[洛谷P4592][TJOI2018]异或

题目大意：有一棵$n$个点的树，第$i$个点权值为$w_i$，有两种操作：

1. $1\;x\;y：$询问节点$x$的子树中与$y$异或结果的最大值
2. $2\;x\;y\;z：$询问路径$x$到$y$上点与$z$异或结果最大值

题解：树剖，然后就可以把树上问题转化为序列上的问题，可持久化$Trie$即可

卡点：树剖判断条件错

C++ Code：

#include <algorithm>
#include <cstdio>
#define maxn 100010
#define M 30
#define N (maxn * (M + 2))

int head[maxn], cnt;
struct Edge {
	int to, nxt;
} e[maxn << 1];
inline void addedge(int a, int b) {
	e[++cnt] = (Edge) {b, head[a]}; head[a] = cnt;
	e[++cnt] = (Edge) {a, head[b]}; head[b] = cnt;
}

namespace Trie {
	int V[N], nxt[N][2], idx;
	void insert(int &rt, int x, int dep) {
		nxt[++idx][0] = nxt[rt][0], nxt[idx][1] = nxt[rt][1], V[idx] = V[rt] + 1, rt = idx;
		if (!~dep) return ;
		insert(nxt[rt][x >> dep & 1], x, dep - 1);
	}
	int query(int L, int R, int x) {
		int res = 0;
		for (int i = M; ~i; i--) {
			int tmp = x >> i & 1;
			if (V[nxt[R][!tmp]] - V[nxt[L][!tmp]]) L = nxt[L][!tmp], R = nxt[R][!tmp], res |= 1 << i;
			else L = nxt[L][tmp], R = nxt[R][tmp];
		}
		return res;
	}
}
int n, m, w[maxn];

int rt[maxn];

int dfn[maxn], idx, dep[maxn], sz[maxn];
int fa[maxn], son[maxn], top[maxn];
void dfs1(int u) {
	sz[u] = 1;
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (v != fa[u]) {
			fa[v] = u;
			dep[v] = dep[u] + 1;
			dfs1(v);
			if (!son[u] || sz[son[u]] < sz[v]) son[u] = v;
			sz[u] += sz[v];
		}
	}
}
void dfs2(int u) {
	dfn[u] = ++idx;
	Trie::insert(rt[idx] = rt[idx - 1], w[u], M);
	int v = son[u];
	if (v) top[v] = top[u], dfs2(v);
	for (int i = head[u]; i; i = e[i].nxt) {
		int v = e[i].to;
		if (!dfn[v]) {
			top[v] = v;
			dfs2(v);
		}
	}
}

int query(int x, int y, int z) {
	int res = 0;
	while (top[x] != top[y]) {
		if (dep[top[x]] < dep[top[y]]) std::swap(x, y);
		res = std::max(res, Trie::query(rt[dfn[top[x]] - 1], rt[dfn[x]], z));
		x = fa[top[x]];
	}
	if (dep[x] > dep[y]) std::swap(x, y);
	res = std::max(res, Trie::query(rt[dfn[x] - 1], rt[dfn[y]], z));
	return res;
}
int main() {
	scanf("%d%d", &n, &m);
	for (int i = 1; i <= n; i++) scanf("%d", w + i);
	for (int i = 1, a, b; i < n; i++) {
		scanf("%d%d", &a, &b);
		addedge(a, b);
	}
	dfs1(1);
	top[1] = 1;
	dfs2(1);
	while (m --> 0) {
		int op, x, y, z;
		scanf("%d%d%d", &op, &x, &y);
		if (op == 1) {
			printf("%d\n", Trie::query(rt[dfn[x] - 1], rt[dfn[x] + sz[x] - 1], y));
		} else {
			scanf("%d", &z);
			printf("%d\n", query(x, y, z));
		}
	}
	return 0;
}