Problem:
Score : 600 points

Problem Statement

Given are N points (x_i, y_i) in a two-dimensional plane.

Find the minimum radius of a circle such that all the points are inside or on it.

Constraints

2 \leq N \leq 50

0 \leq x_i \leq 1000

0 \leq y_i \leq 1000

The given N points are all different.

The values in input are all integers.

Input

Input is given from Standard Input in the following format:

N
x_1 y_1
:
x_N y_N

Output

Print the minimum radius of a circle such that all the N points are inside or on it.

Your output will be considered correct if the absolute or relative error from our answer is at most 10^{-6}.

Sample Input 1

2
0 0
1 0

Sample Output 1

0.500000000000000000

Both points are contained in the circle centered at (0.5,0) with a radius of 0.5.

Sample Input 2

3
0 0
0 1
1 0

Sample Output 2

0.707106781186497524

Sample Input 3

10
10 9
5 9
2 0
0 0
2 7
3 3
2 5
10 0
3 7
1 9

Sample Output 3

6.726812023536805158

If the absolute or relative error from our answer is at most 10^{-6}, the output will be considered correct.

Solution in C++14 (GCC 5.4.1) (judge verdict: Wrong Answer)

#include <bits/stdc++.h>
#define int long long
using namespace std;
main(){
    int n;
    scanf("%lld",&n);
    int x[n], y[n];
    for (int i =0 ; i < n; i++){
        scanf("%lld%lld",&x[i],&y[i]);
        x[i] *= 2;
        y[i] *= 2;
    }
    int ans = 1000000000000000;
    for (int i = 0; i < n; i++){
        for (int j = 0; j < n; j++){
            int mx = (x[i]+x[j])/2, my = (y[i]+y[j])/2;
            int X = x[i]-x[j], Y = y[i]-y[j];
            int rad = (X*X)+(Y*Y);
            //printf("%d %d %d %d %d\n",i,j,rad,mx,my);
            for (int k = 0; k < n; k++){
                int dx = (x[k]-mx), dy = (y[k]-my);
                int dist = (dx*dx)+(dy*dy);
                //printf("%d %d, ",k,dist);
                if (dist*4 > rad){
                    goto done;
                }
            }
            //printf("CAN\n");
            ans = min(ans,rad);
            done:;
        }
    }
    //printf("%lld\n",ans);
    long double kk = (long double)ans;
    kk = pow(kk,(long double)1.00000/2.000000);
    kk /= 4.00000;
    printf("%.9f",(double)kk);
}