POJ 1804(最小相邻数移动)

题目大意:对乱序列相邻2数移动,使得用最小步数使其有序

解法:归并排序

定理:

一个乱序序列的逆序数 = 在只允许相邻两个元素交换的条件下,得到有序序列的交换次数

Program P1804;
const
   maxn=1000;
Var
   tt,i,j,k,n,ans:longint;
   a,le,re:array[1..maxn] of longint;

procedure mergesort(l,r:longint);
var
   i,j,k,mid:longint;

begin
   if l=r then exit;
   mid:=(l+r) shr 1;
   mergesort(l,mid);
   mergesort(mid+1,r);
   for i:=l to mid do le[i-l+1]:=a[i];
   le[mid+2-l]:=maxlongint;
   for i:=mid+1 to r do re[i-mid]:=a[i];
   re[r+1-mid]:=maxlongint;
   i:=1;j:=1;
   for k:=l to r do
   begin
      if (le[i]<=re[j]) then
      begin
         a[k]:=le[i];
         inc(i);
      end
      else
      begin
         a[k]:=re[j];
         inc(j);
         inc(ans,mid-l+1-(i-1));
      end;
   end;
end;
Begin
   readln(tt);
   for k:=1 to tt do
   begin
      writeln('Scenario #',k,':');
      read(n);
      for i:=1 to n do read(a[i]);
      ans:=0;
      mergesort(1,n);
      writeln(ans);

      writeln;
   end;
end.