The XOR-longest path

The XOR-longest path

Time limit:2000 ms		Memory limit:65536 K
Total submissions:3875		Accepted:850

Description

In an edge-weighted tree, the XOR-length of a pathPIs defined as the XOR sum of the weights of edges onP:

Operation 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 integerN(1 <=N<= 100000), the followingN-1 lines each contains three IntegersU(0 <=U<N),V(0 <=V<N),W(0 <=W<2 ^ 31), which means there is an edge between nodeUAndVOf LengthW.

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 limit 4)

This problem is caused by the absence of multi-group data. Wa has been around for a long time. It is relatively simple, but a xor B = a xor c xor B XOR C is used, this should be a common solution or topic skill.

#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//ACtypedef 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;}

 

 

The XOR-longest path