ACM and STL-DNA Sorting

ACM and STL

DNA Sorting

Time Limit:1000MS Memory Limit:10000K

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

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!!

//By LingCh

//2005-3-19

Analysis

This is a sort problem. As stated, we may first calculate the inversions of each string, and then sort the strings use their inversion as the sort key.

Source

Memory:88K Time:0MS

Language:C++ Result:Accepted

· #include<vector>

· #include<algorithm>

· #include<iostream>

·

· using namespace std;

·

· int n,m;

·

· class Mystring

· {

· public:

· int length;

· int inv;

· char str[50];

·

· Mystring()

· :length(0)

· ,inv(0)

· {

· str[0]=0;

· }

·

· Mystring& operator = (const Mystring& s);

·

· friend istream& operator >>(istream &is,Mystring &istr);

· friend ostream& operator <<(ostream &os,Mystring &ostr);

· };

·

· istream& operator >>(istream &is,Mystring &istr)

· {

· int i,j;

· istr.length=n;

· for(i=0;i<istr.length;i++)

· is>>istr.str[i];

· istr.str[istr.length]=0;

·

//here calculate the inversion of this string

· istr.inv=0;

· for(i=0;i<istr.length-1;i++)

· {

· for(j=i+1;j<istr.length;j++)

· {

· if(istr.str[i]>istr.str[j])

· istr.inv++;

· }

· }

· return is;

· }

·

· Mystring& Mystring::operator = (const Mystring& sr)

· {

· if (&sr != this)

· {

· copy(&sr.str[0],&sr.str[length],this->str);

· inv=sr.inv;

· length=sr.length;

· }

· return *this;

· }

·

· ostream& operator <<(ostream &os,Mystring &ostr)

· {

· int i;

· for(i=0;i<ostr.length;i++)

· os<<ostr.str[i];

· return os;

· }

·

· //this predicate use the inversion of the string as the sort key

· bool MyPre(Mystring s1,Mystring s2)

· {

· if(s1.inv<s2.inv)

· return true;

· else

· return false;

· }

·

· int main()

· {

· int i;

· cin>>n>>m;

·

· vector<Mystring> ss;

· for(i=0;i<m;i++)

· {

· Mystring s;

· cin>>s;

· ss.push_back(s);

· }

·

· sort(ss.begin(),ss.end(),MyPre);

·

· for(i=0;i<m;i++)

· cout<<ss[i]<<endl;

·

· //system("pause");

· }