BZOJ3261: 最大异或和(可持久化trie树)

题意

题目链接

sol

设$sum[i]$表示$1 - i$的异或和

首先把每个询问的$x \oplus sum[n]$就变成了询问前缀最大值

可持久化trie树维护前缀xor，建树的时候维护一下每个节点被遍历了多少次

注意设置好偏移量，不然询问区间为$[1, 1]$的时候可能挂掉

#include<bits/stdc++.h>
using namespace std;
const int maxn = 6e5 + 10;
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-')f =- 1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int n, m, a[maxn], sum[maxn], cnt[maxn * 25], ch[maxn * 25][2], tot, root[maxn];
void insert(int i, int x) {
    x = (sum[i] = sum[i - 1] ^ x);
    int p = root[i - 1], now = (root[i] = ++tot);
    for(int i = 23; i >= 0; i--) {
        bool nxt = x >> i & 1;
        ch[now][nxt ^ 1] = ch[p][nxt ^ 1];
        now = ch[now][nxt] = ++tot; p = ch[p][nxt];
        cnt[now] = cnt[p] + 1;
    }
}
int query(int l, int r, int x) {
    r = root[r], l = root[l];
    int ans = 0;
    for(int i = 23; i >= 0; i--) {
        int nxt = x >> i & 1;
        if(cnt[ch[r][nxt ^ 1]] - cnt[ch[l][nxt ^ 1]] > 0) ans += 1 << i, r = ch[r][nxt ^ 1], l = ch[l][nxt ^ 1];
        else r = ch[r][nxt], l = ch[l][nxt];
    }
    return ans;
}
int main() {
    n = read(); m = read();
    for(int i = 1; i <= n; i++) a[i] = read(), insert(i, a[i]);
    char ss[4];
    for(int i = 1; i <= m; i++) {
        scanf("%s", ss + 1);
        if(ss[1] == 'a') 
            n++, a[n] = read(), insert(n, a[n]);
        else {
            int l = read() - 1, r = read() - 1, val = read();
            printf("%d\n", query(l - 1, r, val ^ sum[n]));
        }
    }
}
/*
5 5
2 6 4 3 6
a 1 
q 3 5 4 
a 4 
q 5 7 0 
q 3 6 6 
*/