需要一种数据结构来处理带有标签集的元素，并能高效地查找其子集的所有元素

Sub-linear is not possible in the worst case because if the given set of tags contains ALL tags, then ALL elements must be returned. But here's an algorithm that can be sub-linear in the average case.

For each element, use a 128-bit integer (2 x 64-bit integers if you have to) to represent the 100 or so possible tags (i.e. each bit represents a different tag: 1 if the element has the tag, and 0 otherwise.)

Then sort the array based on the 128-bit integer representation.

Given a list of tags, turn it into the 128-bit integer representation so that you can use bit-wise operations to determine whether an element has a subset of the given tags or not. (Use OR operation, followed by a count of the number of 1s in the result.)

The actual speed up (other than the speed up by using bitwise operations) is explained below. For this, let us simplify the input down to 8 bits.

Suppose the list of tags were given as 01001100.

Then you can search for 00000100 in the list of elements in O(logn) time because the list is sorted, and grab a list of elements until 00000101 is reached. Then you can jump to 00001000, again in O(logn) time, and search there until you reach 00001101, then jump to 01000000, and so on.

The end result is that you've jumped over a significant number of elements, at the cost of spending O(tlogn) for the jumps, where t is the number of tags in the input set of tags.

You can take this idea a bit further by jumping more. i.e. jump to 00000100, then jump to 00001000, then to 00001100, then to 01000000, then to 01000100, and so on. This results in spending O(2^tlogn) time for jumping around, but this maybe worth it if t << n and 2^t is manageable.