题目链接：
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Problem Description
Zero and One are good friends who always have fun with each other. This time, they decide to do something on a tree which is a kind of graph that there is only one path from node to node. First, Zero will give One an tree and every node in this tree has a value. Then, Zero will ask One a series of queries. Each query contains three parameters: x, y, z which mean that he want to know the maximum value produced by z xor each value on the path from node x to node y (include node x, node y). Unfortunately, One has no idea in this question. So he need you to solve it.
Input
There are several test cases and the cases end with EOF. For each case:
The first line contains two integers n(1<=n<=10^5) and m(1<=m<=10^5), which are the amount of tree’s nodes and queries, respectively.
The second line contains n integers a[1..n] and a[i](0<=a[i]<2^{16}) is the value on the ith node.
The next n–1 lines contains two integers u v, which means there is an connection between u and v.
The next m lines contains three integers x y z, which are the parameters of Zero’s query.
The first line contains two integers n(1<=n<=10^5) and m(1<=m<=10^5), which are the amount of tree’s nodes and queries, respectively.
The second line contains n integers a[1..n] and a[i](0<=a[i]<2^{16}) is the value on the ith node.
The next n–1 lines contains two integers u v, which means there is an connection between u and v.
The next m lines contains three integers x y z, which are the parameters of Zero’s query.
Output
For each query, output the answer.
Sample Input
3 2 1 2 2 1 2 2 3 1 3 1 2 3 2
Sample Output
3 0
Source
题意：给一棵带权树，每次询问树上一条链上(x 到 y)的所有权值xor z的最大值。
思路：可持久化字典树+lca
不能用lca的倍增算法，会t，
关于lca可以看这个：https://www.cnblogs.com/zhouzhendong/p/7256007.html
/* data:2018.04.26 author:gswycf link:http://acm.hdu.edu.cn/showproblem.php?pid=4825 accout:tonysave */ #define ll long long #define IO ios::sync_with_stdio(false); #define maxn 100005 #include<stdio.h> #include<iostream> #include<algorithm> #include<math.h> #include<string.h> #include<vector> using namespace std; class Node{ public: int cnt,ls,rs; }; int n,m,cnt,aa,bb; Node tr[32*maxn]; int weight[maxn];int root[maxn];int deep[maxn],f[maxn][31]; vector<int> g[maxn]; inline void init() { memset(tr,0,sizeof(tr)); //memset(weight,0,sizeof(weight)); memset(root,0,sizeof(root)); deep[1]=0; cnt=0; for(int i=0;i<=n;i++) g[i].clear(); } int in(int pre,int x,int deep) { int num=++cnt; tr[num]=tr[pre]; tr[num].cnt=tr[pre].cnt+1; if(deep<0)return num; if(!((x>>deep)&1))tr[num].ls=in(tr[pre].ls,x,deep-1); else tr[num].rs=in(tr[pre].rs,x,deep-1); return num; } int query(int l,int r,int x,int deep) { if(deep<0)return 0; if(!((x>>deep)&1)) { if(tr[tr[r].rs].cnt>tr[tr[l].rs].cnt)return (1<<deep)+query(tr[l].rs,tr[r].rs,x,deep-1); else return query(tr[l].ls,tr[r].ls,x,deep-1); } else { if(tr[tr[r].ls].cnt>tr[tr[l].ls].cnt)return (1<<deep)+query(tr[l].ls,tr[r].ls,x,deep-1); else return query(tr[l].rs,tr[r].rs,x,deep-1); } } void bfs(int node,int fa) { root[node]=in(fa,weight[node],16); f[node][0]=fa; for(int i=0;i<g[node].size();i++) { if(g[node][i]!=fa) { deep[g[node][i]]=deep[node]+1; bfs(g[node][i],node); } } } inline void init2() { for(int j=1;(1<<j)<=n;j++) for(int i=1;i<=n;i++) f[i][j]=f[f[i][j-1]][j-1]; } int lca(int a,int b,int c) { aa=a,bb=b; if(deep[a]>deep[b])swap(a,b); int d=deep[b]-deep[a]; for(int i=0;i<30;i++) if((1<<i)&d)b=f[b][i]; if(a==b)return a; for(int i=29;i>=0;i--) { if(f[a][i]!=f[b][i]) a=f[a][i],b=f[b][i]; } b=f[b][0]; return b; } int main() { int a,b,c; while(~scanf("%d%d",&n,&m)) { init(); for(int i=1;i<=n;i++)scanf("%d",&weight[i]); for(int i=1;i<n;i++) { scanf("%d%d",&a,&b); g[a].push_back(b); g[b].push_back(a); } bfs(1,0);init2(); for(int i=1;i<=m;i++) { scanf("%d%d%d",&a,&b,&c); int k=lca(a,b,c); printf("%d\n",max(query(root[k-1],root[a],c,16),query(root[k-1],root[b],c,16))); } } }