DNA Sorting

1903: DNA Sorting   Time Limit(Common/Java):1000MS/10000MS     Memory Limit:65536KByte Total Submit: 607            Accepted:347 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 (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. Sample Input 10 6 AACATGAAGG TTTTGGCCAA TTTGGCCAAA GATCAGATTT CCCGGGGGGA ATCGATGCAT Sample Output CCCGGGGGGA AACATGAAGG GATCAGATTT ATCGATGCAT TTTTGGCCAA TTTGGCCAAA

#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<algorithm>
using namespace std;
struct node
{
	char str[51];
	int num;//用结构体，num用于记数 
}a[101];
bool cmp(node a,node b)
{
	return a.num < b.num;
}
void Count(int In,int n,int m)
{
	int i,j,measure = 0;
	for(i = 0; i < n - 1; i++)
	{
		if(a[In].str[i] == 'A')
			continue;
		for(j = i + 1; j < n; j++)
		{
			if(a[In].str[i] > a[In].str[j])
				measure++; //左边 大于右边加一 
		}
	}
	a[In].num = measure;
}

int main()
{
	int n,m,i,j,k;
	scanf("%d%d",&n,&m);
	{
		for(i = 0; i < m; i++)
		{
			scanf("%s",a[i].str);
			Count(i,n,m);
		}
		sort(a,a+m,cmp);	
		for(i = 0; i < m ; i++)
			printf("%s\n",a[i].str);
	}
}