POJ 3764 The XOR Longest Path

Trie求最大XOR

设w[x]表示从跟节点到x的路径上所有权值的xor，显然有：

w[x] = w[fa] xor weight(x, fa)

所以我们可以先dfs一次树，把所有点的w都预处理出来。。
根据xor的性质可知，x xor x = 0，所以问题就变成了求w[x] xor w[y]的最大值（因为相同路径都抵消了）
所以我们把w值用trie维护即可

#include <iostream>
#include <cstring>
#include <cstdio>
// 为啥不能用万能头呢poj。。。
#define INF 0x3f3f3f3f
using namespace std;
typedef long long ll;
inline int lowbit(int x){ return x & (-x); }
inline int read(){
    int X = 0, w = 0; char ch = 0;
    while(!isdigit(ch)) { w |= ch == '-'; ch = getchar(); }
    while(isdigit(ch)) X = (X << 3) + (X << 1) + (ch ^ 48), ch = getchar();
    return w ? -X : X;
}
inline int gcd(int a, int b){ return a % b ? gcd(b, a % b) : b; }
inline int lcm(int a, int b){ return a / gcd(a, b) * b; }
template<typename T>
inline T max(T x, T y, T z){ return max(max(x, y), z); }
template<typename T>
inline T min(T x, T y, T z){ return min(min(x, y), z); }
template<typename A, typename B, typename C>
inline A fpow(A x, B p, C yql){
    A ans = 1;
    for(; p; p >>= 1, x = 1LL * x * x % yql)if(p & 1)ans = 1LL * x * ans % yql;
    return ans;
}
const int N = 100005;
int trie[6000000][2], tot, cnt, head[N], w[N];
struct Edge{
    int v, next, w;
}edge[N<<2];

void addEdge(int a, int b, int w){
    edge[cnt].v = b;
    edge[cnt].w = w;
    edge[cnt].next = head[a];
    head[a] = cnt ++;
}

void dfs(int s, int fa){
    for(int i = head[s]; i != -1; i = edge[i].next){
        int u = edge[i].v;
        if(u != fa){
            w[u] = w[s]^edge[i].w;
            dfs(u, s);
        }
    }
}

void insert(int num){
    int cur = 1;
    for(int i = 31; i >= 0; i --){
        int p = (num >> i) & 1;
        if(trie[cur][p] == 0) trie[cur][p] = ++tot;
        cur = trie[cur][p];
    }
}

int search(int num){
    int cur = 1, ret = 0;
    for(int i = 31; i >= 0; i --){
        int p = (num >> i) & 1;
        if(trie[cur][p^1] == 0) cur = trie[cur][p];
        else ret += (1 << i), cur = trie[cur][p^1];
    }
    return ret;
}

int main(){

    int n;
    while(scanf("%d", &n) != EOF){
        cnt = 0;
        memset(head, -1, sizeof head);
        memset(trie, 0, sizeof trie);
        memset(w, 0, sizeof w);
        for(int i = 0; i < n - 1; i++){
            int a = read(), b = read(), c = read();
            addEdge(a, b, c), addEdge(b, a, c);
        }
        dfs(0, 0);
        int ans = 0;
        tot = 1;
        for(int i = 0; i < n; i++){
            insert(w[i]);
            ans = max(ans, search(w[i]));
        }
        printf("%d\n", ans);
    }
    return 0;
}