P4551 最长异或路径 题解

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;
}

\[时间复杂度为:\ O(n\log{V_{max}}) \]