题意
给定一棵根为 \(1\) 的有根树，以及字符串 \(S\)。
- \(x, h\) 求 \(x\) 的子树内，深度为 \(h\) 的节点的字符能否重排为一个回文串。
Sol
不难发现，回文串显然至多有一个字符出现奇数个。
所以我们对于每种字符随机附权值，维护前缀异或值。
查询时枚举 \(26\) 种为奇数的情况，这是显然平凡的。
Code
#include <iostream>
#include <algorithm>
#include <cstdio>
#include <array>
#include <random>
#include <vector>
#include <ctime>
#include <cassert>
#define int long long
using namespace std;
#ifdef ONLINE_JUDGE
#define getchar() (p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++)
char buf[1 << 23], *p1 = buf, *p2 = buf, ubuf[1 << 23], *u = ubuf;
#endif
int read() {
int p = 0, flg = 1;
char c = getchar();
while (c < '0' || c > '9') {
if (c == '-') flg = -1;
c = getchar();
}
while (c >= '0' && c <= '9') {
p = p * 10 + c - '0';
c = getchar();
}
return p * flg;
}
string read_() {
string ans;
char c = getchar();
while (c < 'a' || c > 'z')
c = getchar();
while (c >= 'a' && c <= 'z') {
ans += c;
c = getchar();
}
return ans;
}
void write(int x) {
if (x < 0) {
x = -x;
putchar('-');
}
if (x > 9) {
write(x / 10);
}
putchar(x % 10 + '0');
}
const int N = 5e5 + 5, M = 1e6 + 5;
namespace G {
array <int, N> fir;
array <int, M> nex, to;
int cnt;
void add(int x, int y) {
cnt++;
nex[cnt] = fir[x];
to[cnt] = y;
fir[x] = cnt;
}
}
string s;
array <int, 28> cur;
array <vector <int>, N> isl;
namespace Hpt {
using G::fir; using G::nex; using G::to;
array <int, N> siz, dep, fa, son;
void dfs1(int x) {
siz[x] = 1;
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x]) continue;
dep[to[i]] = dep[x] + 1;
fa[to[i]] = x;
dfs1(to[i]);
siz[x] += siz[to[i]];
if (siz[to[i]] > siz[son[x]]) son[x] = to[i];
}
}
array <int, N> dfn, top, idx;
int cnt;
void dfs2(int x, int Mgn) {
cnt++;
dfn[x] = cnt;
idx[cnt] = x;
top[x] = Mgn;
isl[dep[x]].push_back(dfn[x]);
if (son[x]) dfs2(son[x], Mgn);
for (int i = fir[x]; i; i = nex[i]) {
if (to[i] == fa[x] || to[i] == son[x]) continue;
dfs2(to[i], to[i]);
}
}
}
namespace Bit {
array <int, N> edge;
int lowbit(int x) {
return x & -x;
}
void modify(int x, int y, int n) {
while (x <= n) {
edge[x] ^= y;
x += lowbit(x);
}
return;
}
int query(int x) {
int ans = 0;
while (x) {
ans ^= edge[x];
x -= lowbit(x);
}
return ans;
}
}
array <int, N> prl;
signed main() {
int n = read(), q = read();
for (int i = 2; i <= n; i++) {
int x = read();
G::add(i, x), G::add(x, i);
}
s = " " + read_();
mt19937_64 rnd(time(0));
for (int i = 1; i <= 26; i++)
cur[i] = rnd() % 147744151;
Hpt::dfs1(1), Hpt::dfs2(1, 0);
int cnt = 0;
for (int i = 0; i <= n; i++) {
for (auto x : isl[i]) {
cnt++;
Bit::modify(cnt, cur[s[Hpt::idx[x]] - 'a' + 1], n);
}
prl[i] += prl[i - 1] + isl[i].size();
}
while (q--) {
int x = read(), y = read() - 1;
int l = Hpt::dfn[x], r = Hpt::dfn[x] + Hpt::siz[x] - 1;
if (!isl[y].size()) {
puts("Yes");
continue;
}
int st = lower_bound(isl[y].begin(), isl[y].end(), l) - isl[y].begin() + 1,
ed = 0;
auto it = upper_bound(isl[y].begin(), isl[y].end(), r);
ed = it - isl[y].begin();
st += prl[y - 1], ed += prl[y - 1];
int tp = Bit::query(ed) ^ Bit::query(st - 1);
/* for (auto k : isl[y]) */
/* write(k), putchar(32); */
/* puts(""); */
/* write(st), putchar(32); */
/* write(ed), puts(""); */
bool flg = 0;
for (int i = 0; i <= 26; i++) {
if (tp ^ cur[i]) continue;
flg = 1;
}
if (flg) puts("Yes");
else puts("No");
}
return 0;
}