[POJ1007] DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 108298		Accepted: 43370

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

---

水题~

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

struct date
{
    string st;
    int num;
}a[105];
int n, m;
bool cmp(date x, date y){return x.num < y.num;}
signed main()
{
    cin >> n >> m;
    for (register int i = 1 ; i <= m ; i ++)
    {
        cin >> a[i].st;
        for (register int j = 0 ; j < n ; j ++)
        {
            for (register int k = j + 1 ; k < n ; k ++)
            {
                if (a[i].st[j] > a[i].st[k]) a[i].num ++;
            }
        }
    }
    sort (a + 1, a + 1 + m , cmp);
    for (register int i = 1 ; i <= m ; i ++)
        cout<<a[i].st<<endl;
    return 0;
}