HDU 3007 最小圆覆盖 计算几何

思路：

随机增量法

（好吧这数据范围并不用）

//By SiriusRen
#include <cmath>
#include <cstdio>
#include <algorithm>
using namespace std;
typedef double db;
int n;
struct P{db x,y,r;P(){}P(db X,db Y){x=X,y=Y;}}p[],ans;
P operator-(P a,P b){return P(a.x-b.x,a.y-b.y);}
db dis(P a){return sqrt(a.x*a.x+a.y*a.y);}
bool ck(P a){return dis(a-ans)<=ans.r;}
P get(db x1,db y1,db x2,db y2,db x3,db y3){
    db a1=*(x2-x1),b1=*(y2-y1),c1=x2*x2+y2*y2-x1*x1-y1*y1;
    db a2=*(x3-x2),b2=*(y3-y2),c2=x3*x3+y3*y3-x2*x2-y2*y2;
    return P((c1*b2-c2*b1)/(a1*b2-a2*b1),(a1*c2-a2*c1)/(a1*b2-a2*b1));
}
int main(){
    while(scanf("%d",&n)&&n){
        for(int i=;i<=n;i++)scanf("%lf%lf",&p[i].x,&p[i].y);
        random_shuffle(p+,p++n),ans.x=ans.y=ans.r=;
        for(int i=;i<=n;i++)if(!ck(p[i])){
            ans.x=ans.y=ans.r=;
            for(int j=;j<i;j++)if(!ck(p[j])){
                ans=P((p[i].x+p[j].x)/,(p[i].y+p[j].y)/);
                ans.r=dis(ans-p[j]);
                for(int k=;k<j;k++)if(!ck(p[k])){
                    ans=get(p[i].x,p[i].y,p[j].x,p[j].y,p[k].x,p[k].y);
                    ans.r=dis(ans-p[j]);
                }
            }
        }
        printf("%.2lf %.2lf %.2lf\n",ans.x,ans.y,ans.r);
    }
}