POJ1007 DNA Sorting

本文探讨了排序算法在DNA序列排序中的应用,通过计算序列的逆序对数量来衡量序列的排序程度,并实现了从最排序到最混乱的DNA序列排序算法。

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DNA Sorting

Time Limit: 1000MS    
Memory Limit: 10000K
Total Submissions: 83488    
Accepted: 33609
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. 

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 (0 < 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''. Since two strings can be equally sorted, then output them according to the orginal order.

Sample Input

```
10 6 
AACATGAAGG 

TTTTGGCCAA 

TTTGGCCAAA 

GATCAGATTT 

CCCGGGGGGA 

ATCGATGCAT
```

Sample Output

```
CCCGGGGGGA 

AACATGAAGG 

GATCAGATTT 

ATCGATGCAT 

TTTTGGCCAA 

TTTGGCCAAA
```


Author:Gannyee 

解题思路:

    将输入的字符串数组用for来比较characterString[i] 的 howGoodOfSort[i]大小,howGoodOfSort[i] 按从小到大的顺序;在进行howGoodOfSort[i]的排序时,characterString[i]也需要跟着howGoodOfSort[i]一同交换i,这样后面characterString[i]输出才会正确。

java 代码:

```
import java.util.Scanner;
public class poj1007 {
        public static void main(String[] args){
                Scanner input = new Scanner(System.in);

                int totalOfEachString = input.nextInt();
                int totalOfLine = input.nextInt();
                int howGoodOfSort[] = new int[totalOfLine];
                String characterString[] = new String[totalOfLine];

                for(int i=0;i < totalOfLine;i++){
                        characterString[i] = input.next();
                        howGoodOfSort[i] = 0;
                        for(int j = 0;j < totalOfEachString;j ++){
                                for(int k = j;k < totalOfEachString;k ++){
                                        if(characterString[i].charAt(j) > characterString[i].charAt(k ))
                                                howGoodOfSort[i] ++;
                                }
                        }
                }

                for(int i =0;i < totalOfLine;i++){
                        for(int j =i;j < totalOfLine;j++){
                                if(howGoodOfSort[i] > howGoodOfSort[j]){
                                        int swap;
                                        String swap1;
                                        swap = howGoodOfSort[i];
                                        howGoodOfSort[i] = howGoodOfSort[j];
                                        howGoodOfSort[j] = swap;
                                        swap1 = characterString[i];
                                        characterString[i] = characterString[j];
                                        characterString[j] = swap1;
                                }
                        }
                        System.out.println(characterString[i]);
                }
        }
}
```

注:互相学习,有不对的地方,请指出,共同探讨学习。

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