杭电1379（DNA Sorting）java面向对象编程

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Problem Description

One measure of ``unsortedness'' in a sequence is the number of pairs of entries that are out of order with respect to each other. For instance, in the letter sequence ``DAABEC'', this measure is 5, since D is greater than four letters to its right and E is greater than one letter to its right. This measure is called the number of inversions in the sequence. The sequence ``AACEDGG'' has only one inversion (E and D)--it is nearly sorted--while the sequence ``ZWQM'' has 6 inversions (it is as unsorted as can be--exactly the reverse of sorted).

You are responsible for cataloguing a sequence of DNA strings (sequences containing only the four letters A, C, G, and T). However, you want to catalog them, not in alphabetical order, but rather in order of ``sortedness'', from ``most sorted'' to ``least sorted''. All the strings are of the same length.


This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

Input

The first line contains two integers: a positive integer n (0 < n <= 50) giving the length of the strings; and a positive integer m (1 < m <= 100) giving the number of strings. These are followed by m lines, each containing a string of length n.

 

Output

Output the list of input strings, arranged from ``most sorted'' to ``least sorted''. If two or more strings are equally sorted, list them in the same order they are in the input file.

 

Sample Input

110 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

 

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

面向对象编程，虽然看上去麻烦，但思路清晰。

import java.util.*;class Main{public static void main(String[] args){int n,m,i,j,t;String str;Scanner sc=new Scanner(System.in);t=sc.nextInt();while(t-->0){n=sc.nextInt();m=sc.nextInt();sc.nextLine();StringSort[] strs=new StringSort[m];for(i=0;i<m;i++){str=sc.nextLine();strs[i]=new StringSort(str);}for(i=0;i<m;i++){strs[i].measure();}sort(strs);for(i=0;i<m;i++){System.out.println(strs[i].str);}}}public static void sort(StringSort[] strs){for(int i=0;i<strs.length-1;i++){int minIndex=i;for(int j=i+1;j<strs.length;j++){if(strs[j].count<strs[minIndex].count){minIndex=j;}}if(minIndex!=i){StringSort temp=strs[i];strs[i]=strs[minIndex];strs[minIndex]=temp;}}}}class StringSort{String str;int count;public StringSort(String str){this.str=str;this.count=0;}public void measure(){for(int i=0;i<str.length()-1;i++){for(int j=i+1;j<str.length();j++){if(str.charAt(i)>str.charAt(j)){count++;}}}}}