poj1007题解

题目描述：

DNA Sorting

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 44730		Accepted: 17445

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

 

 

//accepted
#include"iostream"
#include"stdio.h"
#include"string.h"
using namespace std;

typedef struct st_chars
{
 int count;
 char s[50];
}st_chars;

int main()
{
 int n;//串长度
 int m;//串个数
 cin>>n;
 cin>>m;
 st_chars *chars=new st_chars[m];
 int i;
 int j;
 for( i=0;i<m;i++)
 {
  chars[i].count=0;
  cin>>chars[i].s;
 }
 for( j=0;j<m;j++)
 {
  
  for(i=0;i<n-1;i++)
  {
   int k=0;
   while(k<n-1-i)
   {
  if(chars[j].s[i]>chars[j].s[i+k+1]){++chars[j].count;++k;}
    else ++k;
   }
  }
 }
 st_chars temp;
 for(i=0;i<m-1;i++)
 {
  if(chars[i].count<chars[i+1].count)continue;
  else
  {
   temp=chars[i+1];
   j=i;
   while(chars[j].count>temp.count)
   {
   chars[j+1]=chars[j];
    --j;
   }

   chars[j+1]=temp;
  }
 }
 for(i=0;i<m;i++)
  cout<<chars[i].s<<endl;

 return 1;
}