[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+7;
typedef unsigned long long ull;
typedef long long ll;
int dcmp(double x){
    if(fabs(x)<=eps) return 0;
    return x<0?-1:1;
}
struct Point{
    double x,y;
    Point(double x=0,double y=0):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)==0 && dcmp(y-a.y)==0;
    }
};
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[5005];
double r;
Point cp;
bool inCrcle(Point tp){
    return dcmp(dis(tp-cp)-r)<=0;
}
void getCrcle(Point a,Point b,Point c){
    Point p1=b-a,p2=c-a;
    double d=2*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,0,n-1) p[i].read();
    random_shuffle(p,p+n);
    cp=p[0];r=0;
    For(i,1,n-1) {
        if(!inCrcle(p[i])){
            r=0;
            cp=p[i];
            For(j,0,i-1) {
                if(!inCrcle(p[j])) {
                    r=dis(p[j]-p[i])/2;
                    cp=(p[j]+p[i])/2;
                    For(k,0,j-1) {
                        if(!inCrcle(p[k]))
                            getCrcle(p[i],p[j],p[k]);
                    }
                }
            }
        }
    }
    printf("%.2lf %.2lf %.2lf
",cp.x,cp.y,r);
}
int main(){
//    fre("in.txt","r",stdin);
    int t=0;
    while(~scanf("%d",&n) && n)    solve();
    return 0;
}