题意：
给出一系列点，这些点之间的距离都在2附近，求一个单位圆最多能包含几个点。
要点：
极限情况就是2个点在圆的边上，这样就可以求出圆心，再枚举看其他点在不在圆内即可，复杂度是n^3。
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<algorithm>
#define eps 1e-8
using namespace std;
struct node
{double x, y;
}p[310];double dis(node a, node b)
{return sqrt((a.x - b.x)*(a.x - b.x) + (a.y - b.y)*(a.y - b.y));
}
void get_center(node a, node b, node& center)//已知两点和半径求圆心
{node mid;mid.x = (a.x + b.x) / 2.0;mid.y = (a.y + b.y) / 2.0;double angle = atan2(a.x - b.x, b.y - a.y);//注意这里double d = sqrt(1 - dis(mid, a)*dis(mid, a));center.x = mid.x + d*cos(angle);center.y = mid.y + d*sin(angle);//printf("%lf %lf\n", center.x, center.y);
}int main()
{int n, i, j, k;while (scanf("%d", &n), n){for (i = 1; i <= n; i++)scanf("%lf%lf", &p[i].x,&p[i].y);int ans = 1;for(i=1;i<=n;i++)for (j = i + 1; j <= n; j++){if (dis(p[i], p[j]) > 2.0) continue;node center;get_center(p[i], p[j],center);int cnt = 0;for (k = 1; k <= n; k++){if (dis(center, p[k]) < 1.0 + eps)cnt++;}ans = max(cnt, ans);}printf("%d\n", ans);}return 0;
}