B.Boundary

給定一個二維平面，以及n個點，求一個過圓心的圓，問最多有多少個點在圓上。

思路：

列舉C (n,2) 個點 ，從而得到圓心。若在在一個圓上，必然圓心也是同一個，圓心同一個也必在圓上。

要注意，列舉一個迴圈就計算一次，否則兩次再算要考慮容斥，太麻煩了。

#include<iostream>
#include<algorithm>
#include<bitset>
//#include<unordered_map>
#include<fstream>
#include<iomanip>
#include
 
<string>
#include<cmath>
#include<cstring>
#include<vector>
#include<map>
#include<set>
#include<list>
#include<queue>
#include<stack>
#include<sstream>
#include<cstdio>
#include<ctime>
#include<cstdlib>
#define INF 0x3f3f3f3f
#define
 
 inf 0x7FFFFFFF
#define MOD 1000000007
#define moD 1000000003
#define pii pair<int,int>
#define eps 1e-8
#define equals(a,b) (fabs(a-b)<eps)
#define bug puts("bug")
#define re  register
#define fi first
#define se second
const int maxn = 1e5 + 5;
const double Inf = 10000.0;
const double PI = acos(-1.0);
typedef  
 
long long ll;
using namespace std;

struct Point {
    double x, y;
    Point() {
        x = y = 0;
    }
    Point(double _x, double _y) {
        x = _x;
        y = _y;
    }
};

bool operator < (const Point& a, const Point& b)
{
    return a.x < b.x || (a.x == b.x && a.y < b.y);
}

Point get_o(Point a, Point b, Point c) {
    double x1 = a.x, y1 = a.y;
    double x2 = b.x, y2 = b.y;
    double x3 = c.x, y3 = c.y;
    double a1 = 2 * (x2 - x1);
    double b1 = 2 * (y2 - y1);
    double c1 = x2 * x2 + y2 * y2 - x1 * x1 - y1 * y1;
    double a2 = 2 * (x3 - x2);
    double b2 = 2 * (y3 - y2);
    double c2 = x3 * x3 + y3 * y3 - x2 * x2 - y2 * y2;
    double x = (c1 * b2 - c2 * b1) / (a1 * b2 - a2 * b1);
    double y = (a1 * c2 - a2 * c1) / (a1 * b2 - a2 * b1);
    return Point(x, y);
}

bool check(Point a, Point b, Point c) {
    if (equals((b.y - a.y) * (c.x - a.x), (c.y - a.y) * (b.x - a.x))) return 0;
    return 1;
}

bool eq(Point a, Point b) {
    return equals(a.x, b.x) && equals(a.y, b.y);
}

vector<Point> v;
Point pp[maxn];

int main() {
    int n;
    scanf("%d", &n);
    double x, y;
    Point p (0,0);
    for (int i = 0; i < n; i++) {
        scanf("%lf%lf", &x, &y);
        pp[i].x = x, pp[i].y = y;
    }
    if (n <= 2) {
        printf("%d",n);
        return 0;
    }
    int Max = 1;
    for (int i = 0; i < n; i++) {
        v.clear();
        for (int j = i + 1; j < n; j++) {
            if (check(p, pp[i], pp[j]))
                v.push_back(get_o(p, pp[i], pp[j]));
        }
        sort(v.begin(), v.end());
        int cnt = 2;
        for (int i = 1; i < v.size(); i++) {
            if (eq(v[i - 1], v[i])) cnt++, Max = max(Max, cnt);
            else Max = max(Max, cnt), cnt = 2;
        }
    }
    printf("%d", Max);
}