文章标题 HDU 5112- A Curious Matt

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A Curious Matt

Description

There is a curious man called Matt.

One day, Matt’s best friend Ted is wandering on the non-negative half of the number line. Matt finds it interesting to know the maximal speed Ted may reach. In order to do so, Matt takes records of Ted’s position. Now Matt has a great deal of records. Please help him to find out the maximal speed Ted may reach, assuming Ted moves with a constant speed between two consecutive records.

Input

The first line contains only one integer T, which indicates the number of test cases.

For each test case, the first line contains an integer N (2 ≤ N ≤ 10000),indicating the number of records.

Each of the following N lines contains two integers t i and x i (0 ≤ t i, x i ≤ 10 6), indicating the time when this record is taken and Ted’s corresponding position. Note that records may be unsorted by time. It’s guaranteed that all t i would be distinct.

Output

For each test case, output a single line “Case #x: y”, where x is the case number (starting from 1), and y is the maximal speed Ted may reach. The result should be rounded to two decimal places.

Sample Input

2
3
2 2
1 1
3 4
3
0 3
1 5
2 0

Sample Output

Case #1: 2.00
Case #2: 5.00

Hint

In the first sample, Ted moves from 2 to 4 in 1 time unit. The speed 2/1 is maximal.
In the second sample, Ted moves from 5 to 0 in 1 time unit. The speed 5/1 is maximal.

题意：有一些记录点的位置x和在这个位置的时间t，先给你的这些点的时间有可能是乱的，要我们求两个位置之间的最大速度。
分析：以时间为第一关键升序排序，然后从第二个时间点开始算从前一个点到这个点的速度是多少，然后将最大速度记录下来。
代码：

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<vector>
#include<math.h>
#include<map>
#include<queue> 
#include<algorithm>
using namespace std;
const int inf = 0x3f3f3f3f;
struct node {
    int t,x;
    bool operator < (const node &a) const{
        return t<a.t;
    }
};
node v[10005];
int main ()
{
    int t,n;
    scanf ("%d",&t);
    int cnt=1;
    while (t--){
        scanf ("%d",&n);
        for (int i=0;i<n;i++){
            scanf ("%d%d",&v[i].t,&v[i].x);     
        }
        sort (v,v+n);
        double ans=0;
        for (int i=1;i<n;i++){
            double tmp=((double)fabs(v[i].x-v[i-1].x)/(v[i].t-v[i-1].t));//速度
            ans=max(ans,tmp);//取大值
        }
        printf ("Case #%d: %.2f\n",cnt++,ans);
    }
    return 0;
}

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