A Curious Matt

Time Limit: 2000/2000 MS (Java/Others)    Memory Limit: 512000/512000 K (Java/Others)

Total Submission(s): 25    Accepted Submission(s): 25

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 ≤ 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 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 ?rst 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.

Source

2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）

水题。

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
using namespace std;

#define maxn 10010

struct Node {
        int t, x;
} E[maxn];

bool cmp(Node a, Node b) {
        return a.t < b.t;
}

int main() {
        int T, N, i, j;
        double ans;
        scanf("%d", &T);
        for(int cas = 1; cas <= T; ++cas) {
                scanf("%d", &N);
                for(i = 0; i < N; ++i)
                        scanf("%d%d", &E[i].t, &E[i].x);
                sort(E, E + N, cmp);
                for(ans = 0.0, i = 1; i < N; ++i) {
                        ans = max(ans, abs(E[i].x - E[i-1].x) * 1.0 / (E[i].t - E[i-1].t));
                }
                printf("Case #%d: %.2lf\n", cas, ans);
        }
        return 0;
}