因为结论满足单调性，所以用二分。
我们用并查集合并，看看联通块的个数是否小于等于s。

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <cmath>
using namespace std;

const int N = 510;
const double eps = 1e-4;
int n, m, x[N], y[N], fa[N];
double d[N][N];

int findfa(int x) { return x == fa[x] ? x : fa[x] = findfa(fa[x]); }

bool check(double now) {
    for (int i = 1; i <= n; i++) fa[i] = i;
    for (int i = 1; i < n; i++)
        for (int j = i + 1; j <= n; j++)
            if (d[i][j] <= now) fa[findfa(i)] = findfa(j);
    int sum = 0;
    for (int i = 1; i <= n; i++)
        if (fa[i] == i) sum++;
    return sum <= m;
}

int main() {
    int tt; cin >> tt;
    while (tt--) {
        cin >> m >> n;
        for (int i = 1; i <= n; i++)
            scanf("%d%d", &x[i], &y[i]);
        for (int i = 1; i < n; i++)
            for (int j = i + 1; j <= n; j++)
                d[i][j] = (double)sqrt((x[i] - x[j]) * (x[i] - x[j]) + (y[i] - y[j]) * (y[i] - y[j]));
        double l = 0.0, r = 14145.0, mid;
        while (l + eps < r) {
            mid = (l + r) / 2;
            if (check(mid)) r = mid;
            else l = mid;
        }
        printf("%.2lf\n", r);
    }
    return 0;
}