hdu杭电 1379 DNA Sorting

Problem 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.


This problem contains multiple test cases!

The first line of a multiple input is an integer N, then a blank line followed by N input blocks. Each input block is in the format indicated in the problem description. There is a blank line between input blocks.

The output format consists of N output blocks. There is a blank line between output blocks.

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 (1 < 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''. If two or more strings are equally sorted, list them in the same order they are in the input file.

Sample Input

1

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

Sample Output

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

//大意是以上面提到的方式排序，比如DAABEC，D比后面四个字母都要大，E比E后面一个大，所以这个字符串的大小是4+1=5，，安排好序的字符串输出

#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
struct stu
{
	char s[100];
	int cnt;
}arr[100];

int cmp(struct stu a,struct stu b)
{
	//if(a.t==b.t) return 0;
	return a.cnt<b.cnt;//升序 
}
int main()
{
	int m,n;
	int i,j,k,l;
	int t;
	scanf("%d",&t);
	getchar();
	while(t--)
	{
		scanf("%d%d",&m,&n);
		getchar();		
		for(i=0;i<100;++i)
		{
			arr[i].cnt=0;
		}
		
		for(i=0;i<n;++i)
		{
			scanf("%s",arr[i].s);
		}
		for(i=0;i<n;i++)//题目中提到的方式 把得数存到一个数组里 
		{
			for(k=0;k<m;k++)
			{
				for(j=k;j<m;j++)
				{
					if(arr[i].s[k]>arr[i].s[j])
						arr[i].cnt++;
				}
			}
		}
		sort(arr,arr+n,cmp);
		for(i=0;i<n;++i)
		{
			printf("%s\n",arr[i].s);
		}
		
	}
	return 0;
}