区间异或和(Subarray Maximum XOR) 题解

思路分析

按题目要求使用对应数据结构与算法模拟/递推求解，注意数据范围边界与时间复杂度。

参考代码

#include <iostream>
#include <vector>
#include <algorithm>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);
    int n;
    if (!(cin >> n)) return 0;
    vector<int> a(n + 1);
    for (int i = 1; i <= n; ++i) cin >> a[i];
    int max_nodes = (n + 2) * 31;
    vector<int> ch0(max_nodes, 0), ch1(max_nodes, 0);
    int tot = 1;
    auto clear_trie = [&]() {
        for (int i = 1; i <= tot; ++i) {
            ch0[i] = ch1[i] = 0;
        }
        tot = 1;
    };
    auto insert = [&](int x) {
        int u = 1;
        for (int b = 29; b >= 0; --b) {
            int bit = (x >> b) & 1;
            if (bit == 0) {
                if (!ch0[u]) ch0[u] = ++tot;
                u = ch0[u];
            } else {
                if (!ch1[u]) ch1[u] = ++tot;
                u = ch1[u];
            }
        }
    };
    auto query = [&](int x) {
        int u = 1, res = 0;
        for (int b = 29; b >= 0; --b) {
            int bit = (x >> b) & 1;
            if (bit == 0) {
                if (ch1[u]) {
                    res |= (1 << b);
                    u = ch1[u];
                } else {
                    u = ch0[u];
                }
            } else {
                if (ch0[u]) {
                    res |= (1 << b);
                    u = ch0[u];
                } else {
                    u = ch1[u];
                }
            }
        }
        return res;
    };
    vector<int> L(n + 1, 0), R(n + 2, 0);
    insert(0);
    int pref = 0;
    for (int i = 1; i <= n; ++i) {
        pref ^= a[i];
        L[i] = max(L[i - 1], query(pref));
        insert(pref);
    }
    clear_trie();
    insert(0);
    int suff = 0;
    for (int i = n; i >= 1; --i) {
        suff ^= a[i];
        R[i] = max(R[i + 1], query(suff));
        insert(suff);
    }
    long long ans = 0;
    for (int i = 1; i < n; ++i) {
        ans = max(ans, (long long)L[i] + R[i + 1]);
    }
    cout << ans << "\n";
    return 0;
}