【AKOJ】1023-DNA Sorting

DNA Sorting

Time Limit:1000MS  Memory Limit:65536K
Total Submit:16 Accepted:9

原题链接

Description

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

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 

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

Sample Input

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

Source

ahstu@ICPC01

#include
#include

struct st    //定义字符串，与其数值的结构体 
{
	char ss[100];
	int count;
}tt[100];

int co(char *ss)   //计算字符串数值 
{
	int coun=0;
	for(int i=0;ss[i]!='\0';i++)
	{
		for(int j=i+1;ss[j]!='\0';j++)
		{
			if(ss[i]>ss[j])
			{coun++;}
		}
	}
	return coun;
}

void swap(char *s,char *t)  //字符串交换 
{
	char ch;
	for(int i=0;i<100;i++)
	{ch=t[i];t[i]=s[i];s[i]=ch;}

}


int main()
{
	int m,n,i,temp;
	
	scanf("%d%d",&m,&n);
	
	for(i=0;itt[k+1].count)
			{temp=tt[k].count;tt[k].count=tt[k+1].count;tt[k+1].count=temp;swap(tt[k].ss,tt[k+1].ss);}  //数值交换的同时，字符串交换 
		}
		
	}
	
	for(i=0;i