HDU 5112 A Curious Matt

本文介绍了一种通过记录时间点及其对应位置来计算最大瞬时速度的方法。利用排序和简单的数学计算,可以找出两个连续记录间的最大速度变化,进而得出被观测者可能达到的最大速度。

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思路:水题


#include <bits/stdc++.h>
using namespace std;
typedef long long LL;
const int N=10010;
const int INF=0x3f3f3f3f;
int cas=1,T;
int n;
struct node
{
	int t,x;
	bool operator<(const node&a)const
	{
		return t<a.t;
	}
};
node a[N];
int main()
{
	//freopen("1.in","w",stdout);
	//freopen("1.in","r",stdin);
	//freopen("1.out","w",stdout);
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		for(int i=0;i<n;i++) scanf("%d%d",&a[i].t,&a[i].x);
		sort(a,a+n);
		double maxsp=0;
		for(int i=1;i<n;i++) maxsp=max(maxsp,abs(a[i].x-a[i-1].x)/double(a[i].t-a[i-1].t));
		printf("Case #%d: %.2lf\n",cas++,maxsp);
	}
	//printf("time=%.3lf",(double)clock()/CLOCKS_PER_SEC);
	return 0;
}

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. 
         
 


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