POJ 1007

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 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

Source
East Central North America 1998
//696K 0MS
#include <iostream>
#include <string>
#include <vector>
using namespace std;

int main()
{
int m, n;
cin >> n >> m;
vector<string>a(m);
a.reserve(m);
vector<int>t(m);
t.reserve(m);
vector<string>tem(m);
tem.reserve(m);
char temp1;
for (int i = 0; i < m; i++)
{
a.push_back(a[i]);
cin >> a[i];
tem.push_back(tem[i]);
tem[i] = a[i];
}
t[0] = 0;
for (int i = 0; i < m; i++)
{
for (int p = n; p > 0; p--)
{
for (int j = 0; j < n; j++)
{
if (a[i][j] > a[i][j + 1])
{
temp1 = a[i][j];
a[i][j] = a[i][j + 1];
a[i][j + 1] = temp1;
t.push_back(t[i]);
t[i]++;
}
}
}
}
string temp2;
int temp3;
for (int l = m; l > 0; l--)
{
for (int k = 0; k < m; k++)
{
if (t[k] < t[k + 1])
{
temp3 = t[k];
t[k] = t[k + 1];
t[k + 1] = temp3;
temp2 = tem[k];
tem[k] = tem[k + 1];
tem[k + 1] = temp2;
}
}
}
for (int i=m-1; i >=0; i--)
{
cout << tem[i] << endl;
}

return 0;
}