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SubtreeQueries.cpp
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188 lines (168 loc) · 4.91 KB
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//Subtree Queries - https://cses.fi/problemset/task/1137
#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const ll INF = 1e18;
template<class T, class U>
// T -> node, U -> update.
struct Lsegtree {
vector<T> st;
vector<U> lazy;
ll n;
T identity_element; //combine(id, x) = x for all x
U identity_update; //apply(id, x) = x for all x
Lsegtree(ll n, T identity_element, U identity_update) {
this->n = n;
this->identity_element = identity_element;
this->identity_update = identity_update;
st.assign(4*n, identity_element);
lazy.assign(4*n, identity_update);
}
//change this function as required.
//right now this is sum of two nodes.
//we can change it to max(l, r), gcd(l, r), etc..
T combine(T l, T r) {
T ans = (l + r);
return ans;
}
void buildUtil(ll v, ll tl, ll tr, vector<T>&a) {
if (tl == tr) {
st[v] = a[tl];
return;
}
ll tm = (tl + tr) >> 1;
buildUtil(2*v + 1, tl, tm,a);
buildUtil(2*v + 2,tm + 1,tr,a);
st[v] = combine(st[2*v + 1], st[2*v + 2]);
}
//change the following 2 functions, and you're more or less done.
//apply the update to curr, which holds the range [tl, tr].
//this is modification for all elements in the range.
//if we wanted to add, we could change it to curr + (tr - tl + 1)*upd.
T apply(T curr, U upd, ll tl, ll tr) {
if (upd == this->identity_update) {
return curr; //this means no update should be made
}
T ans = (tr - tl + 1) * upd;
return ans;
}
//there may be several updates already, so we need to combine them
//this is modification, so we just replace the old value with new value
//if we wanted addition, we could do ans = old_update + new_update
U combineUpdate(U old_update, U new_update, ll tl, ll tr) {
U ans = old_update;
if (new_update != this->identity_update) {
ans = new_update;
}
return ans;
}
void push_down(ll v, ll tl, ll tr) {
if (lazy[v] == identity_update) return;
st[v] = apply(st[v], lazy[v], tl, tr);
if (2*v + 2 < 4*n) {
ll tm = (tl + tr) >> 1;
lazy[2*v + 1] = combineUpdate(lazy[2*v + 1], lazy[v], tl,tm);
lazy[2*v + 2] = combineUpdate(lazy[2*v + 2], lazy[v],tm + 1, tr);
}
lazy[v] = identity_update;
}
T queryUtil(ll v, ll tl, ll tr, ll l, ll r) {
push_down(v, tl, tr);
if (l > r) {
return identity_element;
}
if (tr < l || tl > r) {
return identity_element;
}
if (l <= tl && r >= tr) {
return st[v];
}
ll tm = (tl + tr) >> 1;
return combine(queryUtil(2*v + 1, tl, tm, l, r), queryUtil(2*v + 2,tm + 1, tr, l, r));
}
void updateUtil(ll v, ll tl, ll tr, ll l, ll r, U upd) {
push_down(v, tl, tr);
if (tr < l || tl > r) return;
if (tl >= l && tr <= r) {
lazy[v] = combineUpdate(lazy[v],upd, tl, tr);
push_down(v, tl, tr);
}
else {
ll tm = (tl + tr) >> 1;
updateUtil(2*v+1, tl,tm, l, r, upd);
updateUtil(2*v+2,tm+1 ,tr, l, r, upd);
st[v] = combine(st[2*v + 1], st[2*v + 2]);
}
}
void build(vector<T>a) {
assert((int)a.size() == n);
buildUtil(0, 0, n - 1, a);
}
//[l, r]
T query(ll l, ll r) {
return queryUtil(0, 0, n - 1, l, r);
}
//[l, r]
void update(ll l,ll r, U upd) {
updateUtil(0, 0, n - 1, l, r, upd);
}
};
void solve() {
int n, q;
cin >> n >> q;
vector<ll> vals(n);
for (int i = 0; i < n; i++) {
cin >> vals[i];
}
vector<vector<int>> g(n);
for (int i = 0; i < n - 1; i++) {
int u, v;
cin >> u >> v;
u--, v--;
g[u].push_back(v);
g[v].push_back(u);
}
vector<ll> a(n);
vector<int> start(n), end(n);
int timer = 0;
auto dfs = [&](auto dfs, int u, int p) -> void {
a[timer] = vals[u];
start[u] = timer++;
for (auto &v: g[u]) {
if (v == p) continue;
dfs(dfs, v, u);
}
end[u] = timer;
};
dfs(dfs, 0, -1);
Lsegtree<ll, ll> st(n, 0, -1);
st.build(a);
for (int i = 0; i < q; i++) {
int op;
cin >> op;
if (op == 1) {
int s, x;
cin >> s >> x;
s--;
st.update(start[s], start[s], x);
} else if (op == 2) {
int s;
cin >> s;
s--;
cout << st.query(start[s], end[s] - 1) << '\n';
} else {
assert(false);
}
}
}
int main() {
ios_base::sync_with_stdio(false);
cin.tie(nullptr);
int T = 1;
// cin >> T;
while (T--) {
solve();
}
return 0;
}