pku2828 平衡树

话说我对朴素的平衡树还是比较拿手的。。某次学了sbt之后就觉得这东西挺好用，恩，确实能解决不少问题。

题目大意：

  有一群人，要排队，额，确切的说是插队，给你一些信息，pos[i]，val[i]代表这个人要插在第pos[i]个位置的后边，他的编号是val[i]，最后让你输出这些人的站队顺序。

简略题解：

  就是维护一棵二叉树，满足左儿子在他前边，右儿子在他后边，最后先序遍历一遍就可以了。

附程序:

program pku_2828; var ctree:array[0..200001] of record l,r,key,size:longint; end; n,tot,root:longint; {-----------------------------} procedure tl(var u:longint); var t:longint; begin t:=ctree[u].r; ctree[u].r:=ctree[t].l; ctree[t].l:=u; ctree[t].size:=ctree[u].size; ctree[u].size:=ctree[ctree[u].l].size+ctree[ctree[u].r].size+1; u:=t; end; {-----------------------------------} procedure tr(var u:longint); var t:longint; begin t:=ctree[u].l; ctree[u].l:=ctree[t].r; ctree[t].r:=u; ctree[t].size:=ctree[u].size; ctree[u].size:=ctree[ctree[u].l].size+ctree[ctree[u].r].size+1; u:=t; end; {-----------------------------------} procedure maintain(var u:longint); begin if ctree[ctree[ctree[u].l].l].size>ctree[ctree[u].r].size then tr(u) else if ctree[ctree[ctree[u].l].r].size>ctree[ctree[u].r].size then begin tl(ctree[u].l);tr(u); end else if ctree[ctree[ctree[u].r].r].size>ctree[ctree[u].l].size then tl(u) else if ctree[ctree[ctree[u].r].l].size>ctree[ctree[u].l].size then begin tr(ctree[u].r);tl(u); end; end; {-----------------------------} procedure insert(var u:longint;k,key:longint); begin if u=0 then begin inc(tot); ctree[tot].size:=1; ctree[tot].key:=key; u:=tot; exit; end; inc(ctree[u].size); if k<ctree[ctree[u].l].size+1 then insert(ctree[u].l,k,key) else insert(ctree[u].r,k-ctree[ctree[u].l].size-1,key); maintain(u); end; {-----------------------------} procedure dfs(u:longint); begin if u=0 then exit; dfs(ctree[u].l); write(ctree[u].key,' '); dfs(ctree[u].r); end; {-----------------------------} procedure main; var i,p,q:longint; begin readln(n); for i:=1 to n do begin readln(p,q); insert(root,p,q); end; dfs(root); writeln; end; {-----------------------------} begin while not eof do begin fillchar(ctree,sizeof(ctree),0); tot:=0; root:=0; main; end; end.