2014ACM/ICPC亚洲区鞍山赛区现场赛1009Osu!

鞍山的签到题，求两点之间的距离除以时间的最大值。直接暴力过的。

A - Osu!

Time Limit:1000MS     Memory Limit:262144KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 5078

Appoint description:  System Crawler  (2014-10-22)

Description

Osu! is a very popular music game. Basically, it is a game about clicking. Some points will appear on the screen at some time, and you have to click them at a correct time. 


Now, you want to write an algorithm to estimate how diffecult a game is. 

To simplify the things, in a game consisting of N points, point i will occur at time t  _i at place (x  _i, y  _i), and you should click it exactly at t  _iat (x  _i, y  _i). That means you should move your cursor from point i to point i+1. This movement is called a jump, and the difficulty of a jump is just the distance between point i and point i+1 divided by the time between t  _i and t  _i+1. And the difficulty of a game is simply the difficulty of the most difficult jump in the game. 

Now, given a description of a game, please calculate its difficulty.

 

Input

The first line contains an integer T (T ≤ 10), denoting the number of the test cases. 

For each test case, the first line contains an integer N (2 ≤ N ≤ 1000) denoting the number of the points in the game.  Then N lines follow, the i-th line consisting of 3 space-separated integers, t  _i(0 ≤ t  _i < t  _i+1 ≤ 10  ⁶), x  _i, and y  _i (0 ≤ x  _i, y  _i ≤ 10  ⁶) as mentioned above.

 

Output

For each test case, output the answer in one line. 

Your answer will be considered correct if and only if its absolute or relative error is less than 1e-9.

 

Sample Input

2

5

2 1 9

3 7 2

5 9 0

6 6 3

7 6 0

10

11 35 67

23 2 29

29 58 22

30 67 69

36 56 93

62 42 11

67 73 29

68 19 21

72 37 84

82 24 98

 

Sample Output

9.2195444573

54.5893762558

Hint

In memory of the best osu! player ever Cookiezi.

 1 #include<cstdio>
 2 #include<algorithm>
 3 #include<cmath>
 4 #include<stdlib.h>
 5 using namespace std;
 6 const int maxn=1000+10;
 7 struct
 8 {
 9     double t;
10     double x;
11     double y
12     ;
13 }Node[maxn];
14 int main()
15 {
16     int T;
17     scanf("%d",&T);
18     while(T--)
19     {
20         int N;
21         double maxd=-1;
22         scanf("%d",&N);
23         for(int i=0;i<N;i++)
24             scanf("%lf%lf%lf",&Node[i].t,&Node[i].x,&Node[i].y);
25         for(int i=0;i<N;i++)
26         {
27             for(int j=i+1;j<N;j++)
28             {
29                 double time=sqrt((Node[i].x-Node[j].x)*(Node[i].x-Node[j].x)+(Node[i].y-Node[j].y)*(Node[i].y-Node[j].y))/fabs(Node[i].t-Node[j].t);
30                 maxd=max(maxd,time);
31             }
32         }
33         printf("%.10lf\n",maxd);
34     }
35     return 0;
36 }