acm pku 1007 DNA Sorting

DNA Sorting

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

 

#include<iostream>
#include<algorithm>

using namespace std;

char dna[100][51];

struct DNA
...{
    char s[50];
    int num;
};
struct sort_f
...{
    bool operator() (DNA d1,DNA d2)
    ...{
        if(d1.num!=d2.num)
            return d1.num<d2.num;
        return strcmp(d1.s,d2.s)<0;
    }
};
int main()
...{
    int n,m,a,c,g,t;
    DNA dna[100];
    cin>>m>>n;
    for(int pi=0;pi<n;++pi)
    ...{
        cin>>dna[pi].s;
        dna[pi].num=0;
    }
    for(int pj=0;pj<n;++pj)
    ...{
        a=0,c=0,g=0,t=0;
        for(int pk=m-1;pk>=0;--pk)
        ...{
            if(dna[pj].s[pk]=='A')
            ...{
                ++a;
            }
            else if(dna[pj].s[pk]=='C')
            ...{
                dna[pj].num+=a;
                ++c;
            }
            else if(dna[pj].s[pk]=='G')
            ...{
                dna[pj].num+=a+c;
                ++g;
            }
            else if(dna[pj].s[pk]=='T')
            ...{
                dna[pj].num+=a+c+g;
                ++t;
            }
        }
    }
    sort_f sf;
    sort(dna,dna+n,sf);
    for(int i=0;i<n;++i)
        cout<<dna[i].s<<endl;
    return 0;
}