POJ 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 96532		Accepted: 38867

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

Source

East Central North America 1998

 

 

#include <string>
#include <iostream>
#include <algorithm>
using namespace std;
struct DNA
{
    string str;
    int r;
}d[105];
bool cmp(const DNA &a,const DNA &b)
{
    return a.r<b.r;
}
int main()
{
    int n,m,num;
    while(cin>>n>>m)
    {
        for(int i=0;i<m;i++)
        {
            cin>>d[i].str;
            num=0;
            for(int j=0;j<n;j++)
            {
                for(int k=j+1;k<n;k++)
                    if(d[i].str[j]>d[i].str[k])
                        num++;
            }
            d[i].r=num;
        }
        sort(d,d+m,cmp);
        for(int i=0;i<m;i++)
            cout<<d[i].str<<endl;
    }
}