poj1007——DNA Sorting

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

题意大概是：设一个字母序列”DAABEC“的逆序数是5，由于D比它右边的4个字母大，而E比它右边的1个字母大。序列”AACEDGG“的逆序数是1。差点儿已经排好序

如今对DNA字符串序列进行分类。然而，分类不是按字母顺序，而是按”排序“的次序，从最多已知排序到最少已知排序

#include<iostream>
#include<cstdio>
#include<algorithm>
using namespace std;
struct aa
{
    char s[55];
    int x;
};
bool cmp(aa m,aa n)
{
    return m.x<n.x;
}
int main()
{
    int m,n,i,j,k;
    char s[10];
    cin>>n>>m;
    aa a[110];
    gets(s);
    for(i=0; i<m; ++i)
    {
        gets(a[i].s);
        a[i].x=0;
        for(j=0; j<n; ++j)
            for(k=j+1; k<n; k++)
            {
                if(a[i].s[j]>a[i].s[k])
                    a[i].x++;
            }
    }
    sort(a,a+m,cmp);
    for(i=0;i<m;++i)
        cout<<a[i].s<<endl;
    return 0;
}