DNA Sorting--hdu1379

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2203    Accepted Submission(s): 1075

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

这个题开始就没读懂，所以也不会做，后来学长讲完后才明白啥意思;

就是比如第一行数据AACATGAAGG     首先定义sum=0， A大于它后面字符的个数为0，sum+=0，第二个同样，第三个C大于它后面字符的个数为3，sum+=3。。。以此类推！

然后按每行数字从小到大输出字符串！！！

 #include<stdio.h>
 #include<string.h>
 #include<algorithm>
 using namespace std;
 struct as
 {
     char s[];
     int m;
 }aa[];
 bool cmp(as s,as t)
 {
     return s.m < t.m;
 }
 int main()
 {
     int n,i,j,k,t,a,b,q;
     scanf("%d",&n);
     while(n--)
     {
         scanf("%d%d",&a,&b);
         getchar();
         for(k=;k<b;k++)
         {

             scanf("%s",aa[k].s);
             {
                 for(t=;t<a;t++)
                 {
                     int sum=;
                     for(i=;i<a-;i++)
                     {
                         for(j=i+;j<a;j++)
                         if(aa[t].s[i]>aa[t].s[j])
                         sum++;
                     }
                     aa[t].m=sum;//将各组总数放在结构体中
                 }

             }
         }
         sort(aa, aa+b, cmp);//排序结构体
         for(i=;i<b;i++)
         printf("%s\n",aa[i].s);
     }
     return ;
 }