<h1 id="heading">题目</h1>
<blockquote>
<h2 id="description">Description</h2>
<blockquote>
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 i at (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.
 
</blockquote>
<h2 id="input">Input</h2>
<blockquote>
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 &lt; t i+1 ≤ 10 6), x i, and y i (0 ≤ x i, y i ≤ 10 6) as mentioned above.
</blockquote>
<h2 id="output">Output</h2>
<blockquote>
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.
</blockquote>
<h2 id="sample-input">Sample Input</h2>
<blockquote>
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
</blockquote>
<h2 id="sample-output">Sample Output</h2>
<blockquote>
9.2195444573 
54.5893762558
</blockquote>
</blockquote>
<h1 id="heading-1">题解</h1>
两个 note 之间的距离与时间差的比值叫做难度 
求整首歌的最大难度
直接按时间排序算一遍,求最大值即可 
尽量多输出小数保证精度
<h1 id="heading-2">代码</h1>
<pre style="background-color:#fff">/*
By:OhYee
Github:OhYee
Blog:http://www.oyohyee.com/
Email:oyohyee@oyohyee.com

かしこいかわいい？
エリーチカ！
要写出来Хорошо的代码哦~
*/
#include &lt;cstdio&gt;
#include &lt;algorithm&gt;
#include &lt;cmath&gt;
#include &lt;cstring&gt;
#include &lt;iomanip&gt;
#include &lt;iostream&gt;
#include &lt;map&gt;
#include &lt;set&gt;
#include &lt;list&gt;
#include &lt;queue&gt;
#include &lt;stack&gt;
#include &lt;string&gt;
#include &lt;vector&gt;
#include &lt;bitset&gt;
#include &lt;functional&gt;

using namespace std;

const int INF = 0x7FFFFFFF;
const double eps = 1e-10;

const int maxn = 1005;

struct Node {
 double t,x,y;
 bool operator &lt; (const Node &amp;rhs)const {
 return t &lt; rhs.t;
 }
 double operator - (const Node &amp;rhs)const {
 return sqrt((x - rhs.x)*(x - rhs.x) + (y - rhs.y)*(y-rhs.y));
 }
};

Node note[maxn];

void Do() {
 int n;
 cin &gt;&gt; n;
 for(int i = 0;i &lt; n;i++)
 cin &gt;&gt; note[i].t &gt;&gt; note[i].x &gt;&gt; note[i].y;
 sort(note,note + n);

 double Max = 0;
 for(int i = 1;i &lt; n;i++)
 Max = max(Max,(note[i] - note[i - 1]) / (note[i].t - note[i - 1].t));

 cout &lt;&lt; fixed &lt;&lt; setprecision(20) &lt;&lt; Max &lt;&lt; endl;
}

int main() {
 cin.tie(0);
 cin.sync_with_stdio(false);

 int T;
 cin &gt;&gt; T;
 while(T--)
 Do();

 return 0;
}
</pre>

HDU 5078.Osu(2014 鞍山赛区现场赛 I)