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[..] of longint;

     v:array[..] 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 ];

     y:=b[(l+r) shr ];

     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-;

       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->) do

       if ((b[q[t]]-b[i])/(a[i]-a[q[t]])<=(b[q[t]]-b[q[t-]])/(a[q[t-]]-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:= to n do

   begin

     readln(a[i],b[i]);

     c[i]:=i;

   end;

   sort(,n);

   push();

   fillchar(v,sizeof(v),false);

   for i:= to n do

     if a[i]<>a[i-] then push(i);

   for i:= to t do

     v[c[q[i]]]:=true;

   for i:= to n do

     if v[i] then write(i,' ');

   writeln;

 end.