POJ 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 82585		Accepted: 33245

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

#include <iostream>
#include<algorithm>
using namespace std;

struct DNA{
	char s[55];
	int count;
}d[150];

bool comp(struct DNA a,struct DNA b){
	return a.count<b.count;
}
int main(){
	int n,m;
	cin>>n>>m;
	for(int i=0;i<m;i++){
		cin>>d[i].s;
		int count=0;
		for(int j=0;j<n-1;j++){
			for(int k=j+1;k<n;k++){
				if(d[i].s[j]>d[i].s[k])
					count++;
			}
		}
		d[i].count=count;
	}
	sort(d,d+m,comp);
	for(int i=0;i<m;i++)
		cout<<d[i].s<<endl;
	return 0;
}