poj1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 60087		Accepted: 23717

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>
using namespace std;
int main()
{
    int n,m;
    int i,j,k;
    char data[101][51];
    int count[101];
    char temp[51];
    int temp_count=0;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<m;i++)
            cin>>data[i];
        for(k=0;k<m;k++)
        {
            count[k]=0;
            for(i=0;i<n;i++)
            {
                for(j=i;j<n;j++)
                {
                    if(data[k][i]>data[k][j])
                        count[k]++;
                }
            }
        }
        for(i=0;i<m;i++)
        {
            for(j=0;j<m-i-1;j++)
            {
                if(count[j]>count[j+1])
                {
                    strcpy(temp,data[j+1]);
                    strcpy(data[j+1],data[j]);
                    strcpy(data[j],temp);
                    temp_count=count[j+1];
                    count[j+1]=count[j];
                    count[j]=temp_count;
                }
            }
        }
        for(i=0;i<m;i++)
        {
            cout<<data[i]<<endl;
        }
    }
    return 0;
}