In my solution, I used sentinel 10001, because according to the statement  0 ≤ x, y ≤ 10000.
I had runtime error (for that reason). when I replaced 10001 with some big number, got AC.
So, I think in Test 12, There is Point with x coordinate 10001.

Edited by author 16.10.2013 15:42

And now WA#13
        Baker's dozen!

Edited by author 25.04.2006 19:43

В слове "подаваться"

На выход должно подвааться единственное целое число
->
На выход должно подаваться единственное целое число

I got WA#13 and I don't understand what's wrong in my program. I try tests from UNKNOWN_LAMER, and my program passed them. Please give me some interesting tests!

Try this test:
2
1 0
2 0

P.S. Though N>2,this can appear in a little big test.

Anyone with the same WA? I think Java is starting to be the problem...

Is there a trick? I must forget something, because i tried 3 different implementations for finding the current convex hull of the points, and all three get WA 10...

My algorithm is to find the convex hull until the numbers of the point on the convex hull is less than 3.
Could anyone tell me what's wrong?or give me some useful test case?

I tried all tests which are posted here and it gives me correct answer, but I have WA #3!!!
Someone help me, please!
Can you post some more tests???

1)
12
1 0
0 1
0 2
2 0
2 1
1 2
1 3
3 1
1 1
2 2
2 3
3 2
Answer:2

2)
6
0 0
4 0
2 0
2 4
1 2
3 2
Answer:1
3)
7
0 0
0 100
100 0
100 100
50 50
50 25
50 75
Answer:1

Good luck!

I'm a newbie, and I don't know why i'm wrong at test #3. Thanks in advanced.
const inp='';
      out='';
var i,c,count,m,n:longint;
    x1,y1,p,x,y:array[1..4005] of integer;
    mark:array[1..4005] of 0..1;
    f:text;

function lower(i,j:integer):boolean;
begin
{ lower:=false;
 if (y[i]=y[1]) and (y[j]=y[1]) then if x[i]<x[j] then lower:=true
 else if(x[i]=x[1]) and (x[j]=x[1]) then if y[i]>y[j] then lower:=true
 else                                                   }
 lower:=(y[i]-y[1])*(x[j]-x[1])<(x[i]-x[1])*(y[j]-y[1]);
end;

procedure swap(i,j:integer);
var t:integer;
begin
 t:=x[i]; x[i]:=x[j]; x[j]:=t;
 t:=y[i]; y[i]:=y[j]; y[j]:=t;
end;

{----------------------------------------------------------------------}
procedure Qsort(left,right:integer);
var l,r,m:integer;
begin
 l:=left; r:=right; m:=(l+r) div 2;
 while l<r do
 begin
       while lower(l,m) do inc(l);
       while lower(m,r) do dec(r);
       if (l<=r) then
        begin
          swap(l,r);
          inc(l); dec(r);
        end;
 end;
 if left<r then qsort(left,r);
 if l<right then qsort(l,right);
end;
{----------------------------------------------------------------------}
procedure input;
var f:text;
    i,j:integer;
begin
 assign(f,inp); reset(f);
 readln(f,n);
 readln(f,x[1],y[1]);
 p[1]:=1;
 for i:=2  to n do
  begin
     readln(f,x[i],y[i]);
  end;
end;
{----------------------------------------------------------------------}
function CCW(i,j,k:integer):integer;
var t,a1,b1,a2,b2:integer;
begin
   a1:=x[j]-x[i]; b1:=y[j]-y[i];
   a2:=x[k]-x[j]; b2:=y[k]-y[j];
   t:=a1*b2-a2*b1;
   if t=0 then ccw:=0 else if t>0 then ccw:=1 else ccw:=-1;
end;
{----------------------------------------------------------------------}
procedure find;
var t,i,j:integer;
begin
 t:=1;
 for i:=2 to n do
     if y[i]<y[t] then t:=i else if y[i]=y[t] then if x[i]>x[t] then t:=i;
 swap(1,t);
 Qsort(2,n);
 j:=2;
 m:=2;p[1]:=1; p[2]:=2;
 for i:=3 to n do
  begin
     while (ccw(p[m-1],p[m],i)<>1) and (m>1) do dec(M);
     inc(m);
     p[m]:=i;
  end;
 fillchar(mark,sizeof(mark),0);
 mark[1]:=1;
 for i:=1 to m do mark[p[i]]:=1;
end;

begin
     input;
     count:=0; c:=n;
     repeat;
          find; c:=0;
          if (m<3) or (ccw(p[1],p[2],p[m])=0) then break;
          inc(count);
          for i:=1 to n do if mark[i]=0 then
           begin
            inc(c); x1[c]:=x[i]; y1[c]:=y[i];
           end;
          y:=y1; x:=x1; n:=c;
     until c<3;
     assign(f,out); rewrite(f);
     writeln(f,count);
     close(f);
end.

Almost everyone have go WA#13 at least once? What is that test?
I can't find any mistakes in my program, but still WA#13 :((

I've got AC but my run time is 0.765s . I saw that some people got AC with little run time so I think there's another solution better than using Graham k times ( k = maximal possible number of fences ).
If it's true , please tell me your solution.
Thank you very much.
My e-mail : hard7771988@yahoo.co.uk