传送门
和主席树的思路差不多，用前缀和的思想保存n颗字典树(当然要动态开点)，用对第i颗字典树上的节点sum[j][0/1]表示前i颗树第j位0/1的个数。对于询问只需要按位贪心就行了。
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e5 + 100;
int out[maxn], cc[maxn], id[maxn], val[maxn];
int root[maxn];
vector<int> G[maxn];
int cnt;
int n, q, m;
void dfs(int u, int fa) {
id[u] = ++cnt;
val[cnt] = u;
for (int i = 0; i < G[u].size(); i++) {
int v = G[u][i];
if (v != fa) dfs(v, u);
}
out[u] = cnt;
}
struct Trie {
int nex[maxn<<5][2], tot, sum[maxn<<5][2];
int newnode() {
nex[tot][0] = nex[tot][1] = 0;
return tot++;
}
void init() {
tot = 1;
nex[0][0] = nex[0][1] = 0;
}
int insert(int x, int id) {
int t = newnode(), res = t;
for (int i = 30; i >= 0; i--) {
int tx = (1<<i)&x;
tx = (tx>>i);
sum[t][tx] = sum[id][tx]+1;
sum[t][!tx] = sum[id][!tx];
nex[t][tx] = newnode();
nex[t][!tx] = nex[id][!tx];
t = nex[t][tx];
id = nex[id][tx];
}
return res;
}
int query(int x, int y, int s) {
int res = 0;
for(int i = 30; i >= 0; i--) {
int t = (1<<i)&s;
t = (t>>i);
if(sum[y][!t]-sum[x][!t]) {
res += (1<<i);
x = nex[x][!t];
y = nex[y][!t];
}
else {
x = nex[x][t];
y = nex[y][t];
}
}
return res;
}
} trie;
void init() {
cnt = 0;
root[0] = 0;
for (int i = 0; i <= n; i++) G[i].clear();
}
int main() {
freopen("in.txt", "r", stdin);
while (scanf("%d%d", &n, &q) == 2) {
init();
trie.init();
for (int i = 1; i <= n; i++) {
scanf("%d", &cc[i]);
}
for (int i = 1; i < n; i++) {
int f;
scanf("%d", &f);
G[f].push_back(i+1);
G[i+1].push_back(f);
}
dfs(1, 0);
for (int i = 1; i <= n; i++) {
root[i] = trie.insert(cc[val[i]], root[i-1]);
}
for (int i = 1; i <= q; i++) {
int u, x;
scanf("%d%d", &u, &x);
printf("%d\n", trie.query(root[id[u]-1], root[out[u]], x));
}
}
return 0;
}