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[][] ;
 const int INF = 0x3f3f3f3f ;

 struct Point
 {
     double x,y;

 }p[];

 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 =  ; i < m ; i++)
             scanf("%lf %lf" , &p[i].x , &p[i].y) ;
         if (n > m)
         {
             printf("-1\n") ;
             continue ;
         }
         int ans = INF ;
         int r ;
         memset(dis ,  , sizeof(dis)) ;
         for (i =  ; i < m ; i++)
         {
             for (j = i+ ; 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-] ;
             if (r <= dis[i][n-])
                 r +=  ;
             if (n < m &&r >= dis[i][n])
                 continue ;
             if (r < ans)
                 ans = r ;
         }
         if (ans == INF)
             printf("-1\n") ;
         else
             printf("%d\n" , ans) ;

     }
     return ;
 }