http://acm.hdu.edu.cn/showproblem.php?pid=5078

现场最水的一道题 连排序都不用，由于说了ti<ti+1

//#pragma comment(linker, "/STACK:102400000,102400000")

#include <cstdio>

#include <cstring>

#include <algorithm>

#include <string>

#include <iostream>

#include <iomanip>

#include <cmath>

#include <map>

#include <set>

#include <queue>

using namespace std;

#define ls(rt) rt*2

#define rs(rt) rt*2+1

#define ll long long

#define ull unsigned long long

#define rep(i,s,e) for(int i=s;i<e;i++)

#define repe(i,s,e) for(int i=s;i<=e;i++)

#define CL(a,b) memset(a,b,sizeof(a))

#define IN(s) freopen(s,"r",stdin)

#define OUT(s) freopen(s,"w",stdout)

const ll ll_INF = ((ull)(-1))>>1;

const double EPS = 1e-8;

const double pi = acos(-1.0);

const int INF = 100000000;

const int MAXN = 1000+20;

double t[MAXN],x[MAXN],y[MAXN];

struct Node {

    double t,x,y;

}a[MAXN];

bool cmp(Node a, Node b){

    return a.t<b.t;

}

int main()

{

    int ncase,n;

    scanf("%d",&ncase);

    while(ncase--)

    {

        scanf("%d",&n);

        double ans=0.0;

        for(int i=0;i<n;i++)

        {

            scanf("%lf%lf%lf",&a[i].t,&a[i].x,&a[i].y);

        }

        for(int i=0;i<n-1;i++)

            ans=max(ans,sqrt( (a[i].x-a[i+1].x)*(a[i].x-a[i+1].x)+(a[i].y-a[i+1].y)*(a[i].y-a[i+1].y) )/(a[i+1].t-a[i].t));

        printf("%.10lf\n",ans);

    }

    return 0;

}