[hdu-3007]Buried memory 最小覆盖圆

大致题意：

　　平面上有n个点，求一个最小的圆覆盖住所有点

　　

　　最小覆盖圆裸题

　　学习了一波最小覆盖圆算法

 #include<cstdio>
 #include<iostream>
 #include<cstring>
 #include<algorithm>
 #include<queue>
 #include<set>
 #include<map>
 #include<stack>
 #include<time.h>
 #include<cstdlib>
 #include<cmath>
 #include<list>
 using namespace std;
 #define MAXN 100100
 #define eps 1e-9
 #define For(i,a,b) for(int i=a;i<=b;i++)
 #define Fore(i,a,b) for(int i=a;i>=b;i--)
 #define lson l,mid,rt<<1
 #define rson mid+1,r,rt<<1|1
 #define mkp make_pair
 #define pb push_back
 #define cr clear()
 #define sz size()
 #define met(a,b) memset(a,b,sizeof(a))
 #define iossy ios::sync_with_stdio(false)
 #define fre freopen
 #define pi acos(-1.0)
 #define inf 1e6+7
 #define Vector Point
 const int Mod=1e9+;
 typedef unsigned long long ull;
 typedef long long ll;
 int dcmp(double x){
     if(fabs(x)<=eps) return ;
     return x<?-:;
 }
 struct Point{
     double x,y;
     Point(double x=,double y=):x(x),y(y) {}
     bool operator < (const Point &a)const{
         if(x==a.x) return y<a.y;
         return x<a.x;
     }
     Point operator - (const Point &a)const{
         return Point(x-a.x,y-a.y);
     }
     Point operator + (const Point &a)const{
         return Point(x+a.x,y+a.y);
     }
     Point operator * (const double &a)const{
         return Point(x*a,y*a);
     }
     Point operator / (const double &a)const{
         return Point(x/a,y/a);
     }
     void read(){
         scanf("%lf%lf",&x,&y);
     }
     void out(){
         cout<<"debug: "<<x<<" "<<y<<endl;
     }
     bool operator == (const Point &a)const{
         return dcmp(x-a.x)== && dcmp(y-a.y)==;
     }
 };
 double Dot(Vector a,Vector b) {
     return a.x*b.x+a.y*b.y;
 }
 double dis(Vector a) {
     return sqrt(Dot(a,a));
 }
 double Cross(Point a,Point b){
     return a.x*b.y-a.y*b.x;
 }
 int n;
 Point p[];
 double r;
 Point cp;
 bool inCrcle(Point tp){
     return dcmp(dis(tp-cp)-r)<=;
 }
 void getCrcle(Point a,Point b,Point c){
     Point p1=b-a,p2=c-a;
     double d=*Cross(p1,p2);
     cp.x=(p2.y*Dot(p1,p1)-p1.y*Dot(p2,p2))/d+a.x;
     cp.y=(p1.x*Dot(p2,p2)-p2.x*Dot(p1,p1))/d+a.y;
     r=dis(a-cp);
 }
 void solve(){
     For(i,,n-) p[i].read();
     random_shuffle(p,p+n);
     cp=p[];r=;
     For(i,,n-) {
         if(!inCrcle(p[i])){
             r=;
             cp=p[i];
             For(j,,i-) {
                 if(!inCrcle(p[j])) {
                     r=dis(p[j]-p[i])/;
                     cp=(p[j]+p[i])/;
                     For(k,,j-) {
                         if(!inCrcle(p[k]))
                             getCrcle(p[i],p[j],p[k]);
                     }
                 }
             }
         }
     }
     printf("%.2lf %.2lf %.2lf\n",cp.x,cp.y,r);
 }
 int main(){
 //    fre("in.txt","r",stdin);
     int t=;
     while(~scanf("%d",&n) && n)    solve();
     return ;
 }