我正在努力解决美国陆军工程兵团的挤奶问题问题陈述如下：https://train.usaco.org/usacoprob2?S=milk2&a=n3lMlotUxJ1
给定二维数组形式的一系列间隔，我必须找到最长间隔和不发生挤奶的最长间隔。
例如，给定数组[[500,1200],[200,900],[100,1200]]，有连续挤奶时的最长间隔为1100，没有休息时间时的最长间隔为0。
我试过看使用字典是否会减少运行时间，但我没有太大的成功。
f = open('milk2.in', 'r')
w = open('milk2.out', 'w')
#getting the input
farmers = int(f.readline().strip())
schedule = []
for i in range(farmers):
schedule.append(f.readline().strip().split())
#schedule = data
minvalue = 0
maxvalue = 0
#getting the minimums and maximums of the data
for time in range(farmers):
schedule[time][0] = int(schedule[time][0])
schedule[time][1] = int(schedule[time][1])
if (minvalue == 0):
minvalue = schedule[time][0]
if (maxvalue == 0):
maxvalue = schedule[time][1]
minvalue = min(schedule[time][0], minvalue)
maxvalue = max(schedule[time][1], maxvalue)
filled_thistime = 0
filled_max = 0
empty_max = 0
empty_thistime = 0
#goes through all the possible items in between the minimum and the maximum
for point in range(minvalue, maxvalue):
isfilled = False
#goes through all the data for each point value in order to find the best values
for check in range(farmers):
if point >= schedule[check][0] and point < schedule[check][1]:
filled_thistime += 1
empty_thistime = 0
isfilled = True
break
if isfilled == False:
filled_thistime = 0
empty_thistime += 1
if (filled_max < filled_thistime) :
filled_max = filled_thistime
if (empty_max < empty_thistime) :
empty_max = empty_thistime
print(filled_max)
print(empty_max)
if (filled_max < filled_thistime):
filled_max = filled_thistime
w.write(str(filled_max) + " " + str(empty_max) + "\n")
f.close()
w.close()
程序运行良好，但我需要减少运行所需的时间。
最佳答案
一个不那么漂亮但更有效的方法是像一个自由列表一样解决这个问题，尽管它有点棘手，因为范围可以重叠。此方法只需要在输入列表中循环一次。
def insert(start, end):
for existing in times:
existing_start, existing_end = existing
# New time is a subset of existing time
if start >= existing_start and end <= existing_end:
return
# New time ends during existing time
elif end >= existing_start and end <= existing_end:
times.remove(existing)
return insert(start, existing_end)
# New time starts during existing time
elif start >= existing_start and start <= existing_end:
# existing[1] = max(existing_end, end)
times.remove(existing)
return insert(existing_start, end)
# New time is superset of existing time
elif start <= existing_start and end >= existing_end:
times.remove(existing)
return insert(start, end)
times.append([start, end])
data = [
[500,1200],
[200,900],
[100,1200]
]
times = [data[0]]
for start, end in data[1:]:
insert(start, end)
longest_milk = 0
longest_gap = 0
for i, time in enumerate(times):
duration = time[1] - time[0]
if duration > longest_milk:
longest_milk = duration
if i != len(times) - 1 and times[i+1][0] - times[i][1] > longest_gap:
longes_gap = times[i+1][0] - times[i][1]
print(longest_milk, longest_gap)
关于python - 如何将O(N * M)优化为O(n ** 2)？，我们在Stack Overflow上找到一个类似的问题：https://stackoverflow.com/questions/57313864/