hdu 3007 最小圆覆盖（模版）

//题目链接：http://acm.hdu.edu.cn/showproblem.php?pid=3007
//求点集最小覆盖圆
//增量法
#include
#include
#include
#define eps 1e-8
const int maxn=10000;
struct point {double x,y;}cen,p[maxn],po;
double r;
int n;

double distance(point p1,point p2)
{
	return sqrt((p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y));
}
double dist2(point p1,point p2)
{
	return (p1.x-p2.x)*(p1.x-p2.x)+(p1.y-p2.y)*(p1.y-p2.y);
}
//直线a1X+ b1Y+c1=0与a2X+b2Y+c2=0的交点
void calc(double a1,double b1,double c1,double a2,double b2,double c2)
{
	cen.y=(c1*a2-c2*a1)/(b1*a2-b2*a1);
	cen.x=(c1*b2-c2*b1)/(a1*b2-b1*a2);
}
void cir()
{
	cen=p[0];r=0;
	int i,j,k;
	for(i=1;ieps)
		{
			cen=p[i];
			r=0;
			for(j=0;j<=i-1;j++)
				if(distance(cen,p[j])-r>eps)
				{
					cen.x=(p[i].x+p[j].x)/2.0;
					cen.y=(p[i].y+p[j].y)/2.0;
					r=distance(cen,p[j]);
					for(k=0;k<=j-1;k++)
						if(distance(cen,p[k])-r>eps)
						{
							calc(p[j].x-p[i].x,
								p[j].y-p[i].y,
								(dist2(p[j],po)-dist2(p[i],po))/2.0,
								p[k].x-p[i].x,
								p[k].y-p[i].y,
								(dist2(p[k],po)-dist2(p[i],po))/2.0
								);
							r=distance(cen,p[k]);
						}
				}
		}
}
int main()
{
	po.x=0;
	po.y=0;
	while(scanf("%d",&n),n)
	{
		int i;
		for(i=0;i