HDU 1379:DNA Sorting

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2329    Accepted Submission(s): 1145

Problem 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. 


This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

 

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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file. 

 

Sample Input

1

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

 

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

 

Source

East Central North America 1998

 

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#include <stdio.h>
#include <algorithm>
#include <iostream>
#include<string.h>
#include<math.h>
using namespace std;
struct DNA
{
    char c[55];
    int data;
} d[105];
int nixu(char *c)
{
    int k=0;
    int n=strlen(c);
    for(int i=0; i<n; i++)
    {
        for(int j=i+1; j<n; j++)
            if(c[i]>c[j])k++;
    }
    return k;
}
void paixu(int m)
{
    for(int i=0; i<m; i++)
        for(int j=0; j<m-i-1; j++)
            if(d[j+1].data<d[j].data)
            {
                DNA a=d[j+1];
                d[j+1]=d[j];
                d[j]=a;
            }
}
int main()
{
    int t;
    cin>>t;
    while(t--)
    {
        int n,m;
        cin>>n>>m;
        getchar();
        for(int i=0; i<m; i++)
        {
            scanf("%s",d[i].c);
            d[i].data=nixu(d[i].c);
        }
        paixu(m);
        for(int i=0; i<m; i++)
            puts(d[i].c);
    }
    return 0;
}