poj 3764 The xor-longest Path (01 Trie）

链接：http://poj.org/problem?id=3764

题面：

The xor-longest Path

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11802		Accepted: 2321

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)

思路;

树上dfs一遍跟区间差不多的写法

实现代码;

#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
#define ll long long
const int M = 4e5+;
int tot;
int ch[*M][],cnt,head[M];
int val[*M],a[M],pre[M],nex[M],dp[M],ans;

struct node{
    int to,next;
    ll w;
}e[M*];

void add(int u,int v,int w){
    e[++cnt].to = v;e[cnt].w = w;e[cnt].next = head[u];head[u] = cnt;
}

void init(){
    tot = ; ans = ; cnt = ;
    ch[][] = ch[][] = ;
    memset(head,,sizeof(head));
}

void ins(ll x){
    int u = ;
    for(int i = ;i >= ;i --){
        int v = (x>>i)&;
        if(!ch[u][v]){
            ch[tot][] = ch[tot][] = ;
            val[tot] = ;
            ch[u][v] = tot++;
        }
        u = ch[u][v];
    }
    val[u] = x;
}

int query(int x){
    int u = ;
    for(int i = ;i >= ;i --){
        int v = (x>>i)&;
        if(ch[u][v^]) u = ch[u][v^];
        else u = ch[u][v];
    }
    return x^val[u];
}

void dfs(int u,int fa,int val){
    ins(val);
    for(int i = head[u];i;i = e[i].next){
        int v = e[i].to;
        if(v == fa) continue;
        ans = max(ans,query(val^e[i].w));
        dfs(v,u,val^e[i].w);
    }
}

int main()
{
    int n,u,v,w;
    while(scanf("%d",&n)!=EOF){
    init();
    for(int i = ;i < n;i++){
        scanf("%d%d%d",&u,&v,&w);
        add(u,v,w); add(v,u,w);
    }
    dfs(,-,);
    printf("%d\n",ans);
    }
}