[POJ3321 Apple Tree]

[题目来源]：POJ Monthly--2007.08.05, Huang, Jinsong

[关键字]：树状数组

[题目大意]：一棵苹果树每个数字上长有1个苹果，有两种操作：1、改变一个树枝上的苹果数；2、询问以某个树枝为根的子树的苹果数量。

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[分析]：此题关键不在数据结构(区间和可以用树状数组或线段树), 主要是如何把树(不一定是二叉树)转化为数组来处理。首先建树(我采用的是邻接表), 然后DFS先根遍历该树, 对每个节点记录其最近序号(按照遍历顺序重新编号)和其子树的最小最大序号(也就是标记一棵树遍历的第一个和最后一个节点的遍历序号st, ed).剩下的就可以是典型的树状数组处理了.如:统计一个节点为根的树中apple数目时, 直接计算到st－, ed的和然后相减即可.

[代码]：

View Code

  1 type
  2   rec = record
  3     y, n: longint;
  4   end;
  5 var
  6   n, m, tot, ans, dat, now: longint;
  7   e: array[0..100100] of rec;
  8   link: array[0..100100] of longint;
  9   c: array[0..100100] of 0..1;
 10   st, ed: array[0..100100] of longint;
 11   tree: array[0..100100] of longint;
 12 
 13 procedure build(root: longint);
 14 var
 15   t, y: longint;
 16 begin
 17   t := link[root];
 18   st[root] := now;
 19   ed[root] := now;
 20   dec(now);
 21   while t > 0 do
 22     begin
 23       y := e[t].y;
 24       build(y);
 25       if ed[y] < ed[root] then ed[root] := ed[y];
 26       t := e[t].n;
 27     end;
 28 end;
 29 
 30 procedure make(x, y: longint);
 31 begin
 32   inc(tot);
 33   e[tot].y := y;
 34   e[tot].n := link[x];
 35   link[x] := tot;
 36 end;
 37 
 38 procedure ins(k, dat: longint);
 39 begin
 40   while k <= n do
 41     begin
 42       tree[k] := tree[k]+dat;
 43       k := k+k and -k;
 44     end;
 45 end;
 46 
 47 function find(k: longint):longint;
 48 var
 49   sum: longint;
 50 begin
 51   sum := 0;
 52   while k >= 1 do
 53     begin
 54       sum := sum+tree[k];
 55       k := k-k and -k;
 56     end;
 57   exit(sum);
 58 end;
 59 
 60 procedure init;
 61 var
 62   i, x, y: longint;
 63 begin
 64   readln(n);
 65   fillchar(c,sizeof(c),0);
 66   tot := 0;
 67   for i := 1 to n-1 do
 68     begin
 69       readln(x,y);
 70       make(x,y);
 71     end;
 72   now := n;
 73   build(1);
 74   for i := 1 to n do
 75     begin
 76       if c[i] = 1 then dat := -1 else dat := 1;
 77       c[i] := c[i]+dat;
 78       ins(st[i],dat);
 79     end;
 80   //for i := 1 to n do writeln(st[i],' ',ed[i]);
 81 end;
 82 
 83 procedure work;
 84 var
 85   i, x: longint;
 86   ch: char;
 87 begin
 88   readln(m);
 89   for i := 1 to m do
 90     begin
 91       read(ch);
 92       case ch of
 93         'C': begin
 94                readln(x);
 95                if c[x] = 1 then dat := -1 else dat := 1;
 96                c[x] := c[x]+dat;
 97                ins(st[x],dat);
 98              end;
 99         'Q': begin
100                readln(x);
101                ans := find(st[x]);
102                ans := ans-find(ed[x]-1);
103                writeln(ans);
104              end;
105       end;
106     end;
107   readln;
108 end;
109 
110 begin
111   init;
112   work;
113 end.