DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 88585		Accepted: 35595

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

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
struct node
{
    char s[55];
    int sum;
}que[110];
bool cmp(node x,node y)
{

    return x.sum<y.sum;
}
int main()
{
    int m,n;
    while(cin>>n>>m)
    {
        memset(que,0,sizeof(que));
        for(int i=0;i<m;i++)
        {
            cin>>que[i].s;
            for(int j=0;j<n-1;j++)
            {
                for(int k=j+1;k<n;k++)
                {
                    if(que[i].s[k]<que[i].s[j])
                    {
                        que[i].sum++;
                    }
                }
            }
        }
        sort(que,que+m,cmp);
        for(int i=0;i<m;i++)
        {
            cout<<que[i].s<<endl;
        }
    }
    return 0;
}

求逆序数从小到达排列；