bzoj1007

其实吧，就是一个半平面交，而且不用考虑转回来的情况，所以只要极角排序然后用栈即可
给的是点斜式，比极角很方便
至于完整版的半平面交还没写过，看到再说吧

var a,b,c,q:array[0..50010] of longint;
    v:array[0..50010] of boolean;
    t,i,n:longint;
procedure swap(var a,b:longint);
  var c:longint;
  begin
    c:=a;
    a:=b;
    b:=c;
  end;

procedure sort(l,r: longint);
  var i,j,x,y: longint;
  begin
    i:=l;
    j:=r;
    x:=a[(l+r) shr 1];
    y:=b[(l+r) shr 1];
    repeat
      while (a[i]<x) or (a[i]=x) and (b[i]>y) do inc(i);
      while (x<a[j]) or (a[j]=x) and (b[j]<y) do dec(j);
      if not(i>j) then
      begin
        swap(a[i],a[j]);
        swap(b[i],b[j]);
        swap(c[i],c[j]);
        inc(i);
        j:=j-1;
      end;
    until i>j;
    if l<j then sort(l,j);
    if i<r then sort(i,r);
  end;

procedure push(i:longint);
  begin
    while (t-1>0) do
      if ((b[q[t]]-b[i])/(a[i]-a[q[t]])<=(b[q[t]]-b[q[t-1]])/(a[q[t-1]]-a[q[t]])) then dec(t)
      else break;  //画图可知，如果栈内最后两条直线L2,L1的交点在新加入的直线的右侧，那么L1一定是不可见的
    inc(t);
    q[t]:=i;
  end;

begin
  readln(n);
  for i:=1 to n do
  begin
    readln(a[i],b[i]);
    c[i]:=i;
  end;
  sort(1,n);
  push(1);
  fillchar(v,sizeof(v),false);
  for i:=2 to n do
    if a[i]<>a[i-1] then push(i);
  for i:=1 to t do
    v[c[q[i]]]:=true;
  for i:=1 to n do
    if v[i] then write(i,' ');
  writeln;
end.