At first: what boundings for sizes of already placed ships?
Can it be 0 or greater than 5?

At second: my solution is not very fast and based on small size and
quantity of already placed ships. Idea is simple: i have two arrays
for verticals and horizontals and sets true if vertical or horizontal
is empty and false otherwise. Then i look for all verticals and
horizontals, trying locate vertical ship in all verticals and
horizontal ship in all horizontals ( if number of decks of placing
ship is 1 then i divide output value by 2 ). If current vertical
(horizontal) line is empty, i just increment total value by "n-
k+1"("m-k+1"), otherwise use full enumeration for line.

I can predict TL, but always got a WA... =(
I searched for bug many times and conclused, that i have bug in
understanding statement of problem. Thank you for reading that.
Could anyone help me? =)

Code below:

{$APPTYPE CONSOLE}
var
  cur, i, j, n, m, l, k, r: longint;
  ships: array[1..100] of
    record
      x, y, d: longint;
      o: char;
    end;
  st: string;
  total: extended;
  curline, vempty, hempty: array[0..30001] of boolean;
{}
begin
{  assign( input, 'p1212.in' ); reset( input );
  assign( output, 'p1212.out' ); rewrite( output );}
  {}
  readln( n, m, l );
  for i := 1 to l do
    begin
      read( ships[i].x, ships[i].y, ships[i].d );
      readln( st );
      if ( ( pos( 'V', st ) = 0 ) and ( pos( 'v', st ) = 0 ) ) then
        ships[i].o := 'H'
      else
        ships[i].o := 'V';
   end;
  read( k );
  {}
  for i := 0 to 30001 do
    begin
      curline[i] := true;
      vempty[i] := true;
      hempty[i] := true;
    end;
  {}
  for i := 1 to l do
    begin
      case ( ships[i].o ) of
        'V':
          begin
            for j := ( ships[i].y - 1 ) to ( ships[i].y + ships
[i].d ) do
              hempty[j] := false;
            vempty[ships[i].x - 1] := false;
            vempty[ships[i].x] := false;
            vempty[ships[i].x + 1] := false;
          end;
        'H':
          begin
            for j := ( ships[i].x - 1 ) to ( ships[i].x + ships
[i].d ) do
              vempty[j] := false;
            hempty[ships[i].y - 1] := false;
            hempty[ships[i].y] := false;
            hempty[ships[i].y + 1] := false;
          end;
      end;
    end;
  {}
  total := 0;
  {}
  for i := 1 to n do
    begin
      if ( hempty[i] ) then
        total := total + m - k + 1
      else
        begin
          for j := 0 to 30001 do
            curline[j] := true;
          curline[0] := false;
          curline[m+1] := false;
          for j := 1 to l do
            begin
              case ( ships[j].o ) of
                'V':
                  begin
                    if ( ( ships[j].y - 1 <= i ) and ( ( ships[j].y +
ships[j].d ) >= i ) ) then
                      begin
                        curline[ships[j].x-1] := false;
                        curline[ships[j].x] := false;
                        curline[ships[j].x+1] := false;
                      end;
                  end;
                'H':
                  begin
                    if ( ( ( ships[j].y - 1 ) = i ) or ( ( ships
[j].y ) = i ) or ( ( ships[j].y + 1 ) = i ) ) then
                      for r := ( ships[j].x - 1 ) to ( ships[j].x +
ships[j].d ) do
                        curline[r] := false;
                  end;
              end;
            end;
          {}
          cur := 0;
          for j := 1 to ( m + 1 ) do
            if ( curline[j] ) then
              inc( cur )
            else
              begin
                if ( cur >= k ) then
                  total := total + cur - k + 1;
                cur := 0;
              end;
        end;
    end;
  {}
  for i := 1 to m do
    begin
      if ( vempty[i] ) then
        total := total + n - k