PKU_1007_DNA Sorting

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 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

Source Code

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

struct strings
{
	string s;
	int sortValue;
};


//返回读入数据串的逆序对
int toSortValue(string s)
{
	int i,j,l,sortValue=0;
	l=s.length();
	for (i=0;i<l-1;i++)
		for (j=i+1;j<l;j++)
			if (s[i]>s[j]) sortValue++;
	return sortValue;
}

int main()
{
	int n,m,k,alen=0,i,j;
	strings a[101];

	cin>>n>>m;
	k=m;
	while (k--)
	{
		string s;
		cin>>s;
		int sortValue=toSortValue(s);
		i=0;
		while ((i<alen)&&(a[i].sortValue<=sortValue))                 //插入排序
			i++;
		for (j=alen-1;j>=i;j--)
		{
			a[j+1].sortValue=a[j].sortValue;
			a[j+1].s=a[j].s;
		}
		a[i].sortValue=sortValue;
		a[i].s=s;
		alen++;
	}

	for (i=0;i<alen;i++)
		cout<<a[i].s<<endl;

	return 0;
}