【POJ2828】Buy Tickets（线段树）

题意：有一个输入序列，每次操作要把b[i]插入到第a[i]个，在第a[i]个后面的要后移，问最后序列。

n<=200000

思路：顺序来只能用splay维护

考虑倒序，对于插入到第K个位置，在线段树二分第K个0的位置，类似于主席树

将其插入后将这个位置修改为已经有数

单点修改

 var t:array[..]of longint;
     a,b,c:array[..]of longint;
     n,i,k:longint;

 procedure pushup(p:longint);
 begin
  t[p]:=t[p<<]+t[p<<+];
 end;

 function query(l,r,k,p:longint):longint;
 var mid:longint;
 begin
  if l=r then exit(l);
  mid:=(l+r)>>;
  if t[p<<]>=k then exit(query(l,mid,k,p<<))
   else exit(query(mid+,r,k-t[p<<],p<<+));
  pushup(p);
 end;

 procedure update(l,r,k,p:longint);
 var mid:longint;
 begin
  if (l=k)and(r=k) then
  begin
   dec(t[p]); exit;
  end;
  mid:=(l+r)>>;
  if k<=mid then update(l,mid,k,p<<)
   else update(mid+,r,k,p<<+);
  pushup(p);
 end;

 procedure build(l,r,p:longint);
 var mid:longint;
 begin
  if l=r then
  begin
   t[p]:=; exit;
  end;
  mid:=(l+r)>>;
  build(l,mid,p<<);
  build(mid+,r,p<<+);
  pushup(p);
 end;

 begin
  assign(input,'poj2828.in'); reset(input);
  assign(output,'poj2828.out'); rewrite(output);
  while not eof do
  begin
   readln(n);
   if n= then break;
   for i:= to n<< do t[i]:=;
   fillchar(c,sizeof(c),);
   for i:= to n do
   begin
    read(a[i],b[i]);
    inc(a[i]);
   end;
   build(,n,);
   for i:=n downto  do
   begin
    k:=query(,n,a[i],);
    c[k]:=b[i];
    update(,n,k,);
   end;
   for i:= to n- do write(c[i],' ');
   write(c[n]);
   writeln;
  end;

  close(input);
  close(output);
 end.