题目链接：最长异或路径
看到树上路径问题，且是异或和这种，先思考树上前缀和转化为前缀和问题。如果我们预处理出 \(pre[curr]\) 表示 \(curr\) 这个点到根的前缀异或值，那么很显然我们路径的两个点 \(u\) 与 \(v\) 的 \(pre[u]\oplus pre[v]\) 和传统的加法的树上前缀和并不一样，因为异或上一个相同的数会消掉，所以 \(LCA\) 所对应的 \(pre[LCA]\) 是会消掉的，不像普通的求和还会重复计数先需要去掉。最后回到题目，观察是点前缀和还是边前缀和，点前缀和显然还得带上 \(LCA\)，边则不用，我们只需要将原问题转化为求 \(pre[u] \oplus pre[v]\) 的最大值即可。如果 \(u=v\) 显然为 \(0\)，这个并不会影响最大值，所以我们预处理出 \(pre\) 数组，全部插入 \(0-1\ Trie\) 中作为 \(pre[u]\) 或者 \(pre[v]\)，在枚举 \(u\) 或者 \(v\) 去算 \(pre\) 异或最值即可。
参照代码
#include <bits/stdc++.h>
// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")
#define isPbdsFile
#ifdef isPbdsFile
#include <bits/extc++.h>
#else
#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>
#endif
using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};
template <typename T>
int disc(T* a, int n)
{
return unique(a + 1, a + n + 1) - (a + 1);
}
template <typename T>
T lowBit(T x)
{
return x & -x;
}
template <typename T>
T Rand(T l, T r)
{
static mt19937 Rand(time(nullptr));
uniform_int_distribution<T> dis(l, r);
return dis(Rand);
}
template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
return (a % b + b) % b;
}
template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
a %= c;
T1 ans = 1;
for (; b; b >>= 1, (a *= a) %= c) if (b & 1) (ans *= a) %= c;
return modt(ans, c);
}
template <typename T>
void read(T& x)
{
x = 0;
T sign = 1;
char ch = getchar();
while (!isdigit(ch))
{
if (ch == '-') sign = -1;
ch = getchar();
}
while (isdigit(ch))
{
x = (x << 3) + (x << 1) + (ch ^ 48);
ch = getchar();
}
x *= sign;
}
template <typename T, typename... U>
void read(T& x, U&... y)
{
read(x);
read(y...);
}
template <typename T>
void write(T x)
{
if (typeid(x) == typeid(char)) return;
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 ^ 48);
}
template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
write(x), putchar(c);
write(c, y...);
}
template <typename T11, typename T22, typename T33>
struct T3
{
T11 one;
T22 tow;
T33 three;
bool operator<(const T3 other) const
{
if (one == other.one)
{
if (tow == other.tow) return three < other.three;
return tow < other.tow;
}
return one < other.one;
}
T3()
{
one = tow = three = 0;
}
T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
{
}
};
template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
if (x < y) x = y;
}
template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
if (x > y) x = y;
}
constexpr int N = 1e5 + 10;
constexpr int T = 31;
int node[N * T][2];
int cnt;
inline void add(const int val)
{
int curr = 0;
forv(i, T, 0)
{
int& nxt = node[curr][val >> i & 1];
if (!nxt) nxt = ++cnt;
curr = nxt;
}
}
inline int query(const int val)
{
int curr = 0, ans = 0;
forv(i, T, 0)
{
const int idx = val >> i & 1;
if (node[curr][idx ^ 1]) ans |= 1 << i, curr = node[curr][idx ^ 1];
else curr = node[curr][idx];
}
return ans;
}
int pre[N];
vector<pii> child[N];
inline void dfs(const int curr, const int fa)
{
for (const auto [nxt,val] : child[curr])
{
if (nxt == fa) continue;
pre[nxt] = pre[curr] ^ val;
dfs(nxt, curr);
}
}
int n, ans;
inline void solve()
{
cin >> n;
forn(i, 1, n-1)
{
int u, v, val;
cin >> u >> v >> val;
child[u].emplace_back(v, val);
child[v].emplace_back(u, val);
}
dfs(1, 0);
forn(i, 1, n) add(pre[i]);
forn(i, 1, n) uMax(ans, query(pre[i]));
cout << ans;
}
signed int main()
{
// MyFile
Spider
//------------------------------------------------------
// clock_t start = clock();
int test = 1;
// read(test);
// cin >> test;
forn(i, 1, test) solve();
// while (cin >> n, n)solve();
// while (cin >> test)solve();
// clock_t end = clock();
// cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}