Question

我想知道什么是最有效（时间和内存）方式来计算子集的数量，其总和小于或等于某个限制。例如，对于集合{1, 2, 4}和3的限制，此类数字应为4（子集为{}, {1}, {2}, {1, 2}）。我尝试在位向量（掩码）中编码子集并以下列方式查找答案（伪代码）：

solve(mask, sum, limit)
  if visited[mask] 
    return
  if sum <= limit
    count = count + 1
  visited[mask] = true
  for i in 0..n - 1
    if there is i-th bit
      sum = sum - array[i]
      mask = mask without i-th bit
      count (mask, sum, limit)
solve(2^n - 1, knapsack sum, knapsack limit)

数组是从零开始的，count可以是全局变量，visited是一个长度为2^n的数组。我知道这个问题具有指数级的复杂性，但是我的想法有更好的方法/改进吗？该算法快速运行n ≤ 24，但我的方法相当暴力，我正在考虑存在一些聪明的方法来找到n = 30的答案。

Answer 1

对空间最有效的是对所有子集的递归遍历，只是保持计数。这将是O(2^n)时间和O(n)内存，其中n是整个集合的大小。

所有已知的解决方案可以在时间上呈指数级，因为您的程序是subset-sum的变体。这已知是NP完整的。但是一个非常有效的DP解决方案如下所示，带有注释的伪代码。

# Calculate the lowest sum and turn all elements positive.
# This turns the limit problem into one with only non-negative elements.
lowest_sum = 0
for element in elements:
    if element < 0:
        lowest_sum += element
        element = -element

# Sort and calculate trailing sums.  This allows us to break off
# the details of lots of ways to be below our bound.
elements = sort elements from largest to smallest
total = sum(elements)
trailing_sums = []
for element in elements:
    total -= element
    push total onto trailing_sums

# Now do dp
answer = 0
ways_to_reach_sum = {lowest_sum: 1}
n = length(answer)
for i in range(0, n):
    new_ways_to_reach_sum = {}
    for (sum, count) in ways_to_reach_sum:
        # Do we consider ways to add this element?
        if bound <= elements[i] + sum:
            new_ways_to_reach_sum[sum] += count

        # Make sure we keep track of ways to not add this element
        if bound <= sum + trailing_sums[i]:
            # All ways to compute the subset are part of the answer
            answer += count * 2**(n - i)
        else:
            new_ways_to_reach_sum[sum] += count
    # And finish processing this element.
    ways_to_reach_sum = new_ways_to_reach_sum

# And just to be sure
for (sum, count) in ways_to_reach_sum:
    if sum <= bound:
        answer += count

# And now answer has our answer!

0-1背包的计数组合

1 个答案: