link

思路：

经典套路，通过dfs序将树上修改转化为线性修改，这样问题就转化为了单点修改，区间查询，用树状数组维护。
类似题

代码：

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<ll, ll>PLL;
typedef pair<int, int>PII;
typedef pair<double, double>PDD;
#define I_int ll
inline ll read()
{
    ll x = 0, f = 1;
    char ch = getchar();
    while(ch < '0' || ch > '9')
    {
        if(ch == '-')f = -1;
        ch = getchar();
    }
    while(ch >= '0' && ch <= '9')
    {
        x = x * 10 + ch - '0';
        ch = getchar();
    }
    return x * f;
}
#define read read()
#define closeSync ios::sync_with_stdio(0);cin.tie(0);cout.tie(0)
#define multiCase int T;cin>>T;for(int t=1;t<=T;t++)
#define rep(i,a,b) for(int i=(a);i<=(b);i++)
#define repp(i,a,b) for(int i=(a);i<(b);i++)
#define per(i,a,b) for(int i=(a);i>=(b);i--)
#define perr(i,a,b) for(int i=(a);i>(b);i--)
ll ksm(ll a, ll b, ll p)
{
    ll res = 1;
    while(b)
    {
        if(b & 1)res = res * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return res;
}
const int inf = 0x3f3f3f3f;
#define PI acos(-1)

const int maxn=1e5+100;

struct BIT{
    ll n,tr[maxn];

    void init(){
        memset(tr,0,sizeof tr);
    }
    ll lowbit(ll x){
        return x&-x;
    }
    void update(ll pos,ll val){
        while(pos<=n){
            tr[pos]+=val;pos+=lowbit(pos);
        }
    }
    ll qask(ll pos){
        ll res=0;
        while(pos){
            res+=tr[pos];pos-=lowbit(pos);
        }
        return res;
    }

};

int n,m,a[maxn];
vector<int>g[maxn];
ll in[maxn],out[maxn],timetmp;
void dfs(int u,int fa){//dfs记录dfs序列
    in[u]=++timetmp;
    for(int i=0;i<g[u].size();i++){
        int j=g[u][i];
        if(j==fa) continue;
        dfs(j,u);
    }
    out[u]=timetmp;
}

int main()
{
    n=read;
    rep(i,1,n-1){
        int u=read,v=read;
        g[u].push_back(v);
        g[v].push_back(u);
    }
    dfs(1,-1);
    BIT t;
    t.n=1e5;t.init();
    rep(i,1,n) t.update(in[i],1),a[i]=1;
    m=read;
    rep(i,1,m){
        char op[2];int x;
        cin>>op;x=read;
        if(op[0]=='C'){
           if(a[x]) t.update(in[x],-1);
           else t.update(in[x],1);
           a[x]=!a[x];
        }
        else{
            printf("%lld\n",t.qask(out[x])-t.qask(in[x]-1));
        }
    }
    return 0;
}

/*

**/