这是我之前的问题的后续。我仍然觉得这是一个非常有趣的问题,因为有一种算法值得更多关注,所以我将其发布在这里。
来自Wikipedia:对于每个 xi 为正且以相同常数为界的情况,Pisinger 找到了一种线性时间算法。
有一篇不同的论文似乎描述了相同的算法,但对我来说有点难以阅读,所以请 - 有人知道如何将第 4 页(balsub
)中的伪代码翻译成工作实现吗?
以下是我迄今为止收集的一些建议:
http://www.diku.dk/~pisinger/95-6.ps(论文)
https://stackoverflow.com/a/9952759/1037407
http://www.diku.dk/hjemmesider/ansatte/pisinger /codes.html
PS:我并不真的坚持这个算法,所以如果你知道任何其他类似性能的算法,请随时在下面提出建议。
编辑
这是 oldboy 在下面发布的代码的 Python 版本:
class view(object):
def __init__(self, sequence, start):
self.sequence, self.start = sequence, start
def __getitem__(self, index):
return self.sequence[index + self.start]
def __setitem__(self, index, value):
self.sequence[index + self.start] = value
def balsub(w, c):
'''A balanced algorithm for Subset-sum problem by David Pisinger
w = weights, c = capacity of the knapsack'''
n = len(w)
assert n > 0
sum_w = 0
r = 0
for wj in w:
assert wj > 0
sum_w += wj
assert wj <= c
r = max(r, wj)
assert sum_w > c
b = 0
w_bar = 0
while w_bar + w[b] <= c:
w_bar += w[b]
b += 1
s = [[0] * 2 * r for i in range(n - b + 1)]
s_b_1 = view(s[0], r - 1)
for mu in range(-r + 1, 1):
s_b_1[mu] = -1
for mu in range(1, r + 1):
s_b_1[mu] = 0
s_b_1[w_bar - c] = b
for t in range(b, n):
s_t_1 = view(s[t - b], r - 1)
s_t = view(s[t - b + 1], r - 1)
for mu in range(-r + 1, r + 1):
s_t[mu] = s_t_1[mu]
for mu in range(-r + 1, 1):
mu_prime = mu + w[t]
s_t[mu_prime] = max(s_t[mu_prime], s_t_1[mu])
for mu in range(w[t], 0, -1):
for j in range(s_t[mu] - 1, s_t_1[mu] - 1, -1):
mu_prime = mu - w[j]
s_t[mu_prime] = max(s_t[mu_prime], j)
solved = False
z = 0
s_n_1 = view(s[n - b], r - 1)
while z >= -r + 1:
if s_n_1[z] >= 0:
solved = True
break
z -= 1
if solved:
print c + z
print n
x = [False] * n
for j in range(0, b):
x[j] = True
for t in range(n - 1, b - 1, -1):
s_t = view(s[t - b + 1], r - 1)
s_t_1 = view(s[t - b], r - 1)
while True:
j = s_t[z]
assert j >= 0
z_unprime = z + w[j]
if z_unprime > r or j >= s_t[z_unprime]:
break
z = z_unprime
x[j] = False
z_unprime = z - w[t]
if z_unprime >= -r + 1 and s_t_1[z_unprime] >= s_t[z]:
z = z_unprime
x[t] = True
for j in range(n):
print x[j], w[j]