Apple Tree DFS＋树状数组

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 toN 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 forku 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 forkx, 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

首先，这是一棵树，且不是二叉树，然后是区间修改和单点查询，所以要把这棵树转化为一维树状数组，对这棵树进行先序遍历也就是DFS，得到每个结点的管辖范围（我看网上大佬称为时间戳）用数组l和r存储

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
#define lowbit(x) (x & (-x))

using namespace std;
const int N = 1e5 + 3;

vector< vector<int> > edge(N);  //直接写vector<int> edge[N]，会TLE...
int l[N], r[N], c[N];
bool vis[N];
int n, key = 1;

void DFS(int u)				    //DFS获取时间戳
{
    l[u] = key;
    for(int i = 0; i < edge[u].size(); i++)
    {
        key++;
        DFS(edge[u][i]);
    }
    r[u] = key;
}
void Update(int k, int v)
{
    for(int i = k; i <= n; i += lowbit(i))
        c[i] += v;
}
int GetSum(int k)
{
    int sum = 0;
    for(int i = k; i > 0; i -= lowbit(i))
        sum += c[i];
    return sum;
}
int main()
{
    int u, v, m, x;
    char s[2];
    scanf("%d", &n);
    for(int i = 1; i < n; i++)
    {
        scanf("%d%d", &u, &v);
        edge[u].push_back(v);
    }
    memset(c, 0, sizeof(c));
    memset(vis, 1, sizeof(vis));
    DFS(1);
    for(int i = 1; i <= n; i++)//一开始树上都有苹果
        Update(i, 1);
    scanf("%d", &m);
    while(m--)
    {
        scanf("%s%d", s, &x);
        if(s[0] == 'Q')
        {
            printf("%d\n", GetSum(r[x]) - GetSum(l[x] - 1));
        }
        else
        {
            if(vis[x]) 
                Update(l[x], -1);
            else 
                Update(l[x], 1);
            vis[x] = !vis[x];  //状态相反
        }
    }
    return 0;
}