The [official editorial](http://usaco.org/current/data/sol_cowland_gold_feb19.html) to this problem is not asymptotically optimal and involves an advanced concept that's infrequent in USACO Gold. I managed to derive a relatively "low-tech" and faster solution, which I will describe below.↵
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### [Problem link](http://usaco.org/index.php?page=viewproblem2&cpid=921)↵
↵
----------------------------------------------------------------------------------------------------------------------↵
↵
Solution↵
------------------↵
↵
for the sake of convenience, let $\sum\limits_u^v$ denote the XOR of all labels on the path from node $u$ to node $v$.↵
↵
**Subtask 1**↵
↵
We want a way to efficiently calculate the XOR of every label along an arbitrary path without modifications. Note that after rooting the tree at an arbitrary root $r$, if we split a path by the LCA of its endpoints, this can be computed alongside binary-jumping LCA, but that approach can't be easily transferred to subtask 2.↵
↵
The key observation is that, since XOR is an involution, if we take the XOR of $\sum\limits_u^r$ and $\sum\limits_v^r$, nodes that are on both paths will not contribute. By definition, these are exactly the common ancestors of $u$ and $v$, and almost all of them are nodes that we wouldn't want to count in the query!↵
↵
Since the LCA is the only common ancestor of $u,v$ on the path between them, we can simply return $(\sum\limits_u^r\oplus\sum\limits_v^r)\oplus e_{\text{LCA(u,v)}}$ for each query. The "prefix sums" of labels can be pre-computed in $O(n)$ using a simple DFS, so the overall complexity of this subtask is $O(n+b(n)+q\cdot (1+c(n))$, where $b(n),c(n)$ are the pre-computation and query times of our LCA method, respectively.↵
↵
**Subtask 2**↵
↵
Going further along the idea to keep track of paths sourcing from the root, for some arbitrary node $x$, which prefix sums will have to be updated if it is modified? The answer is exactly every node that has $x$ as an ancestor. This is the subtree of $x$ by definition, and subtree modifications are right up the alley of the Euler Tour technique!↵
↵
If we establish an Euler Tour of the tree, we can effectively modify subtrees in $O(\log(n))$ with any data structure that supports range modification, such as a lazy propagation segment tree. However, note that when retrieving the answer, we only need to query two individual nodes, so it's sufficient to use a slightly modified version of an iterative segment tree due to [user:Al.Cash,2022-09-07], where a node keeps track of every update ever done over its corresponding segment, instead of the value of its corresponding segment; that way, to query a specific index, we can traverse along the path from it to the root of the segment tree, and their XOR is the current value of that index. A more detailed explanation of iterative segment trees and this specific trick can be found [here](https://mirror.codeforces.com/blog/entry/18051).↵
↵
↵
Every technique used here can be found in multiple past USACO Gold problems (excluding range modification/point query, but that is a fairly natural extension of vanilla segment trees IMO), hence I consider my solution canonical to the gold division.↵
↵
Time complexity: $O(n+b(n)+q\log(n))$.↵
↵
----------------------------------------------------------------------------------------------------------------------↵
↵
My C++ code, which comfortably AC's, is shown below. Credits to [user:Al.Cash,2022-09-07] once again for the segment tree implementation.↵
↵
<spoiler summary="Click">↵
The DFS2 method implicitly keeps track of a reverse topological sort, and retrieves the original tag for each node by XOR-ing its prefix sum with its parent's prefix sum in reverse topological order.↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
using namespace std;↵
↵
#define MAXN 100000↵
#define MAXLOG 18↵
int t[2*MAXN],st[MAXN],en[MAXN],n,q,timer,val[MAXN],up[MAXN][MAXLOG],l2;↵
vector<int> adj[MAXN];↵
↵
↵
void dfs(int node, int parent) {↵
st[node] = timer++;↵
up[node][0] = parent;↵
for (int i = 1; i <= l2; ++i){↵
up[node][i] = up[up[node][i-1]][i-1];↵
}↵
for (int i : adj[node]) {↵
if (i != parent) {↵
val[i] ^= val[node];↵
dfs(i, node);↵
}↵
}↵
en[node] = timer;↵
}↵
↵
void dfs2(int node, int parent) {↵
for (int i : adj[node]) {↵
if (i != parent) {↵
dfs2(i,node);↵
}↵
}↵
if (node) val[node] ^= val[parent];↵
}↵
↵
void modify(int l, int r, int value) {↵
for (l += n, r += n; l < r; l >>= 1, r >>= 1) {↵
if (l&1) t[l++] ^= value;↵
if (r&1) t[--r] ^= value;↵
}↵
}↵
↵
↵
int query(int p) {↵
int res = 0;↵
for (p += n; p > 0; p >>= 1) res ^= t[p];↵
return res;↵
}↵
↵
↵
bool is_ancestor(int u, int v) {↵
return (st[u] <= st[v]) && (en[u] >= en[v]);↵
}↵
↵
↵
int lca(int u, int v) {↵
if (is_ancestor(u, v))↵
return u;↵
if (is_ancestor(v, u))↵
return v;↵
for (int i = l2; i >= 0; --i) {↵
if (!is_ancestor(up[u][i], v))↵
u = up[u][i];↵
}↵
return up[u][0];↵
}↵
↵
↵
int main() {↵
freopen("cowland.in", "r", stdin);↵
freopen("cowland.out", "w", stdout);↵
cin >> n >> q;↵
l2 = (int)ceil(log2(n));↵
for (int i = 0; i < n; i++) cin >> val[i];↵
for (int i = 1; i < n; i++){↵
int a,b;↵
cin >> a >> b;↵
adj[--a].push_back(--b); adj[b].push_back(a);↵
}↵
dfs(0,0);↵
for (int i = 0; i < n; i++) t[n + st[i]] = val[i];↵
dfs2(0,0);↵
while(q--){↵
int type; cin >> type;↵
if (type == 1){↵
int i,v;↵
cin >> i >> v;↵
int delta = val[--i]^v;↵
val[i] = v;↵
modify(st[i], en[i], delta);↵
}↵
else {↵
int i,j;↵
cin >> i >> j;↵
int ret = query(st[--i]) ^ query(st[--j]);↵
ret ^= val[lca(i,j)];↵
cout << ret << endl;↵
}↵
}↵
}↵
~~~~~↵
</spoiler>↵
↵
Thank you for reading, and remember to praise the cows.↵
↵
----------------------------------------------------------------------------------------------------------------------↵
↵
### [Problem link](http://usaco.org/index.php?page=viewproblem2&cpid=921)↵
↵
----------------------------------------------------------------------------------------------------------------------↵
↵
Solution↵
------------------↵
↵
for the sake of convenience, let $\sum\limits_u^v$ denote the XOR of all labels on the path from node $u$ to node $v$.↵
↵
**Subtask 1**↵
↵
We want a way to efficiently calculate the XOR of every label along an arbitrary path without modifications. Note that after rooting the tree at an arbitrary root $r$, if we split a path by the LCA of its endpoints, this can be computed alongside binary-jumping LCA, but that approach can't be easily transferred to subtask 2.↵
↵
The key observation is that, since XOR is an involution, if we take the XOR of $\sum\limits_u^r$ and $\sum\limits_v^r$, nodes that are on both paths will not contribute. By definition, these are exactly the common ancestors of $u$ and $v$, and almost all of them are nodes that we wouldn't want to count in the query!↵
↵
Since the LCA is the only common ancestor of $u,v$ on the path between them, we can simply return $(\sum\limits_u^r\oplus\sum\limits_v^r)\oplus e_{\text{LCA(u,v)}}$ for each query. The "prefix sums" of labels can be pre-computed in $O(n)$ using a simple DFS, so the overall complexity of this subtask is $O(n+b(n)+q\cdot (1+c(n))$, where $b(n),c(n)$ are the pre-computation and query times of our LCA method, respectively.↵
↵
**Subtask 2**↵
↵
Going further along the idea to keep track of paths sourcing from the root, for some arbitrary node $x$, which prefix sums will have to be updated if it is modified? The answer is exactly every node that has $x$ as an ancestor. This is the subtree of $x$ by definition, and subtree modifications are right up the alley of the Euler Tour technique!↵
↵
If we establish an Euler Tour of the tree, we can effectively modify subtrees in $O(\log(n))$ with any data structure that supports range modification, such as a lazy propagation segment tree. However, note that when retrieving the answer, we only need to query two individual nodes, so it's sufficient to use a slightly modified version of an iterative segment tree due to [user:Al.Cash,2022-09-07], where a node keeps track of every update ever done over its corresponding segment, instead of the value of its corresponding segment; that way, to query a specific index, we can traverse along the path from it to the root of the segment tree, and their XOR is the current value of that index. A more detailed explanation of iterative segment trees and this specific trick can be found [here](https://mirror.codeforces.com/blog/entry/18051).↵
↵
↵
Every technique used here can be found in multiple past USACO Gold problems (excluding range modification/point query, but that is a fairly natural extension of vanilla segment trees IMO), hence I consider my solution canonical to the gold division.↵
↵
Time complexity: $O(n+b(n)+q\log(n))$.↵
↵
----------------------------------------------------------------------------------------------------------------------↵
↵
My C++ code, which comfortably AC's, is shown below. Credits to [user:Al.Cash,2022-09-07] once again for the segment tree implementation.↵
↵
<spoiler summary="Click">↵
The DFS2 method implicitly keeps track of a reverse topological sort, and retrieves the original tag for each node by XOR-ing its prefix sum with its parent's prefix sum in reverse topological order.↵
↵
~~~~~↵
#include <bits/stdc++.h>↵
using namespace std;↵
↵
#define MAXN 100000↵
#define MAXLOG 18↵
int t[2*MAXN],st[MAXN],en[MAXN],n,q,timer,val[MAXN],up[MAXN][MAXLOG],l2;↵
vector<int> adj[MAXN];↵
↵
↵
void dfs(int node, int parent) {↵
st[node] = timer++;↵
up[node][0] = parent;↵
for (int i = 1; i <= l2; ++i){↵
up[node][i] = up[up[node][i-1]][i-1];↵
}↵
for (int i : adj[node]) {↵
if (i != parent) {↵
val[i] ^= val[node];↵
dfs(i, node);↵
}↵
}↵
en[node] = timer;↵
}↵
↵
void dfs2(int node, int parent) {↵
for (int i : adj[node]) {↵
if (i != parent) {↵
dfs2(i,node);↵
}↵
}↵
if (node) val[node] ^= val[parent];↵
}↵
↵
void modify(int l, int r, int value) {↵
for (l += n, r += n; l < r; l >>= 1, r >>= 1) {↵
if (l&1) t[l++] ^= value;↵
if (r&1) t[--r] ^= value;↵
}↵
}↵
↵
↵
int query(int p) {↵
int res = 0;↵
for (p += n; p > 0; p >>= 1) res ^= t[p];↵
return res;↵
}↵
↵
↵
bool is_ancestor(int u, int v) {↵
return (st[u] <= st[v]) && (en[u] >= en[v]);↵
}↵
↵
↵
int lca(int u, int v) {↵
if (is_ancestor(u, v))↵
return u;↵
if (is_ancestor(v, u))↵
return v;↵
for (int i = l2; i >= 0; --i) {↵
if (!is_ancestor(up[u][i], v))↵
u = up[u][i];↵
}↵
return up[u][0];↵
}↵
↵
↵
int main() {↵
freopen("cowland.in", "r", stdin);↵
freopen("cowland.out", "w", stdout);↵
cin >> n >> q;↵
l2 = (int)ceil(log2(n));↵
for (int i = 0; i < n; i++) cin >> val[i];↵
for (int i = 1; i < n; i++){↵
int a,b;↵
cin >> a >> b;↵
adj[--a].push_back(--b); adj[b].push_back(a);↵
}↵
dfs(0,0);↵
for (int i = 0; i < n; i++) t[n + st[i]] = val[i];↵
dfs2(0,0);↵
while(q--){↵
int type; cin >> type;↵
if (type == 1){↵
int i,v;↵
cin >> i >> v;↵
int delta = val[--i]^v;↵
val[i] = v;↵
modify(st[i], en[i], delta);↵
}↵
else {↵
int i,j;↵
cin >> i >> j;↵
int ret = query(st[--i]) ^ query(st[--j]);↵
ret ^= val[lca(i,j)];↵
cout << ret << endl;↵
}↵
}↵
}↵
~~~~~↵
</spoiler>↵
↵
Thank you for reading, and remember to praise the cows.↵
