【BZOJ1208】宠物收养所（平衡树，splay）

题意：见题面

思路：因为每个时刻要么全是人要么全是宠物，所以可以一棵splay解决

维护单点插入，单点删除，求前驱，求后继即可

 var t:array[..,..]of longint;
     num,fa:array[..]of longint;
     n,cnt,t1,t2,i,root,p,x,y:longint;
     ans:int64;

 procedure rotate(x:longint;var k:longint);
 var y,z,l,r:longint;
 begin
  y:=fa[x]; z:=fa[y];
  if t[y,]=x then l:=
   else l:=;
  r:=-l;
  if y=k then k:=x
   else
    if t[z,]=y then t[z,]:=x
     else t[z,]:=x;
  fa[x]:=z; fa[y]:=x; fa[t[x,r]]:=y;
  t[y,l]:=t[x,r]; t[x,r]:=y;
 end;

 procedure splay(x:longint;var k:longint);
 var y,z:longint;
 begin
  while x<>k do
  begin
   y:=fa[x]; z:=fa[y];
   if y<>k then
   begin
    if (t[y,]=x)xor(t[z,]=y) then rotate(x,k)
     else rotate(y,k);
   end;
   rotate(x,k);
  end;
 end;

 procedure ins(var k:longint;x,last:longint);
 begin
  if k= then
  begin
   inc(cnt);
   k:=cnt; num[k]:=x;
   fa[k]:=last; splay(k,root);
   exit;
  end;
  if x<num[k] then ins(t[k,],x,k)
   else ins(t[k,],x,k);
 end;

 procedure del(x:longint);
 var k:longint;
 begin
  splay(x,root);
  if t[x,]*t[x,]= then root:=t[x,]+t[x,]
   else
   begin
    k:=t[x,];
    while t[k,]> do k:=t[k,];
    t[k,]:=t[x,]; fa[t[x,]]:=k;
    root:=t[x,];
   end;
  fa[root]:=;
 end;

 procedure pred(k,x:longint);
 begin
  if k= then exit;
  if num[k]<=x then
  begin
   t1:=k; pred(t[k,],x);
  end
   else pred(t[k,],x);
 end;

 procedure succ(k,x:longint);
 begin
  if k= then exit;
  if num[k]>=x then
  begin
   t2:=k; succ(t[k,],x);
  end
   else succ(t[k,],x);
 end;

 begin
  assign(input,'bzoj1208.in'); reset(input);
  assign(output,'bzoj1208.out'); rewrite(output);
  readln(n);
  for i:= to n do
  begin
   readln(x,y);
   if root= then begin p:=x; ins(root,y,); end
    else if p=x then ins(root,y,)
     else
     begin
      t1:=-; t2:=-;
      pred(root,y); succ(root,y);
      if t1=- then
      begin
       ans:=ans+num[t2]-y; ans:=ans mod ; del(t2);
      end
       else if t2=- then
       begin
        ans:=ans+y-num[t1]; ans:=ans mod ; del(t1);
       end
        else
         if y-num[t1]>num[t2]-y then
         begin
          ans:=ans+num[t2]-y; ans:=ans mod ; del(t2);
         end
          else
          begin
           ans:=ans+y-num[t1]; ans:=ans mod ; del(t1);
          end;
     end;
  end;
  writeln(ans);
  close(input);
  close(output);
 end.