1007

DNA Sorting

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 100175 Accepted: 40217

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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAATTTGGCCAAA

Source

East Central North America 1998

#include<iostream>#include<algorithm>#include<string>#include<cstdlib>using  namespace  std;typedef  struct{   string dna;   int count;}DNA;DNA  dna[101];int  cmp(const  void  *a,const  void *b){  DNA  *aa=(DNA *)a;  DNA  *bb=(DNA *)b;  return  aa->count-bb->count;};int  main(void){    int n,m;    cin>>n>>m;    for(int  i=0;i<m;i++)    {       cin>>dna[i].dna;       dna[i].count=0;       for (int j= 0; j<n;j++)       {          for(int k=j+1;k<n;k++)          {             if(dna[i].dna[j]>dna[i].dna[k])              ++dna[i].count;          }       }     }       qsort(dna,m,sizeof(DNA),cmp);       for(int i=0;i<m;i++)       {          cout<<dna[i].dna<<endl;       }       return 0;}