【Trie】The XOR-longest Path

[Link] title:

https://loj.ac/problem/10056

[Title] Italy

Please maximum output XOR tree path two points.

【answer】

The subject, said that after y overall how to do, is simply sobering up.

We know Xor path, we deal with all nodes from the root to the root of the XOR and we want different path or two points.

In fact, the use of the root to two XOR and points. Because LCA to the root of exclusive or twice, so the answer is to retain the exclusive or partial path.

 

In fact, this topic is Xor - pair variant.

Processing of the exclusive-OR value of the position of all the nodes starting from the root node.

 

[Code]

 1 #include<cstdio>
 2 #include<cstring>
 3 #include<algorithm>
 4 using namespace std;
 5 const int N = 1e5 + 100 ;
 6 int Son[N*31][2];
 7 typedef struct Edge{
 8     int To , next , w ;
 9 }Edge ;
10 Edge e[N<<1];
11 int Head[N],cnt,Dis[N],idx;
12 void Init(){
13     memset( Head , -1 ,sizeof Head );
14     cnt = idx = 0 ;
15 }
16 void Add_edge ( int u, int v ,int w ){
17     e[cnt] = Edge { v , Head[u] , w };
18     Head[u] = cnt ++ ;
19 }
20 
21 void dfs(int u,int Fa,int w ){
22     Dis[u] = w ;
23     for(int i = Head[u] ; ~i ; i = e[i].next){
24         int To = e[i].To;
25         if( To == Fa ) continue ;
26         dfs( To , u , w ^ e[i].w );
27     }
28 }
29 
30 void Insert ( int x ){
31     int p = 0 ;
32     for (int i=30;~i;i--){
33         int t = x >> i & 1 ;
34         if( !Son[p][t] )
35             Son[p][t] = ++idx ;
36         p = Son[p][t] ;
37     }
38 }
39 int Query(int x ){
40     int p = 0 ,res = 0;
41     for(int i=30;~i;i--){
42         int t = x >> i & 1 ;
43         if( Son[p][t^1] ){
44             res += 1 << i ;
45             p = Son[p][t^1];
46         }else{
47             p = Son[p][t] ;
48         }
49     }
50     return res ;
51 }
52 int main()
53 {
54     Init();
55     int n ;
56     scanf("%d",&n);
57     for(int i=1,u,v,w;i<n;i++){
58         scanf("%d%d%d",&u,&v,&w);
59         Add_edge( u , v , w );
60         Add_edge( v , u , w );
61     }
62     dfs( 1 , -1 , 0 ) ;
63     /*
64     for(int i=1;i<=n;i++){
65         printf("###%d###\n",Dis[i]);
66     }
67     */
68     for(int i=1;i<=n;i++){
69         Insert( Dis[i] );
70     }
71     int res = 0;
72     for(int i=1;i<=n;i++){
73         res = max( res , Query(Dis[i] ) );
74     }
75     printf("%d\n",res) ;
76     return 0 ;
77 }