line v)            - returns 0 if does not intersect
 *                                  returns 1 if non-standard intersection
 *                                  returns 2 if intersects
 * linecrossseg(Line v)           - line and seg
 * linecrossline(Line v)          - 0 if parallel
 *                                  1 if coincides
 *                                  2 if intersects
 * crosspoint(Line v)             - returns intersection point
 * dispointtoline(Point p)        - distance from point p to the line
 * dispointtoseg(Point p)         - distance from p to the segment
 * dissegtoseg(Line v)            - distance of two segment
 * lineprog(Point p)              - returns projected point p on se line
 * symmetrypoint(Point p)         - returns reflection point of p over se
 *
 */
struct Line{
	Point s,e;
	Line(){}
	Line(Point _s,Point _e){
		s = _s;
		e = _e;
	}
	bool operator ==(Line v){
		return (s == v.s)&&(e == v.e);
	}
	//`根据一个点和倾斜角angle确定直线,0<=angle<pi`
	Line(Point p,double angle){
		s = p;
		if(sgn(angle-pi/2) == 0){
			e = (s + Point(0,1));
		}
		else{
			e = (s + Point(1,tan(angle)));
		}
	}
	//ax+by+c=0
	Line(double a,double b,double c){
		if(sgn(a) == 0){
			s = Point(0,-c/b);
			e = Point(1,-c/b);
		}
		else if(sgn(b) == 0){
			s = Point(-c/a,0);
			e = Point(-c/a,1);
		}
		else{
			s = Point(0,-c/b);
			e = Point(1,(-c-a)/b);
		}
	}
	void input(){
		s.input();
		e.input();
	}
	void adjust(){
		if(e < s)swap(s,e);
	}
	//求线段长度
	double length(){
		return s.distance(e);
	}
	//`返回直线倾斜角 0<=angle<pi`
	double angle(){
		double k = atan2(e.y-s.y,e.x-s.x);
		if(sgn(k) < 0)k += pi;
		if(sgn(k-pi) == 0)k -= pi;
		return k;
	}
	//`点和直线关系`
	//`1  在左侧`
	//`2  在右侧`
	//`3  在直线上`
	int relation(Point p){
		int c = sgn((p-s)^(e-s));
		if(c < 0)return 1;
		else if(c > 0)return 2;
		else return 3;
	}
	// 点在线段上的判断
	bool pointonseg(Point p){
		return sgn((p-s)^(e-s)) == 0 && sgn((p-s)*(p-e)) <= 0;
	}
	//`两向量平行(对应直线平行或重合)`
	bool parallel(Line v){
		return sgn((e-s)^(v.e-v.s)) == 0;
	}
	//`两线段相交判断`
	//`2 规范相交`
	//`1 非规范相交`
	//`0 不相交`
	int segcrossseg(Line v){
		int d1 = sgn((e-s)^(v.s-s));
		int d2 = sgn((e-s)^(v.e-s));
		int d3 = sgn((v.e-v.s)^(s-v.s));
		int d4 = sgn((v.e-v.s)^(e-v.s));
		if( (d1^d2)==-2 && (d3^d4)==-2 )return 2;
		return (d1==0 && sgn((v.s-s)*(v.s-e))<=0) ||
			(d2==0 && sgn((v.e-s)*(v.e-e))<=0) ||
			(d3==0 && sgn((s-v.s)*(s-v.e))<=0) ||
			(d4==0 && sgn((e-v.s)*(e-v.e))<=0);
	}
	//`直线和线段相交判断`
	//`-*this line   -v seg`
	//`2 规范相交`
	//`1 非规范相交`
	//`0 不相交`
	int linecrossseg(Line v){
		int d1 = sgn((e-s)^(v.s-s));
		int d2 = sgn((e-s)^(v.e-s));
		if((d1^d2)==-2) return 2;
		return (d1==0||d2==0);
	}
	//`两直线关系`
	//`0 平行`
	//`1 重合`
	//`2 相交`
	int linecrossline(Line v){
		if((*this).parallel(v))
			return v.relation(s)==3;
		return 2;
	}
	//`求两直线的交点`
	//`要保证两直线不平行或重合`
	Point crosspoint(Line v){
		double a1 = (v.e-v.s)^(s-v.s);
		double a2 = (v.e-v.s)^(e-v.s);
		return Point((s.x*a2-e.x*a1)/(a2-a1),(s.y*a2-e.y*a1)/(a2-a1));
	}
	//点到直线的距离
	double dispointtoline(Point p){
		return fabs((p-s)^(e-s))/length();
	}
	//点到线段的距离
	double dispointtoseg(Point p){
		if(sgn((p-s)*(e-s))<0 || sgn((p-e)*(s-e))<0)
			return min(p.distance(s),p.distance(e));
		return dispointtoline(p);
	}
	//`返回线段到线段的距离`
	//`前提是两线段不相交，相交距离就是0了`
	double dissegtoseg(Line v){
		return min(min(dispointtoseg(v.s),dispointtoseg(v.e)),min(v.dispointtoseg(s),v.dispointtoseg(e)));
	}
	//`返回点p在直线上的投影`
	Point lineprog(Point p){
		return s + ( ((e-s)*((e-s)*(p-s)))/((e-s).len2()) );
	}
	//`返回点p关于直线的对称点`
	Point symmetrypoint(Point p){
		Point q = lineprog(p);
		return Point(2*q.x-p.x,2*q.y-p.y);
	}
};
//圆
struct circle{
	Point p;//圆心
	double r;//半径
	circle(){}
	circle(Point _p,double _r){
		p = _p;
		r = _r;
	}
	circle(double x,double y,double _r){
		p = Point(x,y);
		r = _r;
	}
	//`三角形的外接圆`
	//`需要Point的+ /  rotate()  以及Line的crosspoint()`
	//`利用两条边的中垂线得到圆心`
	//`测试：UVA12304`
	circle(Point a,Point b,Point c){
		Line u = Line((a+b)/2,((a+b)/2)+((b-a).rotleft()));
		Line v = Line((b+c)/2,((b+c)/2)+((c-b).rotleft()));
		p = u.crosspoint(v);
		r = p.distance(a);
	}
	//`三角形的内切圆`
	//`参数bool t没有作用，只是为了和上面外接圆函数区别`
	//`测试：UVA12304`
	circle(Point a,Point b,Point c,bool t){
		Line u,v;
		double m = atan2(b.y-a.y,b.x-a.x), n = atan2(c.y-a.y,c.x-a.x);
		u.s = a;
		u.e = u.s + Point(cos((n+m)/2),sin((n+m)/2));
		v.s = b;
		m = atan2(a.y-b.y,a.x-b.x) , n = atan2(c.y-b.y,c.x-b.x);
		v.e = v.s + Point(cos((n+m)/2),sin((n+m)/2));
		p = u.crosspoint(v);
		r = Line(a,b).dispointtoseg(p);
	}
	//输入
	void input(){
		p.input();
		scanf("%lf",&r);
	}
	//输出
	void output(){
		printf("%.2lf %.2lf %.2lf\n",p.x,p.y,r);
	}
	bool operator == (circle v){
		return (p==v.p) && sgn(r-v.r)==0;
	}
	bool operator < (circle v)const{
		return ((p<v.p)||((p==v.p)&&sgn(r-v.r)<0));
	}
	//面积
	double area(){
		return pi*r*r;
	}
	//周长
	double circumference(){
		return 2*pi*r;
	}

};
int main()
{
	int n;
	scanf("%d",&n);
	
	for(int i=1;i<=n;i++)
	{
		p[i].input();
	}
	if(n<=1)
	{
		printf("%d\n",n);
		return 0;
	}
	int ans=1;
	for(int i=1;i<=n;i++)
	{
		map<Point,int> mp;
		for(int j=i+1;j<=n;j++)
		{
			if(Line(p[i],p[j]).parallel(Line(p[i],Point(0,0))))continue;//三点可能共线，判断三点共线不要用角度，用两条直线平行
			circle c=circle(p[i],p[j],Point(0,0));
			mp[c.p]++;
			ans=max(ans,mp[c.p]+1);
			
			
		}
	}
	printf("%d\n",ans);
}