POJ - 1007 DNA Sorting

[align=center]DNA Sorting[/align]

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 80800		Accepted: 32513

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
East Central North America 1998

题意：就是求出字符串的逆序数，然后排序输出，简单题。

#include "iostream"
#include "cstdio"
#include "cstring"
#include "cmath"
#include "algorithm"
using namespace std;
struct opp
{
char str[200];
int k;
}a[200];
int m, n;

int pri(char* s)
{
int i, j;
int sum = 0;
for(i = 0; i <= m-2; i++)//比较m-1次
for(j = i+1; j <= m-1; j++)
{
if(s[i] > s[j]) sum++;
}
return sum;
}
bool com(struct opp a, struct opp b)
{
return a.k < b.k;
}
int main()
{
//freopen("in.txt", "r", stdin);
scanf("%d %d", &m, &n);
int i, j;
for(i = 1; i <= n; i++)
{
scanf("%s", a[i].str);
a[i].k = pri(a[i].str);
}
sort(a+1, a+n+1, com);
for(i = 1; i <= n; i++)
cout<<a[i].str<<endl;
return 0;
}