树（不一定是2叉树）映射成数组特别麻烦，在网上看了好多。本来想用 vector< vector<int> >，但太麻烦。最后用的是左兄弟右孩子的办法。把树建成后，再DFS一遍，记录时间戳『孩子中最小的 ，最大的（就是自己在树状数组中的位置）』。然后就能用树状是数组求和了。
| Time Limit: 2000MS | Memory Limit: 65536K | |
| Total Submissions: 15487 | Accepted: 4595 |
Description
There is an apple tree outside of kaka's house. Every autumn, a lot of apples will grow in the tree. Kaka likes apple very much, so he has been carefully nurturing the big apple tree.
The tree has N forks which are connected by branches. Kaka numbers the forks by 1 to N and the root is always numbered by 1. Apples will grow on the forks and two apple won't grow on the same fork. kaka wants to know how many apples are there in a sub-tree, for his study of the produce ability of the apple tree.
The trouble is that a new apple may grow on an empty fork some time and kaka may pick an apple from the tree for his dessert. Can you help kaka?
Input
The first line contains an integer N (N ≤ 100,000) , which is the number of the forks in the tree.
The following N - 1 lines each contain two integers u and v, which means fork u and fork v are connected by a branch.
The next line contains an integer M (M ≤ 100,000).
The following M lines each contain a message which is either
"C x" which means the existence of the apple on fork x has been changed. i.e. if there is an apple on the fork, then Kaka pick it; otherwise a new apple has grown on the empty fork.
or
"Q x" which means an inquiry for the number of apples in the sub-tree above the fork x, including the apple (if exists) on the fork x
Note the tree is full of apples at the beginning
Output
Sample Input
31 21 33Q 1C 2Q 1
Sample Output
32
Source
1 #include <cstdio> 2 #include <cstring> 3 4 #define MAXN 100005 5 6 using namespace std; 7 8 typedef struct CH 9 { 10 int idx; 11 CH* nxt; 12 }CHILD;//孩子结点 13 14 typedef struct N 15 { 16 int high;///idx 17 int low; 18 CHILD* child; 19 }NOTE; 20 21 NOTE fort[MAXN]; 22 int vis[MAXN]; 23 char apple[MAXN]; 24 int n,cnt; 25 int tree[MAXN]; 26 27 void addNote(int x,int y) 28 { 29 CHILD* p; 30 p=new CHILD; 31 p->idx=y; 32 p->nxt=fort[x].child; 33 fort[x].child=p; 34 } 35 36 void DFS(int t) 37 { 38 CHILD* p; 39 vis[t]=1; 40 fort[t].low=cnt; 41 p=fort[t].child; 42 while(p) 43 { 44 if(!vis[p->idx]) 45 DFS(p->idx); 46 p=p->nxt; 47 } 48 fort[t].high=cnt++; 49 } 50 ///************树状数组******************/// 51 int lowbit(int x) 52 { 53 return x&(-x); 54 } 55 56 int getSum(int x) 57 { 58 int sum=0; 59 while(x>0) 60 { 61 sum+=tree[x]; 62 x-=lowbit(x); 63 } 64 return sum; 65 } 66 67 void Update(int pos,int val) 68 { 69 while(pos<MAXN) 70 { 71 tree[pos]+=val; 72 pos+=lowbit(pos); 73 } 74 } 75 76 int main() 77 { 78 memset(apple,1,sizeof(apple)); 79 80 int n,s,a,b; 81 char ch[10]; 82 scanf("%d",&n); 83 for(int i=1;i<n;i++) 84 { 85 scanf("%d %d",&a,&b); 86 addNote(a,b);//加2遍 87 addNote(b,a); 88 } 89 90 cnt=1; 91 DFS(1); 92 93 for(int i=1;i<=n;i++) 94 tree[i]=lowbit(i); 95 96 scanf("%d",&s); 97 while(s--) 98 { 99 scanf("%s%d",ch,&a); 100 if(ch[0]=='C') 101 { 102 if(apple[a]==1) 103 { 104 apple[a]=0; 105 Update(fort[a].high,-1); 106 } 107 else if(apple[a]==0) 108 { 109 apple[a]=1; 110 Update(fort[a].high,1);//high就是对应树状数组中的结点 111 } 112 } 113 else if(ch[0]=='Q') 114 { 115 int ans; 116 ans=getSum(fort[a].high)-getSum(fort[a].low-1); 117 printf("%d\n",ans); 118 } 119 } 120 121 return 0; 122 }