1 #include<stdio.h>//求出最小生成树之后，再求出第n-m大的值就行了
2 #include<stdlib.h>
3 #include<math.h>
4 #include<string.h>
5 #define N 501
6 #define Maxint 9999999
7
8 double c[N][N],x[N],y[N],dist[N];
9 int s[N], closest[N],n;
10
11 int cmp(const void *a,const void *b)
12 {
13 return *(double *)a > *(double *)b ? 1 : -1;
14 }
15
16 void Prim(int v)
17 {
18 int i,j,k;
19 double min,lowcost[N];
20 s[v] = 1;
21 for(i=1; i<n; i++)
22 {
23 lowcost[i] = c[v][i];
24 closest[i] = v;
25 s[i] = 0;
26 }
27 for(i=1; i<n; i++)
28 {
29 min = Maxint;
30 j = 1;
31 for(k=1; k<n; k++)
32 {
33 if((lowcost[k]<min)&&(!s[k]))
34 {
35 min = lowcost[k];
36 j = k;
37 }
38 }
39 s[j] = 1;
40 lowcost[j] = min;
41 dist[i-1] = min;//存放最小生成树的值，将每个最小的边都存起来
42 for(k = 0; k<n; k++)
43 {
44 if((c[j][k]<lowcost[k])&&(!s[k]))
45 {
46 lowcost[k] = c[j][k];
47 closest[k] = j;
48 }
49 }
50 }
51 }
52
53 int main()
54 {
55 int i,j,m,ncases;
56
57 scanf("%d",&ncases);
58 while( ncases-- )
59 {
60 scanf("%d%d",&m,&n);
61 for(i=0; i<n; i++)
62 {
63 scanf("%lf%lf",&x[i],&y[i]);
64 for(j=0; j<=i; j++)
65 {
66 c[i][j] = sqrt((x[i]-x[j])*(x[i]-x[j])+(y[i]-y[j])*(y[i]-y[j]));
67 c[j][i] = c[i][j];
68 }
69 }
70 Prim(0);
71 qsort(dist,n-1,sizeof(dist[0]),cmp);
72 printf("%.2lf\n",dist[n-m-1]);
73 }
74 // system("pause");
75 return 0;
76 }
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