Nikitosh 和异或(trie树)

题目：

#10051. 「一本通 2.3 例 3」Nikitosh 和异或

解析：

首先我们知道一个性质\(x\oplus x=0\)
我们要求$$\bigoplus_{i = l}^ra_i$$的话，相当于求$$(\bigoplus_{i = 1}^la_i)\oplus (\bigoplus_{i = 1}^ra_i)$$
所以我们维护一个异或前缀和\(sum_i\)
我们用\(l_i\)表示从左往右到第\(i\)位时的区间最大异或和
\(r_i\)表示从右往左到第\(i\)位时的区间最大异或和
显然\(l_i = max\{sum_L\oplus sum_R\}1\leq L<R\leq i\)
\(r_i\)同理
最后枚举求和\(ans=max\{ans,l_i+r_{i+1}\}\)

代码

#include <bits/stdc++.h>
using namespace std;
const int N = 5e6 + 10;

int n, m, num, ans;
int a[N], sum[N], l[N], r[N];

struct node {
	int nx[2];
} e[N]; 

void insert(int x) {
	bitset<35>b(x);
	int rt = 0;
	for (int i = 30; i >= 0; --i) {
		int v = (int)b[i];
		if (!e[rt].nx[v]) e[rt].nx[v] = ++num;
		rt = e[rt].nx[v];
	}
}

int query(int x) {
	bitset<35>b(x);
	int rt = 0, ret = 0;
	for (int i = 30; i >= 0; --i) {
		int v = (int)b[i];
		if (e[rt].nx[v ^ 1]) ret = ret << 1 | 1, rt = e[rt].nx[v ^ 1];
		else ret <<= 1, rt = e[rt].nx[v];
	}
	return ret;
}

int main() {
	scanf("%d", &n);
	for (int i = 1; i <= n; ++i) scanf("%d", &a[i]), sum[i] = sum[i - 1] ^ a[i];
	insert(0);
	for (int i = 1; i <= n; ++i) {
		ans = max(ans, query(sum[i]));
		l[i] = ans;
		insert(sum[i]);
	}
	num = ans = 0;
	memset(e, 0, sizeof e);
	for (int i = n; i >= 1; --i) {
		ans = max(ans, query(sum[i]));
		r[i] = ans;
		insert(sum[i]);
	}
	ans = 0;
	for (int i = 1; i < n; ++i) ans = max(ans, l[i] + r[i + 1]);
	cout << ans;
}