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\n") ;
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]
)
continue ;
if (r < ans)
ans = r ;
}
if (ans == INF)
printf("-1\n") ;
else
printf("%d\n" , ans) ;

}
return 0;
}