poj 3764 字典树 树上任意两点边权异或最大值

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 uand v of length w.

Output

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

Sample Input

40 1 31 2 41 3 6

Sample Output

7

Hint

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

把点到根的值都加到树上求一遍最大异或就行

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#include <vector>using namespace std;int tot,sz;int si;struct Edge{int v, w,next;}e[200010];int head[200010];inline int read(){    int x=0,f=1;char ch=getchar();    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}    return x*f;}int nxt[100010*31][2];int ans;int a[100020];void insert(int x){int p=0;for(int i=30;i>=0;i--){int dd=(x>>i)&1;if(!nxt[p][dd]) nxt[p][dd]=++sz,memset(nxt[sz],0,sizeof(nxt[sz]));p=nxt[p][dd];}}int ask(int x){int p=0;int res=0;for(int i=30;i>=0;i--){int dd=(x>>i)&1;if(nxt[p][1-dd]) p=nxt[p][1-dd],res+=(1<<i);else p=nxt[p][dd];}return res;}void dfs(int u,int fa,int now){a[++tot]=now;for(int i=head[u];~i;i=e[i].next){if(e[i].v==fa) continue;dfs(e[i].v,u,(now^e[i].w));}}void add(int u,int v,int w){e[si].v=v,e[si].next=head[u];e[si].w=w;head[u]=si++;}int main(){int n;while(~scanf("%d",&n)){tot=0,sz=0;si=1;memset(head,-1,sizeof(head));memset(nxt[0],0,sizeof(nxt[0]));ans=0;for(int i=1;i<n;i++){int u,v,w;u=read(),v=read(),w=read();add(u,v,w);add(v,u,w);}dfs(0,-1,0);for(int i=1;i<=tot;i++){ans=max(ans,ask(a[i]));insert(a[i]);}printf("%d\n",ans );}}