poj 1007

DNA Sorting

Time Limit: 1000MS
Memory Limit: 10000K

Total Submissions: 52558
Accepted: 20563

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

求逆序对

归并排序

 
var
 num,len,i,j,l,anss:longint;
 f:array['A'..'Z'] of integer;
 ff:array[1..26] of char;
 ans:array[1..100] of longint;
 a,temp1,temp2:array[0..50] of longint;
 b:array[1..100,0..50] of longint;
 ch:char;
procedure merge(l,r:longint);
 var
  i,j,mid,l1,l2,k:longint;
 begin
  mid:=(l+r)>>1;
  if l<mid then merge(l,mid);
  if mid+1<r then merge(mid+1,r);
  l1:=mid-l+1;
  l2:=r-mid;
  for i:=1 to l1 do
   temp1[i]:=a[l+i-1];
 for i:=1 to l2 do
  temp2[i]:=a[mid+i];
 temp1[l1+1]:=maxlongint;
 temp2[l2+1]:=maxlongint;
 i:=1; j:=1;
 for k:=l to r do
  begin
   if temp1[i]<=temp2[j] then
    begin
     a[k]:=temp1[i];
     inc(i);
    end
   else
   begin
    a[k]:=temp2[j];
    inc(j);
    inc(anss,l1-i+1);
   end;
  end;
end;
procedure qs(r,l:longint);
 var
  i,j,x,y:longint;
  temp:array[0..50] of longint;
 begin
  i:=r; j:=l;
  x:=ans[(i+j) div 2];
  repeat
   while ans[i]<x do inc(i);
   while ans[j]>x do dec(j);
    if i<=j then
     begin
      y:=ans[i];
      ans[i]:=ans[j];
      ans[j]:=y;
      temp:=b[i];
      b[i]:=b[j];
      b[j]:=temp;
      inc(i);
      dec(j);
     end;
  until i>j;
if i<l then qs(i,l);
if j>r then qs(r,j);
end;
begin
 assign(input,'input.in');
 assign(output,'output.out');
 reset(input);
 rewrite(output);
 readln(len,num);
 i:=1;
 for ch:='A' to 'Z' do
  begin
  f[ch]:=i; inc(i);
  end;
 ff[1]:='A';
 ff[7]:='G';
 ff[3]:='C';
 ff[20]:='T';
 for  i:=1 to num do
  begin
   a[0]:=0;
  for  j:=1 to len do
    begin
     read(ch);
     inc(a[0]);
     a[a[0]]:=f[ch];
    end;
    readln;
  b[i]:=a;
  anss:=0;
  merge(1,len);
  ans[i]:=anss;
 end;
qs(1,num);
for i:=1 to num do
 begin
  for j:=1 to len do
   write(ff[b[i,j]]);
  writeln;
 end;
close(input);
close(output);
end.