A fast algorithm for minimum spanning trees when edge lengths are constrained?

牧云@^-^@ 提交于 2019-11-29 13:18:14

Fredman–Willard gave an O(m + n) algorithm for integer edge lengths on the unit-cost RAM.

That's arguably not much of an improvement: without the restriction on edge lengths (i.e., the lengths are an opaque data type that supports only comparisons), Chazelle gave an O(m alpha(m, n) + n) algorithm (alpha is the inverse Ackermann function) and Karger–Klein–Tarjan gave a randomized O(m + n) algorithm.

I don't think Darren's idea leads to a O(m + n + U)-time algorithm. Jarnik ("Prim") does not use its priority queue monotonically, so buckets may be scanned multiple times; Kruskal requires a disjoint-set data structure, which cannot be O(m + n).

With integer edge weights you can use bucketing to achieve a priority queue with worst-case O(1) complexity, but additional O(U) space complexity.

Within the MST algorithms you've mentioned you should be able to replace the comparison based priority queues with this integer structure, and hence remove the O(log(n)) depenedence in the complexity requirements. I expect you'd end up with an overall complexity in the style of O(n + m).

Essentially you setup a set of compressed linked-lists, where each list is indexed by the (integer!) cost associated with that bucket:

struct bucket_list
{
    _cost; // array[0..N-1] holding current cost for each item

    _head; // array[0..U-1] holding index of first item in each bucket

    _next; // array[0..N-1] where _next[i] is the next item 
           // in a list for the ith item

    _prev; // array[0..N-1] where _prev[i] is the last item 
           // in a list for the ith item
};

This structure is based on the fact that each item can only be in a single bucket list at once.

Based on this structure you can achieve worst-case O(1) complexity for these operations:

push(item, cost); // push an item onto the head of the appropriate bucket list

_pop(item, cost); // _pop an item from (anywhere!) within a bucket list

update(item, old_cost, new_cost); // move an item between buckets by combining
                                  // push and _pop

To use this structure as a priority queue you simply maintain an index pointing at the current minimum bucket to scan. When you want to get the next lowest cost item, you simply pop the head item from this bucket. If the bucket is empty you increment your bucket index until you find a non-empty one.

Of course if U becomes very large the extra space complexity, and the possiblity of a sparse distribution of items over the buckets may make this kind of approach unattractive.

易学教程内所有资源均来自网络或用户发布的内容,如有违反法律规定的内容欢迎反馈
该文章没有解决你所遇到的问题?点击提问,说说你的问题,让更多的人一起探讨吧!