XOR longest path values

# Meaning of the title
given a tree of n nodes, the tree has a weight side, choose any two points from the tree, find the two points on the path the maximum value of the exclusive or up

# Explanations

The array D [x] xor denotes the value of the root node to all sides on the right path x, and the structure is a tree, all nodes determined by dfs d
Determine all D array, the node x to node y all XOR is the weight D [x] xor D [y ], if it does not diverge from the root of doubt

If a node is an ancestor node to another node, in accordance with the nature of the XOR can be offset from the root to the common ancestor node path
of the original problem is converted into D [1] ~ [n] D select any two numbers, the maximum seek XORed.

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 const int N=1e5+10;
 4 int n;
 5 int h[N],e[N*2],ne[N*2],w[N*2],idx;
 6 int d[N];
 7 int trie[N*31][2],tot=1;
 8 inline void add(int x,int y,int z){
 9     e[idx]=y;
10     w[idx]=z;
11     ne[idx]=h[x];
12     h[x]=idx++;
13 }
14 inline int dfs(int u,int f,int sum){
15     d[u]=sum;
16     for(int i=h[u];i!=-1;i=ne[i]){
17         int j=e[i];
18         if(j!=f) dfs(j,u,sum^w[i]);
19     }
20 }
21 inline void insert(int x){
22     int p=1;
23     for(int j=30;j>=0;j--){
24         int ch=(x>>j)&1;
25         if(!trie[p][ch])
26             trie[p][ch]=++tot;
27         p=trie[p][ch];
28     }
29 }
30 inline int search(int x){
31     int res=0;
32     int p=1;
33     for(int i=30;i>=0;i--){
34         int ch=x>>i&1;
35         if(trie[p][!ch]) {
36             res += 1 << i;
37             p=trie[p][!ch];
38         }
39         else
40             p=trie[p][ch];
41     }
42     return res;
43 }
44 int main(){
45     ios::sync_with_stdio(0);
46     cin.tie(0);
47     cout.tie(0);
48     memset(h,-1,sizeof h);
49     cin>>n;
50     for(int i=0;i<n;i++){
51         int u,v,w;
52         cin>>u>>v>>w;
53         add(u,v,w);
54         add(v,u,w);
55     }
56     dfs(0,-1,0);
57     int ans=0;
58     for(int i=0;i<n;i++)
59     {
60         insert(d[i]);
61         ans=max(ans,search(d[i]));
62     }
63     cout<<ans;
64 }