DNA Sorting

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3435 Accepted Submission(s): 1661

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
   题目大意：给你一系列DNA序列，根据他们的逆序数进行排序，按逆序数对由小到大输出，逆序数相等的时候，根据输入的前后顺序输出，
   稍微提一下逆序数，就是当i<j时，a[i]>a[j],比如说AGCT,G < C,G和C就是一对逆序数，其实就是求每个字母比他后面字母大的个数
   思路：要用结构体储存字符串和他本身的逆序数，这里面用到了
    stable_sort()，stable_sort()可保证相等元素的原本相对次序在排序后保持不变。
    而sort（）不能保证这一点。
    AC代码：
    #include <string>  
#include <iostream>  
#include <algorithm>  
#include <cstdio>  
using namespace std;  
struct DNA{  
    string str;  
    int count;   //用来记录这个字符串中的逆序数对 
}w[1005];  
bool cmp(DNA x,DNA y){  
    return x.count < y.count;  
}      
int main(){  
    int s, n, t;  
    cin >> t;  
    while(t--){  
        scanf("%d %d",&s,&n);  
        for(int i = 0; i < n; i++){  
            cin>>w[i].str;
            w[i].count = 0;  
            for(int j = 0; j < s; j++)              
                for(int k = j + 1; k < s; k++)                   
                    if(w[i].str[j] > w[i].str[k]) 
            w[i].count++;                             
        }  
        stable_sort(w, w + n, cmp);  
        for(int i = 0; i < n; i++) 
            cout<< w[i].str <<endl;    
    }  
    return 0;  
}