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-;
 double f(const long double &a){return a*a;}
 void inin(int &ret)
 {
     ret=;int f=;char ch=getchar();
     while(ch<''||ch>''){if(ch=='-')f=;ch=getchar();}
     while(ch>=''&&ch<='')ret*=,ret+=ch-'',ch=getchar();
     ret=f?-ret:ret;
 }
 struct xl
 {
     double x,y;
     xl(double x=.,double y=.):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[];
 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=*(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+,di+n+);
     ret=di[];r=.;
     re(i,,n)if(dis(di[i],ret)>r+eps)
     {
         ret=di[i],r=.;
         re(j,,i-)if(dis(di[j],ret)>r+eps)
         {
             ret=(di[i]+di[j])/.;
             r=dis(di[j],ret);
             re(k,,j-)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,,n)di[i].in();
         double r;
         xl c=solve(r);
         printf("%.2f %.2f %.2f\n",c.x,c.y,r);
     }
      return ;
 }