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

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)

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 );
}
}