BZOJ 1336&1337最小圆覆盖

思路：
http://blog.csdn.net/commonc/article/details/52291822
（照着算法步骤写……）
已知三点共圆 求圆心的时候 就设一下圆心坐标(x,y) 解个方程就好了

//By SiriusRen
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
int n;double R,tempx,tempy,tempz,tmpx,tmpy,tmpz;
struct Point{double x,y;}point[100050],Ans;
double Sqr(double x){return x*x;}
double dis(Point a,Point b){return sqrt(Sqr(a.x-b.x)+Sqr(a.y-b.y));}
bool in_circle(Point x){return dis(Ans,x)<=R;}
int main(){
    scanf("%d",&n);
    for(int i=1;i<=n;i++)scanf("%lf%lf",&point[i].x,&point[i].y);
    random_shuffle(point+1,point+n);
    for(int i=1;i<=n;i++)if(!in_circle(point[i])){
        Ans.x=point[i].x,Ans.y=point[i].y,R=0;
        for(int j=1;j<i;j++)if(!in_circle(point[j])){
            Ans.x=(point[i].x+point[j].x)/2;
            Ans.y=(point[i].y+point[j].y)/2;
            R=dis(Ans,point[j]);
            for(int k=1;k<j;k++)if(!in_circle(point[k])){
                tempz=point[j].x-point[i].x;
                tempx=2*(point[i].y-point[j].y)/tempz;
                tempy=(Sqr(point[j].x)+Sqr(point[j].y)-Sqr(point[i].x)-Sqr(point[i].y))/tempz;
                tmpz=point[k].x-point[j].x;
                tmpx=2*(point[j].y-point[k].y)/tmpz;
                tmpy=(Sqr(point[k].x)+Sqr(point[k].y)-Sqr(point[j].x)-Sqr(point[j].y))/tmpz;
                Ans.y=(tmpy-tempy)/(tempx-tmpx);
                Ans.x=(tempx*Ans.y+tempy)/2;
                R=dis(Ans,point[j]);
            }
        }
    }
    printf("%f
%f %f
",R,Ans.x,Ans.y);
}