PKU 1007 DNA Sorting

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 47618		Accepted: 18598

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

code：

]#include <iostream>
#include <algorithm>
using namespace std;
struct node
{
int inver;
char DNA[52];
}dna[105];
bool cmp(node a,node b)
{
return a.inver<b.inver;
}
int inverse(const char * a,int len)
{
int count=0;
int dna[4]={0};
for(int i=len-1;i>=0;i--)
{
switch(a[i])
{
case 'A':
dna[1]++;
dna[2]++;
dna[3]++;
break;
case 'C':
dna[2]++;
dna[3]++;
count+=dna[1];
break;
case 'G':
dna[3]++;
count+=dna[2];
break;
case 'T':
count+=dna[3];
}
}
return count;
}
int main()
{
int n;int m;
int i,j;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i<m;i++)
{
scanf("%s",&dna[i].DNA);
dna[i].inver=inverse(dna[i].DNA,n);
}
sort(dna,dna+m,cmp);
for(i=0;i<m;i++)
{
printf("%s/n",dna[i].DNA);
}
}
return 0;
}