poj 2828 Buy Tickets 线段树

题意/Description：

    Railway tickets were difficult to buy around the Lunar New Year in China, so we must get up early and join a long queue…

    The Lunar New Year was approaching, but unluckily the Little Cat still had schedules going here and there. Now, he had to travel by train to Mianyang, Sichuan Province for the winter camp selection of the national team of Olympiad in Informatics.

    It was one o’clock a.m. and dark outside. Chill wind from the northwest did not scare off the people in the queue. The cold night gave the Little Cat a shiver. Why not find a problem to think about? That was none the less better than freezing to death!

    People kept jumping the queue. Since it was too dark around, such moves would not be discovered even by the people adjacent to the queue-jumpers. “If every person in the queue is assigned an integral value and all the information about those who have jumped the queue and where they stand after queue-jumping is given, can I find out the final order of people in the queue?” Thought the Little Cat.

读入/Input：

   There will be several test cases in the input. Each test case consists of N + 1 lines where N (1 ≤ N ≤ 200,000) is given in the first line of the test case. The next Nlines contain the pairs of values Pos_i and Val_i in the increasing order of i (1 ≤ i ≤ N). For each i, the ranges and meanings of Pos_i and Val_i are as follows:

      Pos_i ∈ [0, i − 1] — The i-th person came to the queue and stood right behind the Pos_i-th person in the queue. The booking office was considered the     0th person and the person at the front of the queue was considered the first person in the queue.

      Val_i ∈ [0, 32767] — The i-th person was assigned the value Val_i.

   There no blank lines between test cases. Proceed to the end of input.

输出/Output：

    For each test cases, output a single line of space-separated integers which are the values of people in the order they stand in the queue.

题解/solution：

    Draw a map，就知道最后一个插入到该位置的人位置是固定的，那么我们可以从后面进行插入操作，pos   o代表o要插入到pos位置，那么就是说pos前面要留出pos个位置，不是pos-1个，因为是从0开始的。
    线段树：ad记录该区间目前还剩ad个空位，每一次ins即插入的时候，如果该节点左儿子ad>=pos，那么只要在左儿子找就可以了否则要在右儿子中找，此时pos改为pos-左儿子ad。

代码/Code：

type  arr=record    l,r:longint;    ad:longint;  end;var  n:longint;  tree:array [0..800001] of arr;  ap,a,ans:array [0..200001] of longint;procedure cre(p,b,e:longint);var  m:longint;begin  with tree[p] do    begin      l:=b; r:=e;      if l=r then        begin          ad:=1;          exit;        end;      m:=(b+e) div 2;      cre(p*2,b,m);      cre(p*2+1,m+1,e);      ad:=tree[p*2].ad+tree[p*2+1].ad;    end;end;procedure ins(p,pos,o:longint);var  m:longint;begin  with tree[p] do    begin      dec(ad);      if l=r then ans[l]:=o else        if pos<=tree[p*2].ad then ins(p*2,pos,o) else          ins(p*2+1,pos-tree[p*2].ad,o);    end;end;procedure main;var  i:longint;begin  while not eof do    begin      readln(n);      cre(1,1,n);      for i:=1 to n do        readln(ap[i],a[i]);      for i:=n downto 1 do        ins(1,ap[i]+1,a[i]);      for i:=1 to n do        write(ans[i],' ');      writeln;    end;end;begin  main;end.