我正在尝试解决 USACO 的奶牛问题。问题陈述在这里:https://train.usaco.org/usacoprob2?S=milk2&a=n3lMlotUxJ1
给定二维数组形式的一系列间隔,我必须找到最长的间隔和没有发生挤奶的最长间隔。
例如。给定数组 [[500,1200],[200,900],[100,1200]]
,最长间隔为 1100,因为有连续挤奶,最长间隔为 0,因为有连续挤奶没有休息时间。
我尝试过使用字典是否会减少运行时间,但没有取得太大成功。
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()
该程序运行良好,但我需要减少运行时间。
最佳答案
一个不太漂亮但更有效的方法是像空闲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。此方法只需要循环输入列表一次。
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/