bzoj2716

感人肺腑pascal过不去系列（可能是自己弱，因为有pascal过去了毕竟）
那个这种平面点还有一种处理方法是kd tree，太弱了不会有时间学一下
我还是用了cdq分治，首先肯定要把绝对值这个不和谐的东西去掉
然后就变成了4个部分，这样就非常好维护了，然后还是cdq分治的一般步骤
有优化建议的欢迎指教……

 const inf=;
 type node=record
        x,y,p,z:longint;
      end;

 var b,a:array[..] of node;
     g,q,c,next,last:array[..] of longint;
     v:array[..] of boolean;
     ans:array[..] of longint;
     mx,t,i,j,my,n,m,tot,p:longint;

 function max(a,b:longint):longint;
   begin
     if a>b then exit(a) else exit(b);
   end;

 function min(a,b:longint):longint;
   begin
     if a>b then exit(b) else exit(a);
   end;

 function lowbit(x:longint):longint;
   begin
     exit(x and (-x));
   end;

 procedure add(x,w:longint);
   begin
     while x<=p do
     begin
       if not v[x] then
       begin
         inc(tot);
         q[tot]:=x;
         v[x]:=true;
       end;
       c[x]:=max(c[x],w);
       x:=x+lowbit(x);
     end;
   end;

 function ask(x:longint):longint;
   begin
     ask:=-inf;
     while x> do
     begin
       ask:=max(ask,c[x]);
       x:=x-lowbit(x);
     end;
   end;

 procedure work(s,t,dx,dy:longint);
   var i,k,y:longint;
   begin
     tot:=;
     i:=s;
     while i<>t do
     begin
       k:=dx*b[i].x+dy*b[i].y;
       if dy=- then y:=g[my-b[i].y+] else y:=g[b[i].y];
       if b[i].z= then
         add(y,k)
       else ans[b[i].p-n]:=min(ans[b[i].p-n],abs(k-ask(y)));
       i:=i+dx;
     end;
     for i:= to tot do
     begin
       c[q[i]]:=-inf;
       v[q[i]]:=false;
     end;
   end;

 procedure cdq(l,r:longint);
   var m,l1,l2,t,i,tt:longint;
   begin
     if l=r then exit;
     m:=(l+r) shr ;
     l2:=;  t:=;
     for i:=l to r do
       if (a[i].z=) and (a[i].p<=m) then
       begin
         inc(t);
         b[t]:=a[i];
       end
       else if (a[i].z=) and (a[i].p>m) then
       begin
         inc(t);
         b[t]:=a[i];
         inc(l2);
       end;

     if l2> then
     begin
       work(,t+,,);
       work(t,,-,);
       work(,t+,,-);
       work(t,,-,-);
     end;
     t:=l2; tt:=;
     l1:=l; l2:=m+;
     for i:=l to r do
       if a[i].p<=m then
       begin
         b[l1]:=a[i];
         inc(l1);
         if a[i].z= then inc(tt);
       end
       else begin
         b[l2]:=a[i];
         inc(l2);
       end;

     for i:=l to r do
       a[i]:=b[i];
     if tt> then cdq(l,m);
     if t> then cdq(m+,r);
   end;

 begin
   readln(n,m);
   for i:= to n do
   begin
     readln(b[i].x,b[i].y);
     inc(b[i].x);
     inc(b[i].y);
     b[i].p:=i;
     b[i].z:=;
     my:=max(my,b[i].y);
   end;
   for i:=n+ to n+m do
   begin
     readln(b[i].z,b[i].x,b[i].y);
     inc(b[i].x);
     inc(b[i].y);
     b[i].p:=i;
     my:=max(my,b[i].y);
   end;
   for i:= to n+m do
   begin
     next[i]:=last[b[i].x];
     last[b[i].x]:=i;
     mx:=max(mx,b[i].x);
     v[b[i].y]:=true;
     v[my-b[i].y+]:=true;
   end;
   p:=;
   for i:= to my do
   begin
     c[i]:=-inf;
     if v[i] then
     begin
       inc(p);
       g[i]:=p;
       v[i]:=false;
     end;
   end;
   fillchar(ans,sizeof(ans),);
   for i:= to mx do  //这里只需要以x为第一关键字即可，因为计算贡献的时候会从四个方向算，总有一个会算到
   begin
     j:=last[i];
     while j<> do
     begin
       inc(t);
       a[t]:=b[j];
       j:=next[j];
     end;
   end;
   t:=;
   for i:= to n+m do   //先处理初始值对询问的影响
     if (a[i].z=) or (a[i].p<=n) then
     begin
       inc(t);
       b[t]:=a[i];
     end;

   work(,t+,,);
   work(t,,-,);
   work(,t+,,-);
   work(t,,-,-);
   t:=n;
   for i:= to n+m do
     if a[i].p>n then
     begin
       inc(t);
       b[t]:=a[i];
     end;

   for i:=n+ to n+m do
     a[i]:=b[i];

   cdq(n+,n+m);  //分治修改询问序列
   for i:= to m do
     if ans[i]<inf then writeln(ans[i]);
 end.