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

Sample Output

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 ;

 }

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POJ3764 The xor-longest Path

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