A Curious Matt（2014ACM/ICPC亚洲区北京站-A）

A Curious Matt

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)
Total Submission(s): 0    Accepted Submission(s): 0

Problem 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：T组测试数据，给出n组数据，t为时间，s为此时间下的位置，求最大速度。

 思路：先把时间从小到大排序，然后计算。注意精度（sad）

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <iostream>
using namespace std;
struct node
{
	int t;
	int s;
}a[10010];
int cmp(struct node x,struct node y)
{
	return x.t<y.t;
}

int main()
{
	int T,n,i,j;
	double maxx;
	scanf("%d",&T);
	for(j=1;j<=T;j++)
	{
		scanf("%d",&n);
		for(i=0;i<n;i++)
		{
			scanf("%d %d",&a[i].t,&a[i].s);
		}
		sort(a,a+n,cmp);
		maxx=0.00;
		for(i=0;i<n-1;i++)
		{
		   if((double)(abs(a[i+1].s-a[i].s)/(a[i+1].t-a[i].t))>maxx)
                {
                    maxx=(double)abs(a[i+1].s-a[i].s)/abs(a[i+1].t-a[i].t);
                }
		}
		printf("Case #%d: ",j);
        printf("%.2lf\n",maxx);
	}
	return 0;
}