Boundary(几何)

题意：用原点和给出的任意两点作圆，问所作的圆最多可以同时通过几个点。

做法：过三点做出所有圆，因为原点在圆上，则半径一定，所以如果圆心相同，那么这几个圆就是相同的。三点中有一点确定是原点，另外两遍遍历所有的点的组合即可，耗时O(n²),然后找出哪个圆心出现最多次，记为num，假设该圆过了ans个点，每次和圆确定一个圆需要从ans个点中取出两个点，这不就是排列组合那味了吗？所以C(ans,2)=mum,即ans*(ans-1)/2=num.

#include<cstdio>
#include <iostream>
#include <cstring>
#include <deque>
#include<algorithm>
#include<cmath>
#include<vector>
using namespace std;

using namespace std;
double Pointx, Pointy;
struct Point
{
    double x, y;
}point[2010];
void get_center(Point a, Point b, Point c)
{
    double fm1 = 2 * (a.y - c.y) * (a.x - b.x) - 2 * (a.y - b.y) * (a.x - c.x);
    double fm2 = 2 * (a.y - b.y) * (a.x - c.x) - 2 * (a.y - c.y) * (a.x - b.x);

    if (fm1 == 0 || fm2 == 0)//判断是否三点一线
    {
        Pointx = Pointy = -1;
        return;
    }
    double fz1 = a.x * a.x - b.x * b.x + a.y * a.y - b.y * b.y;
    double fz2 = a.x * a.x - c.x * c.x + a.y * a.y - c.y * c.y;
    Pointx = (fz1 * (a.y - c.y) - fz2 * (a.y - b.y)) / fm1;
    Pointy = (fz1 * (a.x - c.x) - fz2 * (a.x - b.x)) / fm2;
}
vector<pair<double, double>> vec;
int main()
{
    int n;
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lf %lf", &point[i].x, &point[i].y);
    }
    Point O;
    O.x = O.y = 0;
    for (int i = 1; i <= n; i++) {
        for (int j = i + 1; j <= n; j++) {
            get_center(O, point[i], point[j]);
            if (Pointx == Pointy && Pointx==-1)
                continue;
            vec.push_back({ Pointx, Pointy });
        }
    }
    if (vec.size() == 0) {
        printf("1\n");
        return 0;
    }
    sort(vec.begin(), vec.end());
    int sum = 1;
    int num = 1;
    pair<double, double> now = vec[0];
    for (int i = 1; i < vec.size(); i++) {
        if (vec[i] == now) ++num;
        else {
            now = vec[i];
            sum = max(sum, num);
            num = 1;
            continue;
        }
        sum = max(sum, num);
    }

    for (int i = 1; i <= n; i++) {
        if (i * (i - 1) == sum * 2) {
            printf("%d\n", i);
            return 0;
        }
    }
    return 0;
}

借鉴于（虽说是借鉴，但感觉差别不是很大）