其实那个oj我不知道叫什么名字
这道题,看完就知道要二分答案了
然后变成判定性问题,其实就是把点分组
假设现在得到的答案是cost,那么如果2个点的垂直距离>2*cost,则这2个点不能分在一组上

/*************************************************************************
    > File Name: code.cpp
    > Author: ALex
   
    > Created Time: 2015年05月01日 星期五 18时01分11秒
 ************************************************************************/

#include <functional>
#include <algorithm>
#include <iostream>
#include <fstream>
#include <cstring>
#include <cstdio>
#include <cmath>
#include <cstdlib>
#include <queue>
#include <stack>
#include <map>
#include <bitset>
#include <set>
#include <vector>

using namespace std;

const double pi = acos(-1.0);
const int inf = 0x3f3f3f3f;
const double eps = 1e-15;
typedef long long LL;
typedef pair <int, int> PLL;

struct POINT {
    int x, y;
}point[10010];


int cmp(POINT a, POINT b) {
    return a.x < b.x;
}

int main() {
    int t;
    scanf("%d", &t);
    while (t--) {
        int n, k;
        scanf("%d%d", &n, &k);
        for (int i = 1; i <= n; ++i) {
            scanf("%d%d", &point[i].x, &point[i].y);
        }
        sort(point + 1, point + n + 1, cmp);
        double l = 0, r = (double)inf, mid;
        int use = 0;
        double ans;
        while (r - l > 1e-6) {
            mid = (l + r) / 2;
            use = 1;
            int mins = point[1].y, maxs = point[1].y;
            for (int i = 1; i <= n; ++i) {
                mins = min(mins, point[i].y);
                maxs = max(maxs, point[i].y);
                if (maxs - mins > 2 * mid) {
                    ++use;
                    mins = point[i].y;
                    maxs = point[i].y;
                }
            }
            if (use > k) {
                l = mid;
            }
            else {
                ans = mid;
                r = mid;
            }
        }
        printf("%.1f\n", ans);
    }
    return 0;
}