There is about nmax*nmax*smax=40 844 804 iterations and time must be about 0.1,0.2 sec. But program works >0.5 sec.
How I can to optimize my code?

const nmax=101;
      kmax=4;
      smax=4004;
      type myint=0..smax+1;
var
 ans:array[1..nmax,1..nmax] of boolean;
 count,s,t:myint;
 n,m,i,j,k,x,y,ii,jj:myint;
 command:array[1..nmax,1..nmax,1..kmax,1..2] of byte;
 memory:array[1..nmax,1..nmax,1..smax] of boolean;
 list:array[1..smax] of byte;

begin
 readln(n,m);

 for k:=1 to kmax do begin
  for i:=1 to n do begin
   for j:=1 to m do read(command[i,j,k,1],command[i,j,k,2]);
  end;
 end;

 read(s);
 for i:=1 to s do read(list[i]);

 for t:=1 to s do begin
  for i:=1 to n do begin
   for j:=1 to m do memory[i,j,t]:=false;
  end;
 end;

 count:=0;
 for i:=1 to n do begin
 for j:=1 to m do begin
  ans[i,j]:=false;
 end;
 end;

 for i:=1 to n do begin
 for j:=1 to m do begin
  t:=1;
  ii:=i;
  jj:=j;
  while(t<=s)and(not memory[ii,jj,t]) do begin
   memory[ii,jj,t]:=true;
   x:=command[ii,jj,list[t],1];
   y:=command[ii,jj,list[t],2];
   ii:=x;
   jj:=y;
   inc(t);
  end;
  if(t>s)and(not ans[ii,jj]) then begin
   inc(count);
   ans[ii,jj]:=true;
  end;
 end;
 end;

 writeln(count);
 for i:=1 to n do begin
 for j:=1 to m do begin
  if(ans[i,j]) then writeln(i,' ',j);
 end;
 end;
end.

Edited by author 04.01.2017 23:24
Now AC.
When I used a lot of memory, I got TL32. When I used a little memory, I got AC.Perhaps the cache helped.

Edited by author 23.06.2017 19:24

I really can't solve this task?
Alwaus got TLE on test 39 :-(
Can someone give me hint or ...

You sure When not disbalance on Pascal and C++ on this Problem?

Link for you
http://acm.timus.ru/rating.aspx?space=1&num=1386&count=16

P.S. I got AC

I got WA on test 3.
I just run over all positions from which player can start game and then perform list of instruction.

#include<stdio.h>

#define FOR(i,n) for(i=0;i<(n);i++)

int s,i,j,n,m,k,ci,cj,cnt;
int a[4][110][110][2];
bool ans[110][110];
int c[4010];

int main(void)
{
    scanf("%d %d",&n,&m);
    FOR(k,4)
        FOR(i,n)FOR(j,m)
        {
            scanf("%d %d",&a[k][i][j][0],&a[k][i][j][1]);
            a[k][i][j][0]--;
            a[k][i][j][1]--;
        }
    scanf("%d",&s);
    FOR(k,s){scanf("%d",&c[k]);c[k]--;}
    memset(ans,0,sizeof(ans));
    FOR(i,n)FOR(j,m)
    {
        ci=i;cj=j;
        FOR(k,s)
        {
            ci=a[c[k]][ci][cj][0];
            cj=a[c[k]][ci][cj][1];
        }
        ans[ci][cj]=1;
    }
    cnt=0;
    FOR(i,n)FOR(j,m)if(ans[i][j])cnt++;
    printf("%d\n",cnt);
    FOR(i,n)FOR(j,m)if(ans[i][j])printf("%d %d\n",i+1,j+1);
    return 0;
}