hud3007 Buried memory

题目链接

最小圆覆盖

并不知道为什么是O(n)的，而且要随机化点的顺序

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<string>
#include<cmath>
#include<ctime>
#include<queue>
#include<stack>
#include<map>
#include<set>
#define rre(i,r,l) for(int i=(r);i>=(l);i--)
#define re(i,l,r) for(int i=(l);i<=(r);i++)
#define Clear(a,b) memset(a,b,sizeof(a))
#define inout(x) printf("%d",(x))
#define douin(x) scanf("%lf",&x)
#define strin(x) scanf("%s",(x))
#define LLin(x) scanf("%lld",&x)
#define op operator
#define CSC main
typedef unsigned long long ULL;
typedef const int cint;
typedef long long LL;
using namespace std;
const double eps=1e-8;
double f(const long double &a){return a*a;}
void inin(int &ret)
{
    ret=0;int f=0;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=1;ch=getchar();}
    while(ch>='0'&&ch<='9')ret*=10,ret+=ch-'0',ch=getchar();
    ret=f?-ret:ret;
}
struct xl
{
    double x,y;
    xl(double x=0.,double y=0.):x(x),y(y){}
    xl op + (const xl &rhs){return xl(x+rhs.x,y+rhs.y);}
    xl op / (const double &rhs){return xl(x/rhs,y/rhs);}
    void in(){douin(x),douin(y);}
};
double dis(const xl &a,const xl &b)
{
    return sqrt(f(a.x-b.x)+f(a.y-b.y));
}
xl di[555];
int n;
xl outoftriangle(const xl &A,const xl &B,const xl &C)
{
    double bx=B.x-A.x,by=B.y-A.y;
    double cx=C.x-A.x,cy=C.y-A.y;
    double area=2*(bx*cy-by*cx);
    double x=(cy*(f(bx)+f(by))-by*(f(cx)+f(cy)))/area+A.x;
    double y=(bx*(f(cx)+f(cy))-cx*(f(bx)+f(by)))/area+A.y;
    return xl(x,y);
}
xl solve(double &r)
{
    xl ret;
    random_shuffle(di+1,di+n+1);
    ret=di[1];r=0.;
    re(i,2,n)if(dis(di[i],ret)>r+eps)
    {
        ret=di[i],r=0.;
        re(j,1,i-1)if(dis(di[j],ret)>r+eps)
        {
            ret=(di[i]+di[j])/2.;
            r=dis(di[j],ret);
            re(k,1,j-1)if(dis(di[k],ret)>r+eps)
            {
                ret=outoftriangle(di[i],di[j],di[k]);
                r=dis(di[i],ret);
            }
        }
    }
    return ret;
}
int main()
{
    while(scanf("%d",&n)&&n)
    {
        re(i,1,n)di[i].in();
        double r;
        xl c=solve(r);
        printf("%.2f %.2f %.2f
",c.x,c.y,r);
    }
     return 0;
}