POj 3321 树状数组苹果树

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
For every inquiry, output the correspond answer per line.
Sample Input
3
1 2
1 3
3
Q 1
C 2
Q 1
Sample Output
3
2

树状数组的中级应用，求一段子树，都觉得和树状数组沾不上边，
但是嘛，有数就有子树有顺序，有顺序就可以用线段数和树状数组了，今天看一个人说的很好，基本上线段树能做到的树状数组也能做到，但是一些最大最小值的还是线段树比较好。
这道题是求一段一段子树的和，所以应该有那么一个数组存取的是每一颗子树（子树就相当于树状数组里面的结点）的和，然后需要两个数组存取的是每棵子树的左右边界也就是每个结点的左右边界，那么就可以用用树状数组了，还有用一个数组表示树状数组，那么怎么得到每个结点的左右边界呢，用深搜，用vector很完美。
创建一个二维vector

#include <cstdio>
#include <string>
#include <cstring>
#include <iostream>
#include <vector>
#define mem(a) memset(a,0,sizeof(a));
#define lowbit(x) ((x)&(-x))
using namespace std;
typedef vector<int> vd; //很方便
const int maxn=500010;
vector<vd> Edge(maxn);  //创建一个二维vector
int Right[maxn],Left[maxn]; //每个子树的边界
int TreeArray[maxn];   //树状数组
int Fork[maxn];        //表示当前树有没有苹果。
int key;
int n;
void init()
{
  mem(Right);mem(Left);
  mem(TreeArray);mem(Fork);
  for(int i=0;i<Edge.size();i++) Edge[i].clear();   
}
void DFS(int node)
{
    Left[node]=key;  //左边界
    for(int i=0;i<Edge[node].size();i++)
    {
        key+=1;
        DFS(Edge[node][i]);//每一个有边的结点都是一棵树。。这样构建很强
    }
    Right[node]=key; //右边界
}
void update(int x,int val)
{
    while(x<maxn)
    {
        TreeArray[x]+=val;
        x+=lowbit(x);
    }
}
int getsum(int x)
{
    int sum=0;
    while(x>0)
    {
        sum+=TreeArray[x];
        x-=lowbit(x);
    }
    return sum;
}
void ReadDataAndDo()
{
    for(int i=1;i<n;i++)
    {
        int a,b;
        scanf("%d%d",&a,&b);
        Edge[a].push_back(b);     //直接push，得到邻接表
    }
    key=1; DFS(1);    //用1开始深搜
    int M;
    scanf("%d%*c",&M);
    char ch;
    int d;
    for(int i=1;i<=n;i++)
    {
        Fork[i]=1;
        update(i,1);
    }
    for(int i=0;i<M;i++)
    {
        scanf(" %c%d%*c",&ch,&d);
        if(ch=='Q')
        {
            printf("%d\n",getsum(Right[d])-getsum(Left[d]-1));//两个相减得去区间和。。
        }
        else
        {
            if(Fork[d]) update(Left[d],-1);
            else update(Left[d],1);
            Fork[d]=!Fork[d];
        }
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
    init();
    ReadDataAndDo();
    }
}