Codeforces Round #815 (Div. 2) D2 Xor-Subsequence (hard version)

原题链接

\(A>B\)，总是有二进制下从高到低的前\(k\)位相等，第\(k+1\)位\(A\)是\(1\)，\(B\)是\(0\)
本题中\(A=a_i\oplus j\)，\(B=a_j\oplus i\)，这里有一个很奇妙的性质（手玩或者打表也可以发现）：

\(a_i\oplus j\)和\(a_j\oplus i\)的前\(k\)位相等，等价于\(a_i\oplus i\)和\(a_j\oplus j\)的前\(k\)位相等

然后这题就可以做了，只要对\(a_i\oplus i\)在之前建立的字典树中从高位枚举，出现分支再进行讨论即可（\(a_i\)和\(i\)的当前位我们知道了，所以\(a_i\oplus j>a_j\oplus i\)成立只要按\(a_j\)和\(j\)的当前位分类讨论即可）

总结：位运算，异或性质，字典树上dp

参考博客

#pragma GCC optimize(2)
#include<bits/stdc++.h>
using namespace std;

#define fr first
#define se second
#define et0 exit(0);
#define rep(i, a, b) for(int i = (int)(a); i <= (int)(b); i ++)
#define rrep(i, a, b) for(int i = (int)(a); i >= (int)(b); i --)
#define IO ios::sync_with_stdio(false),cin.tie(0),cout.tie(0);

typedef long long LL;
typedef pair<int, int> PII;
typedef pair<LL, LL> PLL;
typedef unsigned long long ULL;

const int INF = 0X3f3f3f3f, N = 3e5 + 10, MOD = 1e9 + 7;
const double eps = 1e-7, pi = acos(-1);

int idx;
int a[N];
int tr[N * 40][2], dp[N * 40][2];

void insert(int x, int y) {
	int p = 0;
	rrep (i, 30, 0) {
		int u = x >> i & 1, v = y >> i & 1;
		if (!tr[p][u ^ v]) tr[p][u ^ v] = ++idx;
		p = tr[p][u ^ v];
	}
}

int query(int x, int y) {
	int p = 0, q = 0, mx = 0;
	rrep (i, 30, 0) {
		int u = x >> i & 1, v = y >> i & 1;
		q = tr[p][!(u ^ v)], p = tr[p][u ^ v];
		if (q) mx = max(mx, dp[q][v]);
		if (!p) break;
	}
	return mx + 1;
}

void update(int x, int y, int mx) {
	int p = 0;
	rrep (i, 30, 0) {
		int u = x >> i & 1, v = y >> i & 1;
		p = tr[p][u ^ v];
		dp[p][u] = max(dp[p][u], mx);
	}
}

void work() {
	int n;
	cin >> n;
	rep (i, 0, n - 1) cin >> a[i];
	
	int res = 0;
	rep (i, 0, n - 1) {
		int mx = query(a[i], i);
		res = max(res, mx);
		insert(a[i], i);
		update(a[i], i, mx);
	}
	
	cout << res << endl;
	
	rep (i, 0, idx) tr[i][0] = tr[i][1] = dp[i][0] = dp[i][1] = 0;
	idx = 0;
}

signed main() {
	IO
	
	int test = 1;
	cin >> test;

	while (test--) {
		work();
	}

	return 0;
}