Maximum XOR Sum 系列问题

时间：2019-09-16 09:31:57 阅读：74 评论：0 收藏：0 [点我收藏+]

标签：ict ack oss number over nts algorithm mask pre

给定 $n$ 个两两不同的正整数 $a_1, a_2, \dots, a_n$，$a_i < 2^k$ 。

Problem 1（经典问题）

求 $a_i \xor a_j$ 的最大值，$ 1\le i, j \le n $ 。

解法

字典树

Problem 2

从 $n$ 个数中任选出一些数，求异或和的最大值。

Let the length of a number be the number of digits needed to write it out in binary, excluding any leading zeros.

Clearly, if all the input numbers had a different length, the problem would have a trivial solution: just iterate over the input numbers in decreasing order by length, choosing each number if and only if XORing it with the maximum so far increases the maximum, i.e., if and only if its leading bit is not set in the current maximum.

The tricky part is when the input may contain multiple numbers with the same length, since then it‘s not obvious which of them we should choose to include in the XOR. What we‘d like to do is reduce the input list into an equivalent form that doesn‘t contain more than one number of the same length.

Conveniently, this is exactly what Gaussian elimination does: it transforms a list of vectors into another list of vectors which have strictly decreasing length, as defined above (that is, into a list which is in echelon form), but which spans the same linear subspace.

The reason this linear algebra algorithm is relevant here is that binary numbers satisfy the axioms of a vector space over the finite field of two elements, a.k.a. GF(2), with the number viewed as vectors of bits, and with XOR as the vector addition operation. (We also need a scalar multiplication operation to satisfy the axioms, but that‘s trivial, since the only scalars in GF(2) are $1$ and $0$.)

The linear subspace spanned by a set of bit vectors (i.e. binary numbers) over GF(2) is then simply the set of vectors obtainable by XORing a subset of them. Thus, if we can use Gaussian elimination to convert our input list into another one, which spans the same subspace, we can solve the problem using this other list and know that it gives the same solution as for the original problem.

Thus, we need to implement Gaussian elimination over GF(2).

// a[i] < (1LL << 60)
long long max_xor_sum(vector<long long> a, int n) {
    long long res = 0;
    int index = 0;
    for (int column = 59; column >= 0; --column) {
        long long mask = 1LL << column;
        for (int row = index; row < n; ++row) {
            if (a[row] & mask) {
                swap(a[row], a[index]);
                for (int row_ = row + 1; row_ < n; ++row_) {
                    if (a[row_] & mask) {
                        a[row_] ^= a[index];
                    }
                }
                if ((res & mask) == 0) {
                    res ^= a[index];
                }
                ++index;
                break;
            }
        }
    }
    return res;
}