HDU4757

【题意】给出一棵树，每个结点有个权值a[i],m个询问，每个询问(x,y,z)，输出x和y路径上a[i]^z的最大值。a[i],z < 2^16

【分析】首先如果区间固定，那么可以直接把区间内的所有点插入trie(从高位插入，每次插入16位)；然后直接trie中询问即可，但是这里区间会改变，其实可以想可持久化线段树一样的方法做，每个结点建一颗trie依赖父亲，然后求出lca之后便在查找的时候用下lca的距离公式：dis = dis[x]+dis[y]-2*dis[lca]；吐槽下自己弱：找了半天bug竟然是lca线段树写搓了(本以为线段树很自信了，唉)。

【AC CODE】1809ms

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <cstring>
#include <cctype>
#include <cmath>
#include <set>
#include <bitset>
//#include <unordered_set>
#include <queue>
#include <stack>
#include <vector>
#include <string>
#include <algorithm>
using namespace std;
typedef long long LL;
#define rep(i,a,n) for(int i = a; i < n; i++)
#define repe(i,a,n) for(int i = a; i <= n; i++)
#define per(i,n,a) for(int i = n; i >= a; i--)
#define clc(a,b) memset(a,b,sizeof(a))
const int INF = 0x3f3f3f3f, MAXN = 200000+10, MAXM = MAXN<<1, BIT = 17;
int head[MAXN],tol,nxt[MAXM],to[MAXM];
int a[MAXN], ch[MAXN*50][2],sum[MAXN*50],all,rt[MAXN];
int cnt, ft[MAXN],d[MAXN<<1],num[MAXN<<1],fa[MAXN];

void insert(int &u, int v, int p)
{
	ch[++all][0] = ch[u][0],ch[all][1] = ch[u][1],sum[all] = sum[u]+1;u = all;
	if(p < 0) return;
	insert(ch[u][(bool)(v&(1<<p))],v,p-1);
}
int trie_query(int x, int y, int lca, int val)
{
	int fac = rt[fa[lca]];
	x = rt[x], y = rt[y], lca = rt[lca];
	int ans = 0;
	per(i,BIT,0)
	{
		bool v = val&(1<<i);
		if(sum[ch[y][v^1]]+sum[ch[x][v^1]]-sum[ch[lca][v^1]]-sum[ch[fac][v^1]])
		{
			ans |= 1<<i;
			x = ch[x][v^1],y = ch[y][v^1],lca = ch[lca][v^1],fac = ch[fac][v^1];
		}
		else if(sum[ch[y][v]]+sum[ch[x][v]]-sum[ch[lca][v]]-sum[ch[fac][v]])
			x = ch[x][v],y = ch[y][v],lca = ch[lca][v],fac = ch[fac][v];
	}
	return ans;
}
inline void add_edge(int u, int v)
{
	nxt[tol] = head[u], to[tol] = v;
	head[u] = tol++;
}
void dfs(int u, int dep)
{
	d[++cnt] = dep;num[cnt] = u;ft[u] = cnt;
	rt[u] = rt[fa[u]];
	insert(rt[u],a[u],BIT);
	for(int i = head[u]; ~i; i = nxt[i])
	{
		int v = to[i];
		if(v == fa[u]) continue;
		fa[v] = u;
		dfs(v,dep+1);
		d[++cnt] = dep,num[cnt] = u;
	}
}
int mi[MAXN<<2],minum[MAXN<<2];
inline int id(int x, int y){return x+y|x!=y;}
void bulid(int x, int y)
{
	if(x == y)
	{
		mi[id(x,y)] = d[x];
		minum[id(x,y)] = x;
		return;
	}
	int m = (x+y)>>1;
	bulid(x,m);
	bulid(m+1,y);
	int u = id(x,y), l = id(x,m),r = id(m+1,y);
	if(mi[l] < mi[r]) mi[u] = mi[l],minum[u] = minum[l];
	else mi[u] = mi[r],minum[u] = minum[r];
}
int query(int x, int y, int ql, int qr)
{
	if(ql <= x && y <= qr) return minum[id(x,y)];
	int m = (x+y)>>1,ans = 0;
	if(ql <= m)
	{
		int p = query(x,m,ql,qr);
		if(d[p] < d[ans]) ans = p;
	}
	if(qr > m)
	{
		int p = query(m+1,y,ql,qr);
		if(d[p] < d[ans]) ans = p;
	}
	return ans;
}
int lca_query(int x, int y)
{
	x = ft[x], y = ft[y];
	if(x > y) swap(x,y);
	return num[query(1,cnt,x,y)];
}

int main()
{
#ifdef SHY
	freopen("d:\\1.txt", "r", stdin);
#endif
	int n,m;
	d[0] = INF;
	while(~scanf("%d %d", &n, &m))
	{
		repe(i,1,n) scanf("%d", &a[i]);
		tol = 0;
		clc(head,-1);
		rep(i,1,n)
		{
			int u,v;
			scanf("%d %d", &u, &v);
			add_edge(u,v);
			add_edge(v,u);
		}
		cnt = all = 0;
		dfs(1,0);
		bulid(1,cnt);
		while(m--)
		{
			int x,y,v;
			scanf("%d %d %d", &x, &y, &v);
			int lca = lca_query(x,y);
			int ans = trie_query(x,y,lca,v);
			printf("%d\n", ans);
		}
	}
	return 0;
}