/*********************************************************************************
纠结了良久,这个题是DLX的重复覆盖问题,以前还没见过,最后套了晓天大牛的模板,水过了。
还是模板题。。。因为最后的答案必然是某两点的距离,所以离散化一下所有距离后二分再套用DLX
重复覆盖模板就能AC了~赞一下神函数unique_copy,这个模板的细节还是要很小心才行,
计数都要从1开始~
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#include <iostream>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <utility>
#include <cstdio>
#include <vector>
#include <cmath>
#include <ctime>
#include <map>
#include <set>
using namespace std;
typedef long long LL;
typedef pair<int, int> PII;
typedef pair<double, double> PDD;
const int INF_INT = 0x3f3f3f3f;
const double oo = 10e9;
const double eps = 10e-7;
const int MAXN = 104;
const int MAXM = MAXN * MAXN;
PII house[MAXN];
int test, n, m;
int dtop, dis[MAXM], laby[MAXN][MAXN];
int nodecnt;
int U[MAXM], L[MAXM], D[MAXM], R[MAXM], row[MAXM], col[MAXM];
int sz[MAXN];
bool vis[MAXN], maze[MAXN][MAXN];
int sqa_dist(const PII &pa, const PII &pb)
{
return (pa.first - pb.first) * (pa.first - pb.first) +
(pa.second - pb.second) * (pa.second - pb.second);
}
void initDLX(int mm)
{
memset(sz, 0, sizeof(sz));
memset(col, -1, sizeof(col));
nodecnt = mm + 1;
for (int i = 0; i <= mm; ++i)
{
L[i] = i - 1;
R[i] = i + 1;
U[i] = D[i] = i;
col[i] = i;
}
L[0] = mm;
R[mm] = 0;
sz[0] = INF_INT;
return ;
}
void insert(int id, int *tt, int c)
{
for (int i = 0; i < c; ++i, ++nodecnt)
{
int x = tt[i];
row[nodecnt] = id;
col[nodecnt] = x;
++sz[x];
U[nodecnt] = x;
D[nodecnt] = D[x];
U[ D[x] ] = nodecnt;
D[x] = nodecnt;
if (0 == i)
{
L[nodecnt] = R[nodecnt] = nodecnt;
}
else
{
L[nodecnt] = nodecnt - 1;
R[nodecnt] = nodecnt - i;
R[nodecnt - 1] = nodecnt;
L[nodecnt - i] = nodecnt;
}
}
return ;
}
void buildDLX(int key)
{
int tt[MAXN], c;
memset(maze, false, sizeof(maze));
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
maze[i][j] = (laby[i][j] <= key);
}
}
for (int i = 1; i <= n; ++i)
{
c = 0;
for (int j = 1; j <= n; ++j)
{
if (maze[i][j])
{
tt[c++] = j;
}
}
insert(i, tt, c);
}
return ;
}
void remove(int x)
{
for (int i = D[x]; i != x; i = D[i])
{
L[ R[i] ] = L[i];
R[ L[i] ] = R[i];
--sz[ col[i] ];
}
return ;
}
void resume(int x)
{
for (int i = U[x]; i != x; i = U[i])
{
L[ R[i] ] = i;
R[ L[i] ] = i;
++sz[ col[i] ];
}
return ;
}
int h()
{
memset(vis, false, sizeof(vis));
int res = 0;
for (int i = R[0]; i != 0; i = R[i])
{
if (vis[ col[i] ])
{
continue ;
}
vis[ col[i] ] = true;
++res;
for (int j = D[i]; j != i; j = D[j])
{
if (col[j] != 0)
{
for (int k = R[j]; k != j; k = R[k])
{
vis[ col[k] ] = true;
}
}
}
}
return res;
}
bool dfs(int step, int key)
{
if (h() + step > m)
{
return false;
}
if (0 == R[0])
{
return true;
}
int id = 0;
for (int i = R[0]; i != 0; i = R[i])
{
if (sz[i] < sz[id])
{
id = i;
}
}
for (int i = D[id]; i != id; i = D[i])
{
remove(i);
for (int j = R[i]; j != i; j = R[j])
{
remove(j);
}
if (dfs(step + 1, key))
{
return true;
}
for (int j = L[i]; j != i; j = L[j])
{
resume(j);
}
resume(i);
}
return false;
}
double bisearch()
{
int low = 0, high = dtop - 1, mid = (low + high) >> 1;
int res = mid, key;
while (low <= high)
{
mid = (low + high) >> 1;
key = dis[mid];
initDLX(n);
buildDLX(key);
if (dfs(0, key))
{
res = mid;
high = mid - 1;
}
else
{
low = mid + 1;
}
}
return sqrt(dis[res] + 0.0);
}
void ace()
{
double ans;
for (scanf("%d", &test); test--; )
{
scanf("%d %d", &n, &m);
for (int i = 1; i <= n; ++i)
{
scanf("%d %d", &house[i].first, &house[i].second);
}
dtop = 0;
for (int i = 1; i <= n; ++i)
{
for (int j = 1; j <= n; ++j)
{
laby[i][j] = dis[dtop++] = sqa_dist(house[i], house[j]);
}
}
sort(dis, dis + dtop);
dtop = unique_copy(dis, dis + dtop, dis) - dis;
ans = bisearch();
printf("%.6lf\n", ans);
}
return ;
}
int main()
{
ace();
return 0;
}