hdu5512

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 ti and xi (0 ≤ ti, xi ≤ 106), 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 ti 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

232 21 13 430 31 52 0

 

Sample Output

Case #1: 2.00Case #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：给你每个时刻的位置，求最大速度。
思路：这就不用说了，直接对时间排个序取最大就可以了。
代码：
#include<bits/stdc++.h>using namespace std;const int maxn=10005;struct node{    double time;    double pos;}a[maxn];bool cmp(const node &a,const node &b){    return a.time<b.time;}int main(){    int T;    scanf("%d",&T);    while(T--)    {        int n;        scanf("%d",&n);        for(int i=1;i<=n;i++) scanf("%lf%lf",&a[i].time,&a[i].pos);        sort(a+1,a+1+n,cmp);        double ans=0;        for(int i=2;i<=n;i++)        {            double sum=abs(a[i].pos-a[i-1].pos)/abs(a[i].time-a[i-1].time);            ans=max(ans,sum);        }        static int t=1;        printf("Case #%d: %.2f\n",t++,ans);    }    return 0;}