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 6AACATGAAGGTTTTGGCCAATTTGGCCAAAGATCAGATTTCCCGGGGGGAATCGATGCAT

Sample Output

CCCGGGGGGAAACATGAAGGGATCAGATTTATCGATGCATTTTTGGCCAA
TTTGGCCAAA
#include<iostream>#include<string>using namespace std;int main(){int n, m;cin >> n >> m;//m表示行  n表示列//char **a = new char *[m];//动态创建行数//for (int i = 0; i < m; i++)//动态创建列数//{//a[i] = new char[n];//}int *countLine = new int[m];string *str = new string[m];for (int i = 0; i < m; i++){cin >> str[i];countLine[i] = 0;}for (int i = 0; i < m; i++)//每一行{for (int k = 0; k < n; k++){char temp = str[i][k];for (int j = k; j < n; j++)//判断这一行的字母有几个小于该字母的{if (temp>str[i][j]){countLine[i]++;}}}}for (int i = 1; i < m; i++){for (int j = 0; j < m-i; j++){if (countLine[j] > countLine[j + 1]){int temp = countLine[j];countLine[j] = countLine[j + 1];countLine[j + 1] = temp;string tempStr = str[j];str[j] = str[j + 1];str[j + 1] = tempStr;}}}for (int i = 0; i < m; i++){cout << str[i] << endl;}return 0;}