给定长度为nn的序列aia_i，有次qq询问(l,r)(l,r)，求在 中选取任意个，使得它们的异或和最大

n≤500000,ai≤106n\le 500000, a_i\le 10^6

CF1100F

Solution

考虑分治，把序列和询问一起分治

类似于树上点分治的做法，每次只计算跨区间的答案

对于当前分治区间，处理出以midmid为中心，分别向两边扩展的前/后缀线性基（[l,mid)[l, mid)的后缀和(mid,r](mid, r]的前缀），询问的时候直接合并起来就行了

总复杂度O(nlog2n)O(n\log^2 n)，好想好写好调也跑得过

Code

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#include <bits/stdc++.h>

#define x first
#define y second
#define y1 Y1
#define y2 Y2
#define mp make_pair
#define pb push_back

using namespace std;

typedef long long LL;
typedef pair <int, int> pii;

template <typename T> inline int Chkmax (T &a, T b) { return a < b ? a = b, 1 : 0; }
template <typename T> inline int Chkmin (T &a, T b) { return a > b ? a = b, 1 : 0; }
template <typename T> inline T read ()
{
	T sum = 0, fl = 1; char ch = getchar();
	for (; !isdigit(ch); ch = getchar()) if (ch == '-') fl = -1;
	for (; isdigit(ch); ch = getchar()) sum = (sum << 3) + (sum << 1) + ch - '0';
	return sum * fl;
}

inline void proc_status ()
{
	ifstream t ("/proc/self/status");
	cerr << string (istreambuf_iterator <char> (t), istreambuf_iterator <char> ()) << endl;
}

const int Maxn = 5e5 + 100, Maxlen = 20;

int N, A[Maxn], M;

struct info {int x, y, id;};

struct Basis
{
	int A[Maxlen + 2];

	inline void init () { memset(A, 0, sizeof A); }

	inline int insert (int x)
	{
		for (int i = Maxlen; ~i; --i)
		{
			if (!(x & (1 << i))) continue;
			if (!A[i])
			{
				for (int j = 0; j < i; ++j) if (x & (1 << j)) x ^= A[j];
				for (int j = i + 1; j <= Maxlen; ++j) if (A[j] & (1 << i)) A[j] ^= x;
				A[i] = x;
				return 1;
			}
			x ^= A[i];
		}
		return 0;
	}

	friend Basis merge (const Basis &a, const Basis &b)
	{
		Basis res = a;
		for (int i = 0; i <= Maxlen; ++i) if (b.A[i]) res.insert(b.A[i]);
		return res;
	}

	inline int query (int ans = 0)
	{
		for (int i = Maxlen; i >= 0; --i) Chkmax(ans, ans ^ A[i]);
		return ans;
	}

}B[Maxn];

int Ans[Maxn];
info Q[Maxn], Q1[Maxn], Q2[Maxn];

inline void Divide (int l, int r, int L, int R)
{
	if (L > R) return ;
	if (l == r)
	{
		for (int i = L; i <= R; ++i) Ans[Q[i].id] = A[l];
		return ;
	}

	int mid = l + r >> 1, len1 = 0, len2 = 0;
	B[mid].init(), B[mid].insert (A[mid]);
	for (int i = mid - 1; i >= l; --i) B[i] = B[i + 1], B[i].insert (A[i]);
	for (int i = mid + 1; i <= r; ++i) B[i] = B[i - 1], B[i].insert (A[i]);

	for (int i = L; i <= R; ++i)
	{
		if (Q[i].x <= mid)
		{
			if (Q[i].y > mid) Ans[Q[i].id] = merge (B[Q[i].x], B[Q[i].y]).query();
			else Q1[++len1] = Q[i];
		}
		else Q2[++len2] = Q[i];
	}

	for (int i = 1; i <= len1; ++i) Q[L + i - 1] = Q1[i];
	for (int i = 1; i <= len2; ++i) Q[L + len1 - 1 + i] = Q2[i];

	Divide (l, mid, L, L + len1 - 1), Divide (mid + 1, r, L + len1, L + len1 + len2 - 1);
}

inline void Solve ()
{
	Divide(1, N, 1, M);
	for (int i = 1; i <= M; ++i) printf("%d\n", Ans[i]);
}

inline void Input ()
{
	N = read<int>();
	for (int i = 1; i <= N; ++i) A[i] = read<int>();
	M = read<int>();
	for (int i = 1; i <= M; ++i) Q[i].x = read<int>(), Q[i].y = read<int>(), Q[i].id = i;
}

int main()
{
#ifndef ONLINE_JUDGE
	freopen("F.in", "r", stdin);
	freopen("F.out", "w", stdout);
#endif
	Input();
	Solve();
	return 0;
}