这个问题很容易想到之间的关系，假设现在所要查询的这条链上有V1、V2、…… VK个点，那么第i个点的贡献在抑或中出现的次数XOR为
于是，如果K为偶数的时候，我们直接求这条链上所有值的抑或XOR和即可，树链剖分就可以很好的维护了。
如果K为奇数的时候，我们需要求第2、4、6、……这几个偶数点的抑或XOR和。如果我们直接在线段树上维护当然是不行的，因为dfs序号是没有这样的独立性的，可能会有两个同时为偶数dfs序的点却是连接在一起的。
所以考虑深度，会发现，深度奇偶不同的点一定是不相连接的，并且中间只隔了一个点。于是，我们这里开两个线段树，维护一下每个奇偶深度的抑或XOR和。
这样，我们查询的时候，只用查询于两个结点的deep的奇偶不同的点的抑或和即可。
因为是奇数个点，所以也代表了两个查询结点一定是在深度上同奇偶。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#include <bitset>
//#include <unordered_map>
//#include <unordered_set>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
#define INF 0x3f3f3f3f3f3f3f3f
#define eps 1e-8
#define HalF (l + r)>>1
#define lsn rt<<1
#define rsn rt<<1|1
#define Lson lsn, l, mid
#define Rson rsn, mid+1, r
#define QL Lson, ql, qr
#define QR Rson, ql, qr
#define myself rt, l, r
#define MP(a, b) make_pair(a, b)
#define MAX_3(a, b, c) max(a, max(b, c))
#define Rabc(x) x > 0 ? x : -x
using namespace std;
typedef unsigned long long ull;
typedef unsigned int uit;
typedef long long ll;
inline int read()
{
int x=0; char c=getchar();
while(c<'0'||c>'9') c=getchar();
while(c>='0'&&c<='9') { x=(x<<1)+(x<<3)+c-'0'; c=getchar(); }
return x;
}
const int maxN = 2e5 + 7;
int N, Q, a[maxN], head[maxN], cnt;
struct Eddge
{
int nex, to;
Eddge(int a=-1, int b=0):nex(a), to(b) {}
}edge[maxN << 1];
inline void addEddge(int u, int v)
{
edge[cnt] = Eddge(head[u], v);
head[u] = cnt++;
}
inline void _add(int u, int v) { addEddge(u, v); addEddge(v, u); }
int siz[maxN], deep[maxN], Wson[maxN], fa[maxN];
void dfs_1(int u, int father)
{
fa[u] = father; deep[u] = deep[father] + 1; Wson[u] = 0; siz[u] = 1;
int maxx = 0;
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == father) continue;
dfs_1(v, u);
siz[u] += siz[v];
if(maxx < siz[v])
{
maxx = siz[v];
Wson[u] = v;
}
}
}
int top[maxN], dfn[maxN], tot, rid[maxN];
void dfs_2(int u, int topy)
{
top[u] = topy;
dfn[u] = ++tot; rid[tot] = u;
if(Wson[u]) dfs_2(Wson[u], topy);
for(int i=head[u], v; ~i; i=edge[i].nex)
{
v = edge[i].to;
if(v == fa[u] || v == Wson[u]) continue;
dfs_2(v, v);
}
}
int tree[maxN << 2][3] = {0};
inline void pushup(int rt)
{
tree[rt][2] = tree[lsn][2] ^ tree[rsn][2];
tree[rt][0] = tree[lsn][0] ^ tree[rsn][0];
tree[rt][1] = tree[lsn][1] ^ tree[rsn][1];
}
void buildTree(int rt, int l, int r)
{
if(l == r)
{
tree[rt][deep[rid[l]] & 1] = a[rid[l]];
tree[rt][2] = a[rid[l]];
return;
}
int mid = HalF;
buildTree(Lson); buildTree(Rson);
pushup(rt);
}
void update(int rt, int l, int r, int qx, int val)
{
if(l == r) { tree[rt][deep[rid[l]] & 1] = val; tree[rt][2] = val; return; }
int mid = HalF;
if(qx <= mid) update(Lson, qx, val);
else update(Rson, qx, val);
pushup(rt);
}
int s[3] = {0};
void query(int rt, int l, int r, int ql, int qr)
{
if(ql <= l && qr >= r)
{
s[0] ^= tree[rt][0];
s[1] ^= tree[rt][1];
s[2] ^= tree[rt][2];
return;
}
int mid = HalF;
if(qr <= mid) query(QL);
else if(ql > mid) query(QR);
else { query(QL); query(QR); }
}
inline int _LCA(int u, int v)
{
while(top[u] ^ top[v])
{
if(deep[top[u]] < deep[top[v]]) swap(u, v);
u = fa[top[u]];
}
if(deep[u] > deep[v]) swap(u, v);
return u;
}
inline int Range_Query(int u, int v)
{
int ans = 0, lca = _LCA(u, v), len, op = (deep[u] & 1) ^ 1;
int dis = deep[u] + deep[v] - 2 * deep[lca] + 1; dis &= 1;
if(dis)
{
while(top[u] ^ top[v])
{
if(deep[top[u]] < deep[top[v]]) swap(u, v);
len = deep[u] - deep[top[u]] + 1; len &= 1;
s[0] = s[1] = s[2] = 0;
query(1, 1, N, dfn[top[u]], dfn[u]);
ans ^= s[op];
u = fa[top[u]];
}
if(deep[u] < deep[v]) swap(u, v);
len = deep[u] - deep[v] + 1; len &= 1;
s[0] = s[1] = s[2] = 0;
query(1, 1, N, dfn[v], dfn[u]);
ans ^= s[op];
}
else
{
s[2] = 0;
while(top[u] ^ top[v])
{
if(deep[top[u]] < deep[top[v]]) swap(u, v);
query(1, 1, N, dfn[top[u]], dfn[u]);
u = fa[top[u]];
}
if(deep[u] < deep[v]) swap(u, v);
query(1, 1, N, dfn[v], dfn[u]);
ans = s[2];
}
return ans;
}
inline void init()
{
cnt = tot = 0;
for(int i=1; i<=N; i++) head[i] = -1;
}
int main()
{
N = read(); Q = read();
init();
for(int i=1; i<=N; i++) a[i] = read();
for(int i=1, u, v; i<N; i++)
{
u = read(); v = read();
_add(u, v);
}
deep[0] = 0;
dfs_1(1, 0);
dfs_2(1, 1);
buildTree(1, 1, N);
int op, x, y;
while(Q--)
{
op = read(); x = read(); y = read();
if(op == 1)
{
update(1, 1, N, dfn[x], y);
}
else
{
printf("%d\n", Range_Query(x, y));
}
}
return 0;
}