题目要你支持：
- 更改树上某一点 (x) 权值为 (y)
- 求 (x) 到 (y) 的路径权值异或和
这一看就是树链剖分模板题么
用线段树维护单点修改，区间查询异或
然后树链剖分维护路径就搞定了
#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(s, v, it) for (s::iterator it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define ub upper_bound
#define lb lower_bound
#define pq priority_queue
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(#x".in", "r", stdin),freopen(#x".out", "w", stdout)
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 100010 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int MOD = 1000000007 ;
const double eps = 1e-7 ;
void print(int x) { cout << x << endl ; exit(0) ; }
void PRINT(string x) { cout << x << endl ; exit(0) ; }
void douout(double x){ printf("%lf
", x + 0.0000000001) ; }
vi e[N] ;
int fa[N], dep[N], sz[N], son[N], dfn[N], seq[N], top[N], a[N] ;
int n, m, cnt ;
void dfs1(int x, int fat) {
sz[x] = 1 ;
rep(i, 0, siz(e[x]) - 1) {
int to = e[x][i] ;
if (to == fat) continue ;
fa[to] = x ; dep[to] = dep[x] + 1 ;
dfs1(to, x) ;
if (sz[to] > sz[son[x]]) son[x] = to ;
sz[x] += sz[to] ;
}
}
void dfs2(int x, int tp) {
dfn[x] = ++cnt ; seq[cnt] = x ;
top[x] = tp ;
if (!son[x]) return ;
dfs2(son[x], tp) ;
rep(i, 0, siz(e[x]) - 1) {
int to = e[x][i] ;
if (to == fa[x] || to == son[x]) continue ;
dfs2(to, to) ;
}
}
struct SegTree {
int l, r, v ;
#define ls(x) x << 1
#define rs(x) x << 1 | 1
#define l(x) tr[x].l
#define r(x) tr[x].r
#define v(x) tr[x].v
} tr[N << 2] ;
void pushup(int x) {
v(x) = v(ls(x)) ^ v(rs(x)) ;
}
void build(int x, int l, int r) {
l(x) = l, r(x) = r ;
if (l == r) {
v(x) = a[seq[l]] ;
return ;
}
int mid = (l + r) >> 1 ;
build(ls(x), l, mid) ;
build(rs(x), mid + 1, r) ;
pushup(x) ;
}
void modify(int x, int pos, int c) {
if (l(x) == r(x)) {
v(x) = c ;
return ;
}
int mid = (l(x) + r(x)) >> 1 ;
if (pos <= mid) modify(ls(x), pos, c) ;
else modify(rs(x), pos, c) ;
pushup(x) ;
}
int query(int x, int l, int r) {
if (l <= l(x) && r(x) <= r) return v(x) ;
int mid = (l(x) + r(x)) >> 1, ans = 0 ;
if (l <= mid) ans ^= query(ls(x), l, r) ;
if (mid < r) ans ^= query(rs(x), l, r) ;
return ans ;
}
int Query(int x, int y) {
int fx = top[x], fy = top[y], ans = 0 ;
while (fx != fy) {
if (dep[fx] < dep[fy]) swap(x, y), swap(fx, fy) ;
ans ^= query(1, dfn[fx], dfn[x]) ;
x = fa[fx] ; fx = top[x] ;
}
if (dep[x] > dep[y]) swap(x, y) ;
return ans ^ query(1, dfn[x], dfn[y]) ;
}
signed main(){
freopen("cowland.in", "r", stdin) ;
freopen("cowland.out", "w", stdout) ;
// 支持单点修改，查询路径，也就是树链剖分模板题
scanf("%lld%lld", &n, &m) ;
rep(i, 1, n) scanf("%lld", &a[i]) ;
rep(i, 1, n - 1) {
int a, b ; scanf("%lld%lld", &a, &b) ;
e[a].pb(b) ; e[b].pb(a) ;
}
dfs1(1, -1) ;
dfs2(1, 1) ;
build(1, 1, n) ;
while (m--) {
int op, x, y ; scanf("%lld%lld%lld", &op, &x, &y) ;
if (op == 1) modify(1, dfn[x], y) ;
else printf("%lld
", Query(x, y)) ;
}
return 0 ;
}
/*
写代码时请注意：
1.ll？数组大小，边界？数据范围？
2.精度？
3.特判？
4.至少做一些
思考提醒：
1.最大值最小->二分？
2.可以贪心么？不行dp可以么
3.可以优化么
4.维护区间用什么数据结构？
5.统计方案是用dp？模了么？
6.逆向思维？
*/