Poj 3764 The xor-longest Path(Trie树+xor+贪心)

The xor-longest Path

Time Limit: 2000MS Memory Limit: 65536K

Total Submissions: 6455 Accepted: 1392

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:

{xor}length(p)=\oplus{e \in p}w(e)

⊕ 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)

/*
Trie树+异或+贪心.
异或性质 a^b=(a^c)^(b^c).
首先dfs出根节点到每个节点的xor.
然后a^b=(a^root)^(b^root).
拆成二进制挂在Trie树上跑.
从高位到低位存
贪心此位为1则看0是否存在
正确性显然.
*/
#include<iostream>
#include<cstdio>
#include<cstring>
#define MAXN 100001
using namespace std;
int n,m,dis[MAXN],fa[MAXN],ans,cut,tot,head[MAXN],s[MAXN];
struct edge{int v,next,x;}e[MAXN*2];
struct data{int next[2];}tree[MAXN*16];
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-48,ch=getchar();
    return x*f;
}
void add(int u,int v,int x)
{
    e[++cut].v=v;
    e[cut].x=x;
    e[cut].next=head[u];
    head[u]=cut;
}
void add(int x)
{
    int now=0,xx;
    for(int i=30;i>=0;i--)
    {
        if(x&(1<<i)) xx=1;
        else xx=0;
        if(!tree[now].next[xx]) tot++,tree[now].next[xx]=tot;
        now=tree[now].next[xx];
    }
}
int query(int x)
{
    int now=0,xx,num=0;
    for(int i=30;i>=0;i--)
    {
        if(x&(1<<i)) xx=0;
        else xx=1;
        if(tree[now].next[xx])
        {
            num|=(1<<i);
            now=tree[now].next[xx];
        }
        else now=tree[now].next[!xx];
    }
    return num;
}
void dfs(int u,int f,int xo)
{
    fa[u]=f;s[u]=xo;
    for(int i=head[u];i;i=e[i].next)
    {
        int v=e[i].v;
        if(!fa[v]&&v!=f) dfs(v,u,xo^e[i].x);
    }
}
int main()
{
    int x,y,z;
    while(~scanf("%d",&n))
    {
        ans=0;cut=0;tot=0;
        memset(head,0,sizeof head);
        memset(fa,0,sizeof fa);
        memset(tree,0,sizeof tree);
        for(int i=1;i<=n-1;i++)
        {
            x=read(),y=read(),z=read();
            x++,y++;
            add(x,y,z),add(y,x,z);
        }
        dfs(1,1,0);
        for(int i=1;i<=n;i++) add(s[i]);
        for(int i=1;i<=n;i++) ans=max(ans,query(s[i]));

        printf("%d\n",ans);
    }
    return 0;
}