维护2个操作,树上单点修改,子树求和,
算出dfs序,子树变成区间
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
#include<iomanip>
#include<vector>
#include<string>
#include<queue>
#include<stack>
#include<map>
#include<sstream>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,0x3f,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define MEMx(a,b) memset(a,b,sizeof(a));
#define INF (0x3f3f3f3f)
#define F (1000000007)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define vi vector<int>
#define pi pair<int,int>
#define Pr(kcase,ans) printf("Case #%d: %lld\n",kcase,ans);
#define PRi(a,n) For(i,n-1) cout<<a[i]<<' '; cout<<a[n]<<endl;
#define PRi2D(a,n,m) For(i,n) { \
For(j,m-1) cout<<a[i][j]<<' ';\
cout<<a[i][m]<<endl; \
}
#pragma comment(linker, "/STACK:102400000,102400000")
typedef int ll;
int read()
{
int x=0,f=1; char ch=getchar();
while(!isdigit(ch)) {if (ch=='-') f=-1; ch=getchar();}
while(isdigit(ch)) { x=x*10+ch-'0'; ch=getchar();}
return x*f;
}
#define MAXN (110000+10)
ll C[MAXN];
int n;
int lowbit(int x){return x&(-x);}
void init(int n){
For(i,n) C[i]=0;
}
ll sum(int x) {
ll ret=0; if (x<0) return 0;
for(;x;x-=lowbit(x)) {
ret+=C[x];
}
return ret;
}
void Add(int x,ll d) {
for(;x<=n;x+=lowbit(x)) {
C[x]+=d;
}
}
int t;
int pre[MAXN],post[MAXN];
vector< vector<int> > edges(100005) ;
void dfs(int x,int fa) {
pre[x]=++t;
int sz=edges[x].size();
Rep(i,sz) {
int v=edges[x][i];
if (v!=fa) dfs(v,x);
}
post[x]=t;
}
bool a[MAXN];
int main()
{
// freopen("D.in","r",stdin);
// freopen(".out","w",stdout);
while(cin>>n) {
init(n); For(i,n) Add(i,a[i]=1); t=0;
For(i,n) edges[i].clear();
For(i,n-1) {
int u=read(),v=read();
edges[u].pb(v);
edges[v].pb(u);
}
dfs(1,-1);
int m=read();
while (m--) {
char s[2];
scanf("%s",s);
if (s[0]=='Q') {
int x=read();
printf("%d\n",sum(post[x])-sum(pre[x]-1));
} else {
int x=read();
if (a[x]==0 ) Add(pre[x],1);
else Add(pre[x],-1);
a[x]^=1;
}
}
}
return 0;
}