洛谷P4735 最大异或和 题解 可持久化字典树模板题

题目链接：https://www.luogu.com.cn/problem/P4735

解题思路：

可持久化字典树 模板题。

参考思路：https://www.luogu.com.cn/article/swqlf5n3

示例程序：

#include <bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 5;

struct Node {
    int son[2], cnt;
} tr[maxn*20];
int rt[maxn] = { 1 }, idx = 1;

int cpy_node(int u) {
    tr[++idx] = tr[u];
    return idx;
}

int n, m, a[maxn], sum[maxn];

void ins(int pu, int u, int val) {
    for (int d = 24; d >= 0; d--) {
        int x = (val >> d) & 1;
        tr[u].son[x^1] = tr[pu].son[x^1];
        tr[u].son[x] = cpy_node(tr[pu].son[x]);
        pu = tr[pu].son[x];
        u = tr[u].son[x];
        tr[u].cnt++;
    }
}

int query(int u1, int u2, int val) {
    int ans = 0;
    for (int d = 24; d >= 0; d--) {
        int x = (val >> d) & 1;
        int s1 = tr[u1].son[x^1],
            s2 = tr[u2].son[x^1];
        int cnt = tr[s2].cnt - tr[s1].cnt;
        if (cnt > 0) {
            ans += (1 << d);
            u1 = s1;
            u2 = s2;
        }
        else {
            u1 = tr[u1].son[x];
            u2 = tr[u2].son[x];
        }
    }
    return ans;
}

int main() {
    scanf("%d%d", &n, &m);
    for (int i = 1; i <= n; i++) {
        scanf("%d", a+i);
        sum[i] = sum[i-1] ^ a[i];
    }
    for (int i = 0; i <= n; i++) {
        if (!i) ins(rt[0], rt[0], 0);
        else {
            rt[i] = ++idx;
            ins(rt[i-1], rt[i], sum[i]);
        }
    }
    char op[2];
    while (m--) {
        scanf("%s", op);
        if (op[0] == 'A') {
            n++;
            scanf("%d", a+n);
            sum[n] = sum[n-1] ^ a[n];
            rt[n] = ++idx;
            ins(rt[n-1], rt[n], sum[n]);
        }
        else {
            int l, r, x;
            scanf("%d%d%d", &l, &r, &x);
            int ans = 0;
            if (l == 1) {
                ans = sum[n] ^ x;
                l++;
            }
            if (l <= r)
                ans = max(ans, query(rt[l-2], rt[r-1], sum[n] ^ x));
            printf("%d\n", ans);
        }
    }
    return 0;
}