hiho 1227 找到一个恰好包含n个点的圆 （2015北京网赛 A题）

平面上有m个点，要从这m个点当中找出n个点，使得包含这n个点的圆的半径（圆心为n个点当中的某一点且半径为整数）最小，同
时保证圆周上没有点。

n > m 时要输出-1

样例输入
4
3 2 0 0 1 0 1.2 0
2 2 0 0 1 0
2 1 0 0 1.2 0
2 1 0 0 1 0
样例输出
1
2
1
-1

# include <iostream>
# include <cstdio>
# include <cstring>
# include <algorithm>
# include <string>
# include <cmath>
# include <queue>
# include <list>
# define LL long long
using namespace std ;

double dis[200][200] ;
const int INF = 0x3f3f3f3f ;

struct Point
{
    double x,y;

}p[200];

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

int main()
{
    //freopen("in.txt","r",stdin) ;
    int T ;
    scanf("%d" , &T) ;
    while(T--)
    {
        int m , n ;
        int i , j ;
        scanf("%d %d" , &m , &n) ;
        for (i = 0 ; i < m ; i++)
            scanf("%lf %lf" , &p[i].x , &p[i].y) ;
        if (n > m)
        {
            printf("-1
") ;
            continue ;
        }
        int ans = INF ;
        int r ;
        memset(dis , 0 , sizeof(dis)) ;
        for (i = 0 ; i < m ; i++)
        {
            for (j = i+1 ; j < m ; j++)
            {
                dis[i][j] = dis[j][i] = dist(p[i] , p[j]) ;
            }
            sort(dis[i] , dis[i]+m) ;
            r = (int)dis[i][n-1] ;
            if (r <= dis[i][n-1])
                r += 1 ;
            if (n < m &&r >= dis[i][n])
                continue ;
            if (r < ans)
                ans = r ;
        }
        if (ans == INF)
            printf("-1
") ;
        else
            printf("%d
" , ans) ;

    }
    return 0;
}