POJ3764 The xor-longest Path

 

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 6361		Accepted: 1378

Description

In an edge-weighted tree, the xor-length of a path p is defined as the xor sum of the weights of edges on p:

⊕ is the xor operator.

We say a path the xor-longest path if it has the largest xor-length. Given an edge-weighted tree with n nodes, can you find the xor-longest path? 　

Input

The input contains several test cases. The first line of each test case contains an integer n(1<=n<=100000), The following n-1 lines each contains three integers u(0 <= u < n),v(0 <= v < n),w(0 <= w < 2^31), which means there is an edge between node u and v of length w.

Output

For each test case output the xor-length of the xor-longest path.

Sample Input

4
0 1 3
1 2 4
1 3 6

Sample Output

7

Hint

The xor-longest path is 0->1->2, which has length 7 (=3 ⊕ 4)

Source

trie树+贪心

DFS预处理出每个结点到根的路径的异或和。两点之间路径的异或和等于各自到根的路径的异或和的异或。

将所有的异或和转化成二进制串，建成trie树。

对于每个二进制串，在trie树上贪心选取使异或值最大的路径（尽量通往数值相反的结点），记录最优答案。

↑为了防止匹配到自身的一部分，每个二进制串都应该先查完再插入trie树。

 /*by SilverN*/
 #include<algorithm>
 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<vector>
 using namespace std;
 const int mxn=;
 int read(){
     int x=,f=;char ch=getchar();
     while(ch<'' || ch>''){if(ch=='-')f=-;ch=getchar();}
     while(ch>='' && ch<=''){x=x*+ch-'';ch=getchar();}
     return x*f;
 }
 struct edge{
     int v,nxt,w;
 }e[mxn<<];
 int hd[mxn],mct=;
 void add_edge(int u,int v,int w){
     e[++mct].v=v;e[mct].nxt=hd[u];e[mct].w=w;hd[u]=mct;return;
 }
 int t[mxn*][],cnt=;
 int n,ans=;
 int f[mxn];
 void insert(int x){
     int now=;
     for(int i=;i>=;i--){
         int v=(x>>i)&;
         if(!t[now][v])t[now][v]=++cnt;
         now=t[now][v];
     }
     return;
 }
 void query(int x){
     int now=;
     int res=;
     for(int i=;i>=;i--){
         int v=(x>>i)&;
         if(t[now][v^]){
             v^=;
             res+=(<<i);
         }
         now=t[now][v];
     }
     ans=max(res,ans);
     return;
 }
 void DFS(int u,int fa,int num){
     f[u]=num;
     for(int i=hd[u];i;i=e[i].nxt){
         int v=e[i].v;
         if(v==fa)continue;
         DFS(v,u,num^e[i].w);
     }
     return;
 }
 void init(){
     memset(hd,,sizeof hd);
     memset(t,,sizeof t);
 //    memset(f,0,sizeof f);
     mct=;cnt=;ans=;
     return;
 }
 int main(){
     while(scanf("%d",&n)!=EOF){
         init();
         int i,j,u,v,w;
         for(i=;i<n;i++){
             u=read();v=read();w=read();
             u++;v++;
             add_edge(u,v,w);
             add_edge(v,u,w);
         }
         DFS(,,);
         for(i=;i<=n;i++){
 //            printf("f[%d]:%d\n",i,f[i]);
             query(f[i]);
             insert(f[i]);
         }
         printf("%d\n",ans);
     }
     return ;
 }