DNA Sorting

DNA Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2215 Accepted Submission(s): 1082

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 sequenceDAABEC”, 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 sequenceZWQM” 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'', frommost 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'' toleast 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
本看不懂题目，结果看了别人的解析才知道原来就是看每个字母后面有几个字母比它大，然后再把数目全加起来，按总和的升序排列

#include<stdio.h>
#include <algorithm>
#define max 1001
using namespace std;
struct node
{
    int num;
    char s[max];
}a[max];
bool cmp(node a,node b)
{
    return a.num<b.num;
}
int main()
{
    int i,j,k,z,n,m;
    scanf("%d",&z);
    while(z--)
    {
        scanf("%d%d",&n,&m);
        for(i=0;i<m;i++)
        {
            scanf("%s",&a[i].s);
            a[i].num=0;
            for(j=0;j<n-1;j++)
            {
                for(k=j+1;k<n;k++)
                {
                    if(a[i].s[j]>a[i].s[k])
                    a[i].num++;
                }
            }
        }
        sort(a,a+m,cmp);
        for(i=0;i<m;i++)
        {
            printf("%s\n",a[i].s);
        }   
    }
    return 0;
}