POJ3321 Apple Tree

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 37048		Accepted: 11048

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

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

【代码】

#include <iostream>
using namespace std;

#define maxn 210100 //注意：maxn设得太大会导致RE
int head[maxn], a[maxn], C[maxn], cnt, tot, N;

struct fork
{
    int l, r; //fork代表交叉点，左值为该分叉点的序号（按访问顺序），r为该分叉点以上的最大的序号
}f[maxn];//index为分叉点的编号（题给）

struct edge
{
    int to, next; //to为指向的分叉点序号，next指向下一条树枝
}e[maxn];

void add(int u, int v)
{
    cnt++;
    e[cnt].to = v;
    e[cnt].next = head[u];
    head[u] = cnt;
}

int lowbit(int x)
{
    return x & (-x);
}

void dfs(int x)
{
    //cout << "x: " << x << endl;
    f[x].l = ++tot;
    //cout << "tot: " << tot << endl;
    for (int i = head[x]; i; i = e[i].next)
        dfs(e[i].to);
    f[x].r = tot;
}

void change(int x)
{
    for (int i = x; i <= N; i += lowbit(i)) {  //维护树状数组信息，复杂度O(logN)
        if (a[x] == 1) C[i]++;
        else C[i]--;
    }
}

int sum(int x)
{
    int re = 0;
    for (int i = x; i > 0; i -= lowbit(i))//树状数组求和，复杂度O(logN)
        re += C[i];
    return re;
}

int main()
{
    int M, u, v;
    char query;
    cin >> N;
    for (int i = 1; i < N; i++) {
        cin >> u >> v;
        add(u, v);
    }

    dfs(1);

    for (int i = 1; i <= N; i++) {
        a[i] = 1;
        change(i);
    }

    cin >> M;
    while (M--) {
        cin >> query >> u;
        if (query == 'C') {
            a[f[u].l] = (a[f[u].l] + 1) % 2;//注意：并不是a[u]，因为C是按访问顺序求和的
            change(f[u].l);
        }
        if (query == 'Q') {
            //cout << f[u].r << endl;
            //cout << sum(f[u].r) << endl;
            cout << sum(f[u].r) - sum(f[u].l - 1) << endl;
        }
    }

    return 0;
}