P4551 最长异或路径

求树上两点使得两点路径xor和最大，可以用01字典树维护。
假设1为根，d[i]表示1~i路径上xor和,每次查询x,y路径的xor和就是d[x] xor d[y],因为连续异或两个数字会抵消，相当于1~lca（x，y）的贡献消失了，那剩下的就是x到y的路径异或和
注意01字典树取出x第i位写法

#include<iostream>
#include<cstdio>
#define rep(i,j,k) for(register int i(j);i<=k;++i)
#define drp(i,j,k) for(register int i(j);i>=k;--i)
#define bug cout<<"~~~~~~~~~~~~~"<<'
';
#define bugout(x) cout<<x<<endl;
using std::cin;
using std::cout;
typedef long long lxl;
template<typename T>
inline T  max(T a, T b) {
	return a > b ? a : b;
}
template<typename T>
inline T  min(T a, T b) {
	return a < b ? a : b;
}

inline char gt() {
	static char buf[1 << 21], *p1 = buf, *p2 = buf;
	return p1 == p2 && (p2 = (p1 = buf) + fread(buf, 1, 1 << 21, stdin), p1 == p2) ? EOF : *p1++;
}
template <typename T>
inline void  read(T &x) {
	register char ch = gt();
	x = 0;
	int w(0);
	while(!(ch >= '0' && ch <= '9'))w |= ch == '-', ch = gt();
	while(ch >= '0' && ch <= '9')x = x * 10 + (ch & 15), ch = gt();
	w ? x = ~(x - 1) : x;
}
template <typename T>
inline void out(T x, char cc) {
	if(x < 0) x = -x, putchar('-');
	char ch[20];
	int num(0);
	while(x || !num) ch[++num] = x % 10 + '0', x /= 10;
	while(num) putchar(ch[num--]);
	putchar(cc);
}
const int N = 2e5 + 79;

struct graph {
	int head[N], tot, next[N << 1], ver[N << 1], edge[N << 1];
	inline void add(int a, int b, int c) {
		ver[++tot] = b;
		next[tot] = head[a];
		head[a] = tot;
		edge[tot] = c;
	}
} G;
int d[N], n;
struct Trie {
	int ch[N * 27][2], tot;
    int val[N*27];
	inline void insert(int x) {
		int p =0;
		drp(i,31,0){
			int num((x>>i)&1);//right
			if(!ch[p][num]) ch[p][num]=++tot;
			p=ch[p][num];
		}
         val[p]=x;
	}
	inline int query(int x){
		int p=0;
		drp(i,31,0){
			int num((x>>i)&1 );
			if(ch[p][num^1]) p=ch[p][num^1];
			else p=ch[p][num];
		}
		return val[p];
	}
} T;


inline void dfs(int x, int fa) {
	for(register int i(G.head[x]); i; i = G.next[i]) {
		int y(G.ver[i]), z(G.edge[i]);
		if(y == fa) continue;
		d[y] = d[x] ^ z;
		dfs(y, x);
	}
}


int main() {
//	freopen("path.in", "r", stdin);
//	freopen("path.out", "w", stdout);
	read(n);
	int x, y, z;
	rep(i, 2, n) {
		read(x);
		read(y);
		read(z);
		G.add(x, y, z);
		G.add(y, x, z);
	}
	dfs(1, 0);
	int ans(0);
	T.insert(0); //init
	rep(i, 1, n) {
		ans = max(ans, d[i] ^ T.query(d[i]));
		T.insert(d[i]);
	}
	out(ans,'
');
	return 0;
}