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

Output

Sample Input

3
1 2
1 3
3
Q 1
C 2
Q 1

Sample Output

3
2
題意：給出一顆樹，對某個節(jié)點(diǎn)進(jìn)行操作，如果有蘋果，則摘；沒有，則長個蘋果。求某個節(jié)點(diǎn)擁有的蘋果數(shù)量，每個節(jié)點(diǎn)最多只有一個蘋果。
代碼：

#include<stdio.h>
#include<stdlib.h>
#define?maxn?100020
int?low[maxn],?high[maxn],?n,?c[maxn],?t[maxn];
struct?P
{
????int?pt,?head;
}p[maxn];
struct?N
{
????int?pt,?id;
}node[maxn];
int?th?=?0,?time?=?0;
int?create(int?id)
{
????th++;
????node[th].pt?=?-1,?node[th].id?=?id;
????return?th;
}
int?lowbit(int?x)
{
????return?x?&?(-x);
}
void?update(int?x,?int?y)
{
????while?(x?<=?n)
????{
????????c[x]?+=?y;
????????x?+=?lowbit(x);
????}
}
int?sum(int?x)
{
????int?s?=?0;
????while?(x?>?0)
????{
????????s?+=?c[x];
????????x?-=?lowbit(x);
????}
????return?s;
}
void?dfs(int?root)
{
????int?v?=?p[root].head;
????time++;//添加時間戳?，就可以是問題轉(zhuǎn)化為求區(qū)間和?
????low[root]?=?time;
????for?(v?=?p[root].head;?v?!=?-1;?v?=?node[v].pt)
????{
????????dfs(node[v].id);
????}
????high[root]?=?time;
}
int?main()
{
????int?i,?u,?v,?no,?m,?x,?num;
????scanf("%d",?&n);
????for?(i?=?1;?i?<=?n;?i++)
????{
????????p[i].pt?=?-1,?p[i].head?=?-1;
????????update(i,?1);
????????t[i]?=?1;
????}
????num?=?n;
????while?(--num)
????{
????????scanf("%d%d",?&u,?&v);
????????no?=?create(v);
????????node[p[u].pt].pt?=?no;
????????p[u].pt?=?no;
????????if?(p[u].head?==?-1)
????????{
????????????p[u].head?=?no;
????????}
????}
????dfs(1);
????char?a[2];
????scanf("%d",?&m);
????while?(m--)
????{
????????scanf("%s%d",?a,?&x);
????????if?(a[0]?==?'Q')
????????{
????????????printf("%d\n",?sum(high[x])?-?sum(low[x]?-?1));
????????}
????????else
????????{
????????????if?(t[x])
????????????{
????????????????update(low[x],?-1);
????????????????t[x]?=?0;
????????????}
????????????else
????????????{
????????????????update(low[x],?1);
????????????????t[x]?=?1;
????????????}
????????}
????}
????system("pause");
????return?0;
}
/*
5
1?2
1?3
3?4
3?5
*/