Problem

n個節點的有根數，每個節點擁有一個顏色。邊權為1。
現在有m個查詢，每個問題有兩個整數x和d，表示詢問x的字數中depth$depth$不超過dep[x]+d$dep[x]+d$的所有點中出現來多少種本質不同的顏色。

Solution

每個點的貢獻都是1$1$，那麼若有顏色相同的點對(i,j)$(i,j)$，那麼lca(i,j)$lca(i,j)$的貢獻值就要−1$-1$。於是子樹和即為答案。
dfs$dfs$序相鄰的兩個節點的lca$lca$是最深的。所以我們用當前節點與其相鄰的兩個點做一下即可。如果與它相鄰的確實是兩個點，那麼這兩個節點的lca$lca$處需要貢獻+1$+1$（容斥）。
這個相鄰我們可以給每一個顏色都開一個

Code

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <set>
using namespace std;
#define N 100010
inline char gc() {
    static char now[1<<16], *S, *T;
    if(S == T) {T = (S = now) + fread(now, 1
 
, 1<<16, stdin); if(S == T) return EOF;}
    return *S++;
}
inline int read() {
    int x = 0; char c = gc();
    while(c < '0' || c > '9') c = gc();
    while(c >= '0' && c <= '9') {x = x * 10 + c - 48; c = gc();}
    return x;
}
int c[N], f[N][17], dep[N], head[N], id[N];
int
 
 n, m, cnt;
struct edge {
    int to, next;
}e[N];
inline void ins(int x, int y) {
    e[++cnt].to = y;
    e[cnt].next = head[x];
    head[x] = cnt;
}
inline void init_lca() {
    for(int i = 1; i <= 16; ++i)
        for(int j = 1; j <= n; ++j) {
            f[j][i] = f[f[j][i - 1]][i - 1];
        }
}
inline int lca(int x, int y) {
    if(dep[x] < dep[y]) swap(x, y);
    for(int i = 16; i >= 0; --i) {
        if(dep[f[x][i]] >= dep[y]) {
            x = f[x][i];
        }
    }
    if(x == y) return x;
    for(int i = 16; i >= 0; --i) {
        if(f[x][i] != f[y][i]) {
            x = f[x][i];
            y = f[y][i];
        }
    }
    return f[x][0];
}
int st[N], ed[N], real[N], tim;
void dfs(int x, int d) {
    dep[x] = d; st[x] = ++tim; real[tim] = x;
    for(int i = head[x]; i; i = e[i].next) {
        dfs(e[i].to, d + 1);
    }
    ed[x] = tim;
}
inline bool cmp(int A, int B) {
    return dep[A] < dep[B];
}
set<int> s[N];
set<int>::iterator it;
int rt[N];
#undef N
#define M 5000010
int L[M], R[M], v[M], len;
#undef M
void insert(int x, int &y, int l, int r, int pos, int delta) {
    y = ++len; L[y] = R[y] = 0;
    v[y] = v[x] + delta;
    if(l == r) return ;
    int mid = (l + r)>>1;
    if(pos <= mid) R[y] = R[x], insert(L[x], L[y], l, mid, pos, delta);
    else L[y] = L[x], insert(R[x], R[y], mid + 1, r, pos, delta);
}
int query(int x, int l, int r, int a, int b) {
    if(!x || (a <= l && r <= b)) return v[x];
    int mid = (l + r)>>1;
    int ret = 0;
    if(a <= mid) ret+= query(L[x], l, mid, a, b);
    if(mid + 1 <= b) ret+= query(R[x], mid + 1, r, a, b);
    return ret;
}
inline void work() {
    cnt = 1;
    n = read(); m = read();
    for(int i = 1; i <= n; ++i) {
        c[i] = read();
        s[i].clear();
        memset(f[i], 0, sizeof(f[i]));
        head[i] = 0;
        rt[i] = 0;
        id[i] = i;
    }
    for(int i = 2; i <= n; ++i) {
        f[i][0] = read();
        ins(f[i][0], i);
    }
    init_lca(); tim = 0; dfs(1, 1);
    sort(id+1, id+n+1, cmp);
    len = 0;
    for(int i = 1; i <= n; ++i) {
        int u = id[i], a = 0, b = 0;
        it = s[c[u]].lower_bound(st[u]);
        insert(rt[dep[id[i - 1]]], rt[dep[u]], 1, n, st[u], 1);
        if(it != s[c[u]].end()) {
            b = real[*it];
            insert(rt[dep[u]], rt[dep[u]], 1, n, st[lca(b, u)], -1);
        }
        if(it != s[c[u]].begin()) {
            a = real[*(--it)];
            insert(rt[dep[u]], rt[dep[u]], 1, n, st[lca(a, u)], -1);
        }
        if(a && b) {
            insert(rt[dep[u]], rt[dep[u]], 1, n, st[lca(a, b)], 1);
        }
        s[c[u]].insert(st[u]);
    }
    int lastans = 0;
    for(int i = 1; i <= m; ++i) {
        int x = read() ^ lastans, d = read() ^ lastans;
        lastans = query(rt[min(dep[x] + d, dep[id[n]])], 1, n, st[x], ed[x]);
        printf("%d\n", lastans);
    }
}
int main() {
    int T = read();
    while(T--) work();
    return 0;
}