显然二分答案\(mid\)，然后距离\(\leq mid\)的点对只能放在一个部落里。然后可以并查集\(O(n^2)\)算出有多少个部落。
// luogu-judger-enable-o2
#include<bits/stdc++.h>
#define il inline
#define vd void
typedef long long ll;
il int gi(){
int x=0,f=1;
char ch=getchar();
while(!isdigit(ch)){
if(ch=='-')f=-1;
ch=getchar();
}
while(isdigit(ch))x=x*10+ch-'0',ch=getchar();
return x*f;
}
int n,k,x[1010],y[1010],fa[1010],siz[1010];
il int hd(int x){return fa[x]==x?x:fa[x]=hd(fa[x]);}
il vd Union(int x,int y){
x=hd(x),y=hd(y);
if(x==y)return;
if(siz[x]<siz[y])fa[x]=y,siz[y]+=siz[x];
else fa[y]=x,siz[x]+=siz[y];
}
il int check(double mid){
mid*=mid;
for(int i=1;i<=n;++i)fa[i]=i,siz[i]=1;
for(int i=1;i<=n;++i)
for(int j=i+1;j<=n;++j)
if((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j])<mid)
Union(i,j);
int tot=0;
for(int i=1;i<=n;++i)if(fa[i]==i)++tot;
return tot;
}
int main(){
#ifndef ONLINE_JUDGE
freopen("4047.in","r",stdin);
freopen("4047.out","w",stdout);
#endif
n=gi(),k=gi();
for(int i=1;i<=n;++i)x[i]=gi(),y[i]=gi();
double l=0,r=14143,mid;
while(r-l>1e-4){
mid=(l+r)*0.5;
if(check(mid)<=k)r=mid;
else l=mid;
}
r=14143;
while(r-l>1e-4){
mid=(l+r)*0.5;
if(check(mid)<k)r=mid;
else l=mid;
}
printf("%.2lf\n",l);
return 0;
}