NJPC2017: H - 白黒ツリー
solution
親と子の色が一致してしまっている個数を上手く管理。 重軽分解して無理矢理やればできるはず。 しかしEuler tourして点更新/区間和のsegment treeで十分。 $O((n + q) \log n)$。
Euler tourは行きと帰りで両方追加するもの。 行きの重みと帰りの重みが上手く相殺されるように乗せればよい。 ここ重みは$+1, -1$で算術の和をとれば十分だがLCAして2回の区間和が必要。 重みを乱数で生成し排他的論理和でまとめればLCAなしの1回の区間和で済む。 ただし確率的には衝突しうることに注意。
note
重軽分解をEuler tourで置き換えられる条件は何だろうか。 点更新と$x + x = 0$な操作に関する区間和のsegment treeが乗ってることな気がする。
implementation
#include <bits/stdc++.h>
#define REP(i, n) for (int i = 0; (i) < (int)(n); ++ (i))
#define REP3(i, m, n) for (int i = (m); (i) < (int)(n); ++ (i))
#define REP_R(i, n) for (int i = int(n) - 1; (i) >= 0; -- (i))
using namespace std;
/**
* @brief euler tour
* @arg g must be a tree, directed or undirected
*/
void do_euler_tour(vector<vector<int> > const & g, int root, vector<int> & tour, vector<int> & left, vector<int> & right) {
int n = g.size();
tour.clear();
left.resize(n);
right.resize(n);
function<void (int, int)> go = [&](int x, int parent) {
left[x] = tour.size();
tour.push_back(x);
for (int y : g[x]) if (y != parent) {
go(y, x);
}
right[x] = tour.size();
tour.push_back(x);
};
go(root, -1);
}
/**
* @brief a segment tree, or a fenwick tree
* @tparam Monoid (commutativity is not required)
*/
template <class Monoid>
struct segment_tree {
typedef typename Monoid::underlying_type underlying_type;
int n;
vector<underlying_type> a;
Monoid mon;
segment_tree() = default;
segment_tree(int a_n, underlying_type initial_value = Monoid().unit(), Monoid const & a_mon = Monoid()) : mon(a_mon) {
n = 1; while (n < a_n) n *= 2;
a.resize(2 * n - 1, mon.unit());
fill(a.begin() + (n - 1), a.begin() + ((n - 1) + a_n), initial_value); // set initial values
REP_R (i, n - 1) a[i] = mon.append(a[2 * i + 1], a[2 * i + 2]); // propagate initial values
}
void point_set(int i, underlying_type z) { // 0-based
assert (0 <= i and i <= n);
a[i + n - 1] = z;
for (i = (i + n) / 2; i > 0; i /= 2) { // 1-based
a[i - 1] = mon.append(a[2 * i - 1], a[2 * i]);
}
}
underlying_type range_concat(int l, int r) { // 0-based, [l, r)
assert (0 <= l and l <= r and r <= n);
underlying_type lacc = mon.unit(), racc = mon.unit();
for (l += n, r += n; l < r; l /= 2, r /= 2) { // 1-based loop, 2x faster than recursion
if (l % 2 == 1) lacc = mon.append(lacc, a[(l ++) - 1]);
if (r % 2 == 1) racc = mon.append(a[(-- r) - 1], racc);
}
return mon.append(lacc, racc);
}
};
struct xor_monoid {
typedef uint64_t underlying_type;
uint64_t unit() const { return 0; }
uint64_t append(uint64_t a, uint64_t b) const { return a ^ b; }
};
class solver {
static constexpr int root = 0;
vector<bool> delta;
vector<uint64_t> id;
vector<int> tour, left, right;
segment_tree<xor_monoid> segtree;
public:
solver(int n, vector<int> const & parent, vector<vector<int> > const & children, vector<bool> const & c) {
mt19937_64 gen;
id.resize(n);
delta.resize(n);
REP (x, n) {
id[x] = uniform_int_distribution<uint64_t>()(gen);
if (x != root) {
delta[x] = c[x] == c[parent[x]];
}
}
do_euler_tour(children, root, tour, left, right);
segtree = segment_tree<xor_monoid>(tour.size());
REP (i, tour.size()) {
int x = tour[i];
segtree.point_set(i, delta[x] ? id[x] : 0);
}
}
void query_op(int x) {
if (x == 0) return;
delta[x] = not delta[x];
segtree.point_set( left[x], delta[x] ? id[x] : 0);
segtree.point_set(right[x], delta[x] ? id[x] : 0);
}
bool query_ans(int x, int y) {
if (left[x] > left[y]) swap(x, y);
return segtree.range_concat(left[x] + 1, left[y] + 1) == 0;
}
};
int main() {
// input
int n; cin >> n;
vector<int> parent(n);
vector<vector<int> > children(n);
parent[0] = -1;
REP3 (i, 1, n) {
cin >> parent[i];
-- parent[i];
children[parent[i]].push_back(i);
}
vector<bool> c(n);
REP (i, n) {
int c_i; cin >> c_i;
c[i] = c_i;
}
// solve
solver s(n, parent, children, c);
// output
int q; cin >> q;
while (q --) {
int t; cin >> t;
if (t == 1) {
int u; cin >> u;
-- u;
s.query_op(u);
} else if (t == 2) {
int u, v; cin >> u >> v;
-- u; -- v;
bool ans = s.query_ans(u, v);
cout << (ans ? "YES" : "NO") << endl;
}
}
return 0;
}