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.

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 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.

 1 #include<stdio.h>
 2 #include<string.h>
 3 #include<algorithm>
 4 using namespace std;
 5 struct as 
 6 {
 7     char s[101];
 8     int m;
 9 }aa[200];
10 bool cmp(as s,as t)
11 {
12     return s.m < t.m;
13 }
14 int main()
15 {
16     int n,i,j,k,t,a,b,q;
17     scanf("%d",&n);
18     while(n--)
19     {
20         scanf("%d%d",&a,&b);
21         getchar();
22         for(k=0;k<b;k++)
23         {
24             
25             scanf("%s",aa[k].s);
26             {
27                 for(t=0;t<a;t++)
28                 {
29                     int sum=0;
30                     for(i=0;i<a-1;i++)
31                     {
32                         for(j=i+1;j<a;j++)
33                         if(aa[t].s[i]>aa[t].s[j])
34                         sum++;
35                     }
36                     aa[t].m=sum;//将各组总数放在结构体中
37                 }
38                 
39             }    
40         }
41         sort(aa, aa+b, cmp);//排序结构体
42         for(i=0;i<b;i++)
43         printf("%s\n",aa[i].s);    
44     }
45     return 0;
46 }