DNA Sorting(DNA排序)

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

​

#include <iostream>
#define INF 0xffffff    //定义最大地址

using namespace std;

char str[200][200];   //二维数组写入字符串每一个字符的值
int num[200];     //逆序数的值

int main()
{
  int m,n;
  memset(num,0,sizeof(num));
  cin>>n;
  cin>>m;

  num[0] = INF;   //初始化num[0]为最大地址，方便后来的选择排序法输出字符串数组
  for (int i=1;i<=m;i++)
  {
    cin>>str[i];

    //求出逆序数
    for (int j=0;j<n;j++)
    {
      for (int k=j+1;k<n;k++)
      {
        if (str[i][j] > str[i][k])
        {
          num[i]++;
        }
      }
    }

  }

  int p=0;

  //选择排序法，输出字符串数组
  for (int i=1;i<=m;i++)
  {
    for (int j=1;j<=m;j++)
    {
      if (num[j] < num[p])
      {
        p = j;
      }
    }
    cout<<str[p]<<endl;
    num[p] = INF;   //将当前num[p]置为最大地址，方便下一个循环的比较

  }

  system("pause");
  return 0;

}