我最近一直试图学习递归算法，而且我已陷入死胡同。给定一定数量的列表，每个列表代表从一个商店转到所有其他商店所需的时间，以及包含一系列时间间隔的列表，是否有办法使用递归找到商店之间的所有可能路线？
例如
list_of_shops = [shop1, shop2, shop3, shop4]
# Index for this list and the one below are the same
list_of_time_it_takes = [[0, 2, 1, 4], [2, 0, 1, 4], [2, 1, 0, 4], [1, 2, 3, 0]]
# the 0 indicates the index of the shop. It takes 0 minutes to reach the shop from itself.
list_of_time_intervals = [0, 2, 2]
商店只能访问一次。我们可以看到有3家商店每隔2分钟访问一次，可能的路线是：
shop4＆gt; shop2＆gt; shop1
shop3＆gt; shop1＆gt; SHOP2
因此，我尝试使用以下代码解决上述问题：
shops = [[0, 2, 1, 4, 9], [2, 0, 1, 4, 9], [2, 1, 0, 4, 9], [1, 2, 3, 0, 11], [3, 6, 16, 4, 0]]
times = [0, 2, 2, 4, 11]
list_of_shops = ['shop0', 'shop1', 'shop2', 'shop3', 'shop4']
index_dict = {}
def function(shops_input, times_input, i, index_list):
#print("given i = ", i)
#print("given shops = ", shops_input)
#print("given times = ", times_input)
shops_copy, times_copy = shops_input[:], times_input[:]
pop = times_copy.pop(0)
#print("obtained pop = ", pop)
if shops[i] in shops_copy:
index_list.append(shops.index(shops[i]))
shops_copy.pop(shops_copy.index(shops[i]))
if len(index_list) == len(times):
index_dict[list_of_shops[index_list[0]]] = index_list
print(index_list)
print(index_dict)
if len(times_copy):
try:
function(shops_copy, times_copy, shops[i].index(times_copy[0]), index_list)
except ValueError:
return
def main_function(shops, times):
for i in range(len(shops)):
function(shops, times, i, [])
print("---------end funct---------")
main_function(shops, times)
它在某些情况下有效，但绝不是在所有情况下都适用。它应该根据给定的时间间隔给我所有可能的路线，但是，它似乎在很多情况下都不起作用。
例如，如果我将商店和时间改为：
shops = [[0,1,1,1],[1,0,1,1],[1,1,0,1],[1,1,1,0]]
times = [0, 1, 1]
它给出2个可能性的输出 - ＆gt;从shop2 = [2,0,1]和 - ＆gt;开始从shop3 = [3,0,1]开始。有没有办法让我的算法工作？
答案 0 :(得分：2)
我写了一个小脚本来解决你的问题。首先让我们看一下输出。它是代表树的字典。根元素是将所有东西放在一起。所有其他孩子（或叶子）都是可能的访问，因为您在该节点（或商店）。
{'children': [{'children': [{'children': [{'children': [{'children': [{'children': [],
'shop': 4}],
'shop': 3}],
'shop': 0}],
'shop': 1}],
'shop': 0},
{'children': [{'children': [{'children': [{'children': [{'children': [],
'shop': 4}],
'shop': 3}],
'shop': 1}],
'shop': 0}],
'shop': 1},
{'children': [{'children': [{'children': [{'children': [{'children': [],
'shop': 4}],
'shop': 3}],
'shop': 1}],
'shop': 0}],
'shop': 2},
{'children': [{'children': [{'children': [{'children': [{'children': [],
'shop': 4}],
'shop': 3}],
'shop': 0}],
'shop': 1}],
'shop': 3},
{'children': [], 'shop': 4}],
'shop': 'root'}
Drink beer :)
嗯，这是脚本。如果您有任何疑问，请问：）
shops = [[0,1,1,1],[1,0,1,1],[1,1,0,1],[1,1,1,0]]
times = [0, 1, 1]
"""
Data structure
{
shop: 'root',
children: [
{
shop: 1, # index of the shop
children: [ # shops where you can go from here
{
shop: 2,
children: [...]
},
...
]
},
...
]
}
"""
def get_shop(index):
return shops[index]
def get_next_shop_indexes(shop, time_interval):
next_shops = []
for index, shop_time_interval in enumerate(shop):
if shop_time_interval == time_interval:
next_shops.append(index)
return next_shops
def build_route_branch(branch, shop_index, time_intervals):
shop = get_shop(shop_index)
next_shop_indexes = get_next_shop_indexes(shop, time_intervals[0])
for next_shop_index in next_shop_indexes:
child_branch = {
'shop': next_shop_index,
'children': []
}
branch['children'].append(child_branch)
if len(time_intervals) > 1:
child_branch = build_route_branch(
child_branch,
next_shop_index,
time_intervals[1:]
)
tree = {
'shop': 'root',
'children': []
}
for index, shop in enumerate(shops):
branch = build_route_branch(tree, index, times)
import pprint
pprint.pprint(tree)
print 'Drink beer :)'