Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Welcome To Ask or Share your Answers For Others

Categories

0 votes
159 views
in Technique[技术] by (71.8m points)

algorithm - find all combinations with non-overlapped regions

Within a super-region S, there are k small subregions. The number k can be up to 200. There may be overlap between subregions. I have millions of regions S.

For each super-region, my goal is to find out all combinations in which there are 2 or more non-overlapped subregions.

Here is an example:

Super region: 1-100

Subregions: 1-8, 2-13, 9-18, 15-30, 20-35

Goal:

Combination1: 1-8, 9-18

Combination2: 1-8, 20-35

Combination3: 1-8, 9-18, 20-35

Combination4: 1-8, 15-30

...

See Question&Answers more detail:os

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
Welcome To Ask or Share your Answers For Others

1 Reply

0 votes
by (71.8m points)

Number of subsets might be exponential (max 2^k), so there is nothing wrong to traverse all possible independent subsets with recursion. I've used linear search of the next possible interval, but it is worth to exploit binary search.

def nonovl(l, idx, right, ll):
    if idx == len(l):
        if ll:
            print(ll)
        return

    #find next non-overlapping interval without using l[idx]
    next = idx + 1  
    while next < len(l) and right >= l[next][0]:
        next += 1
    nonovl(l, next, right, ll)

    #find next non-overlapping interval after using l[idx]
    next = idx + 1
    right = l[idx][1]
    while next < len(l) and right >= l[next][0]:
        next += 1
    nonovl(l, next, right, ll + str(l[idx]))

l=[(1,8),(2,13),(9,18),(15,30),(20,35)]
l.sort()
nonovl(l, 0, -1, "")

(20, 35)
(15, 30)
(9, 18)
(9, 18)(20, 35)
(2, 13)
(2, 13)(20, 35)
(2, 13)(15, 30)
(1, 8)
(1, 8)(20, 35)
(1, 8)(15, 30)
(1, 8)(9, 18)
(1, 8)(9, 18)(20, 35)

与恶龙缠斗过久,自身亦成为恶龙;凝视深渊过久,深渊将回以凝视…
OGeek|极客中国-欢迎来到极客的世界,一个免费开放的程序员编程交流平台!开放,进步,分享!让技术改变生活,让极客改变未来! Welcome to OGeek Q&A Community for programmer and developer-Open, Learning and Share
Click Here to Ask a Question

...