题目链接：poj 3764 The xor-longest Path

题目大意：给定一棵树，每条边上有一个权值，找出一条路径，使得路径上权值的亦或和最大。

解题思路：dfs一遍，预处理出每个节点到根节点路径的亦或和rec，那么任意路径均可以表示rec[a] ^ rec[b]，所以问题

就转换成在一些数中选出两个数亦或和最大，那么就建立字典树查询即可。

#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;

const int maxn = 100005 * 32;
const int sigma_size = 2;

struct Tire {
    int sz;
    int g[maxn][sigma_size];

    void init();
    int idx(char ch);
    void insert(int s);
    int find(int s);
}T;

int N, M, E, first[maxn], jump[maxn], link[maxn], val[maxn], rec[maxn];

inline void add_Edge (int u, int v, int w) {
    link[E] = v;
    val[E] = w;
    jump[E] = first[u];
    first[u] = E++;
}

void dfs (int u, int pre, int s) {
    T.insert(s);
    rec[M++] = s;
    for (int i = first[u]; i + 1; i = jump[i]) {
        int v = link[i];
        if (v == pre)
            continue;
        dfs(v, u, s ^ val[i]);
    }
}

int main () {
    while (scanf("%d", &N) == 1) {
        M = E = 0;
        T.init();
        memset(first, -1, sizeof(first));

        int u, v, w;
        for (int i = 1; i < N; i++) {
            scanf("%d%d%d", &u, &v, &w);
            add_Edge(u, v, w);
            add_Edge(v, u, w);
        }
        dfs(0, 0, 0);

        int ans = 0;
        for (int i = 0; i < M; i++)
            ans = max(ans, T.find(rec[i]));
        printf("%d\n", ans);
    }
    return 0;
}

void Tire::init() {
    sz = 1;
    memset(g[0], 0, sizeof(g[0]));
}

int Tire::find(int s) {
    int ret = 0, u = 0;
    for (int i = 30; i >= 0; i--) {
        int v = ((s>>i)&1) ^ 1;

        if (g[u][v])
            ret |= (1<<i);
        else
            v = v^1;
        u = g[u][v];
    }
    return ret;
}

void Tire::insert(int s) {
    int u = 0;

    for (int i = 30; i >= 0; i--) {
        int v = (s>>i)&1;

        if (g[u][v] == 0) {
            memset(g[sz], 0, sizeof(g[sz]));
            g[u][v] = sz++;
        }
        u = g[u][v];
    }
}

poj 3764 The xor-longest Path(字典树)