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.