HDU 5112 A Curious Matt (Java comparator + 快速IO + 控制浮点精度)

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A Curious Matt

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

                                                                                         Total Submission(s): 2395    Accepted Submission(s): 1355

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

  
  
   
   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.

   
   
   
    
  
  

  Source 2014ACM/ICPC亚洲区北京站-重现赛（感谢北师和上交）
题目链接：http://acm.hdu.edu/showproblem.php?pid=5112

import java.util.*;
import java.text.*;
import java.io.*;

public class Main{

    public static int T;
    public static int n;

    static class myScanner {
        BufferedReader br;
        StringTokenizer st;

        public myScanner(InputStream in) {
            br = new BufferedReader(new InputStreamReader(in));
            st = new StringTokenizer("");
        }

        public String nextLine() {
            try {
                return br.readLine();
            } catch(IOException e) {
                return null;
            }
        }

        public boolean hasNext() {
            while (!st.hasMoreTokens()) {
                String s = nextLine();
                if (s == null) {
                    return false;
                }
                st = new StringTokenizer(s);
            }
            return true;
        }

        public String next() {
            hasNext();
            return st.nextToken();
        }

        public int nextInt() {
            return Integer.parseInt(next());
        }

        public double nextDouble() {
            return Double.parseDouble(next());
        }
    }

    static class Data {
        double t, x;
    }

    public static void main(String[] agrs) {
        myScanner in = new myScanner(System.in);
        PrintWriter pw = new PrintWriter(System.out);
        T = in.nextInt();
        for (int ca = 1; ca <= T; ca ++) {
            System.out.print("Case #" + ca + ": ");
            n = in.nextInt();
            Data[] d = new Data[n];
            for (int i = 0; i < n; i ++) {
                d[i] = new Data();
                d[i].t = in.nextDouble();
                d[i].x = in.nextDouble();
            }

            Comparator <Data> cmp = new Comparator<Data>() {
                public int compare(Data a, Data b) {
                    if (a.t > b.t) {
                        return 1;
                    } else if (a.t < b.t) {
                        return -1;
                    } else {
                        if (a.x > b.x) {
                            return 1;
                        } else if (a.x < b.x) {
                            return -1;
                        } else {
                            return 0;
                        }
                    }
                }
            };
            Arrays.sort(d, cmp);
            double ma = 0;
            for (int i = 1; i < n; i ++) {
                ma = Math.max(ma, Math.abs(d[i].x - d[i - 1].x) / (d[i].t - d[i - 1].t));
            }
            pw.println(new DecimalFormat("0.00").format(ma));
            pw.flush();
        }
        pw.close();
    }
}

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