poj 1007 DNA Sorting

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.

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 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.

10 6
AACATGAAGG
TTTTGGCCAA
TTTGGCCAAA
GATCAGATTT
CCCGGGGGGA
ATCGATGCAT

CCCGGGGGGA
AACATGAAGG
GATCAGATTT
ATCGATGCAT
TTTTGGCCAA
TTTGGCCAAA

East Central North America 1998

简单的算逆序数  再按从大到小一次输出    第一次用结构体。。。。。。

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

struct H
{
	int l;
	char s[55];
}a[105];

int cmp(H b,H c)
{
	return b.l<c.l;
}

int main()
{
	int i,j,k,n,m;
	while(cin>>n>>m)
	{
		memset(a,0,sizeof(a));
		for(i=0;i<m;i++)
		{
			scanf("%s",&a[i].s);
			//printf("%d\n",a[i].l);
		}
		for(i=0;i<m;i++)
		{
			for(k=0;k<n;k++)
			{	for(j=k+1;j<n;j++)
				{
					if(a[i].s[k]>a[i].s[j])
						a[i].l++;
					
				}
			//cout<<a[i].l<<" ";
			}
				//cout<<endl;
				
		}
		sort(a,a+m,cmp);
		for(i=0;i<m;i++)
		{
			//printf("%d\n",a[i].l);
			puts(a[i].s);
		}
	}


	return 0;
}