ZOJ 1450 Minimal Circle 最小圆覆盖

//#pragma comment(linker, "/STACK:16777216") //for c++ Compiler
#include <stdio.h>
#include <iostream>
#include <cstring>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <algorithm>
#define ll long long
#define Max(a,b) (((a) > (b)) ? (a) : (b))
#define Min(a,b) (((a) < (b)) ? (a) : (b))
#define Abs(x) (((x) > 0) ? (x) : (-(x)))
using namespace std;

const int INF = 0x3f3f3f3f;
const int MAXN = 220;
const double eps = 1e-9;

struct POINT{
    double x;
    double y;
    POINT() : x(0), y(0) {};
    POINT(double _x_, double _y_) : x(_x_), y(_y_) {};
};

struct CIRCLE{
    POINT p;
    double r;
    CIRCLE() {};
    CIRCLE(POINT _p_, double _r_) : p(_p_), r(_r_) {};
};

double dist(POINT a,POINT b){
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}

CIRCLE calc(POINT p1,POINT p2,POINT p3){//三点的外接圆圆心的函数：
    CIRCLE temp;
    double a,b,c,d,e,f;
    a = p2.x - p1.x;
    b = p2.y - p1.y;
    c = (p2.x * p2.x + p2.y * p2.y - p1.x * p1.x - p1.y * p1.y) / 2;
    d = p3.x - p1.x;
    e = p3.y - p1.y;
    f = (p3.x * p3.x + p3.y * p3.y - p1.x * p1.x - p1.y * p1.y) / 2;
    temp.p.y = (c * d - f * a) / (b * d - e * a);
    temp.p.x = (c * e - f * b) / (a * e - b * d);
    return temp;
}
CIRCLE minC(POINT *p,int n){
    CIRCLE O;
    int i,j,k;
    O.p = p[0];
    O.r = 0;
    for(i=  1; i < n ; i++){
        if(dist(O.p,p[i]) <= O.r + eps) continue;
        O.p = p[i];O.r = 0;
        for(j = 0; j < i; j++){
            if(dist(O.p,p[j]) <= O.r + eps) continue;
            O.p.x = (p[i].x + p[j].x) / 2;
            O.p.y = (p[i].y + p[j].y) / 2;
            O.r = dist(O.p,p[j]);
            for(k = 0; k < j; k++){
                if(dist(O.p,p[k]) <= O.r + eps) continue;
                O = calc(p[i],p[j],p[k]);
                O.r = dist(O.p,p[k]);
            }
        }
    }
    return O;
}

int main(){
    int i, j, n;
    POINT p[200];
    while(EOF != scanf("%d",&n)){
        if(n == 0)  break;
        for(i = 0; i < n; ++i)
            scanf("%lf%lf",&p[i].x,&p[i].y);
        CIRCLE c = minC(p, n);
        printf("%.2f %.2f %.2f\n",c.p.x,c.p.y,c.r);
    }
    return 0;
}