Advertisement Problem

Broadcasting stations need to schedule the ads, so no collision occurs. How does an ads schedule software work? Below you can find the basic algorithm.

The Algorithm

Pascal Source Code

program ad;
uses crt;
var
a:array[1..100,1..100]of integer;
c,d:array[1..100]of integer;
i,n,j,t,k,l,yedek,byk,v,old:integer;
begin
  clrscr;
  writeln('number of advertisements');
  readln(n);
  for i:=1 to n do
   begin
    writeln(i,' advertisement...:');
    write('cost...:');readln(a[1,i]);
    write('validity time...:');readln(a[2,i]);
   end;
  for i:=1 to n-1 do
    begin
     for j:=1 to n-i do
       if(a[2,j]>=a[2,j+1]) then
          for k:=1 to 2 do
           begin
            yedek:=a[k,j];
            a[k,j]:=a[k,j+1];
            a[k,j+1]:=yedek;
           end;
    end;
  writeln('sorted array...:');
  write('cost...:  ');
  for i:=1 to n do write(a[1,i]:4);
  writeln;
  write('validity..:');
  for i:=1to n do write(a[2,i]:4);readkey; writeln;
  c[1]:=a[2,1];
  t:=1;
  for i:=2 to n do
   begin
    if (a[2,i]<>a[2,i-1]) then
      begin
t:=t+1;c[t]:=a[2,i];
end;
   end;
  for j:=t downto 1 do
   begin
    i:=n;
    while (a[2,i]>=c[j]) and (i>1) do
    begin
        i:=i-1;  l:=i;
    end;
    byk:=a[1,l];
    old:=l;
    for v:=l to n do
      begin
          if a[1,v]>byk then
              begin
                byk:=a[1,v];
                old:=v;
              end;
      end;
    d[j]:=byk;
    a[1,old]:=0;
  end;
  for i:=0 to t+1 do  begin
  write(d[i],'  ');writeln;end; readkey;
end.

Sample Execution

number of advertisements

1. advertisement...: cost...:30

validity...:4

2. advertisement...: cost...:44

validity...:3

3. advertisement...: cost...:45

validity...:2

4. advertisement...: cost...:50

validity...:5

5. advertisement...: cost...:60

validity...:8

sorted array...:

cost...: 45 44 30 50 60

validity..: 2 3 4 5 8

1. advertisement...:30

2. advertisement...:45

3. advertisement...:44

4. advertisement...:50

5. advertisement...:60