The xor-longest Path

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 3875		Accepted: 850

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)

　　這道題由於沒看到多組數據，WA了很久，題目比較簡單，但是運用了a xor b= a xor c xor b xor c，這應該是解異或題目常用的技巧。 

#include<iostream>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<ctime>
#include<cmath>
#include<algorithm>
#include<set>
#include<map>
#include<vector>
#include<string>
#include<queue>
#include<stack>
using namespace std;
#ifdef WIN32
#define LL "%I64d"
#else
#define LL "%lld"
#endif
#define MAXN 110000
#define MAXV MAXN
#define MAXE MAXV*2
#define MAXT MAXN*200
//AC
typedef long long qword;

int n,m;
struct trie
{
        int ch[2];
}tree[MAXT];
int toptr=1;
inline void new_node(int &now)
{
        now=++toptr;
        tree[now].ch[0]=tree[now].ch[1]=0;
}
int dbg=0;
void build_trie(qword x)
{
        dbg++;
        int now=1;
        int i;
        for (i=32;i>=0;i--)
        {
                if (!tree[now].ch[(x&(1ll<<i))!=0])
                {
                        new_node(tree[now].ch[(x&(1ll<<i))!=0]);
                }
                now=tree[now].ch[(x&(1ll<<i))!=0];
        }
        if (dbg==175672)
                dbg--;
}

struct Edge
{
        int np;
        qword val;
        Edge *next;
}E[MAXE],*V[MAXV];
int tope=-1;
void addedge(int x,int y,qword z)
{
        E[++tope].np=y;
        E[tope].val=z;
        E[tope].next=V[x];
        V[x]=&E[tope];
}
int fa[MAXN],depth[MAXN],val[MAXN];
void dfs(int now,int v)
{
        Edge *ne;
        build_trie(v);
        val[now]=v;
        for (ne=V[now];ne;ne=ne->next)
        {
                if (ne->np==fa[now])continue;
                fa[ne->np]=now;
                dfs(ne->np,v^ne->val);
        }
}
qword find(qword x)
{
        int now=1;
        int ret=0;
        int i;
        for (i=32;i>=0;i--)
        {
                if (tree[now].ch[(x&(1ll<<i))==0])
                {
                        now=tree[now].ch[(x&(1ll<<i))==0];
                        ret+=1ll<<i;
                }else
                {
                        now=tree[now].ch[(x&(1ll<<i))!=0];
                }
        }
        if (!now)throw 1;
        return ret;
}

int main()
{
        freopen("input.txt","r",stdin);
        //freopen("output.txt","w",stdout);
        int i,j,k;
        qword x,y,z;
        while (~scanf("%d",&n))
        {
                memset(V,0,sizeof(V));
                memset(fa,-1,sizeof(fa));
                toptr=1;
                tope=-1;
                tree[1].ch[0]=tree[1].ch[1]=0;
                for (i=1;i<n;i++)
                {
                        scanf(LL LL LL,&x,&y,&z);
                        addedge(x,y,z);
                        addedge(y,x,z);
                }
                fa[0]=0;
                dfs(0,0);
                qword ans=0;
                for (i=0;i<n;i++)
                {
                        ans=max(ans,find(val[i]));
                }
                printf(LL "\n",ans);
        }
        return 0;
}