In a country popular for train travel, you have planned some train travelling one year in advance. The days of the year that you will travel is given as an arraydays. Each day is an integer from1to365.
Train tickets are sold in 3 different ways:
a 1-day pass is sold for
costs[0]
dollars;
a 7-day pass is sold for
costs[1]
dollars;
a 30-day pass is sold for
costs[2]
dollars.
The passes allow that many days of consecutive travel. For example, if we get a 7-day pass on day 2, then we can travel for 7 days: day 2, 3, 4, 5, 6, 7, and 8.
Return the minimum number of dollars you need to travel every day in the given list ofdays.
Example 1:
Input:
days =
[1,4,6,7,8,20]
, costs =
[2,7,15]
Output:
11
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 1-day pass for costs[0] = $2, which covered day 1.
On day 3, you bought a 7-day pass for costs[1] = $7, which covered days 3, 4, ..., 9.
On day 20, you bought a 1-day pass for costs[0] = $2, which covered day 20.
In total you spent $11 and covered all the days of your travel.
Example 2:
Input:
days =
[1,2,3,4,5,6,7,8,9,10,30,31]
, costs =
[2,7,15]
Output:
17
Explanation:
For example, here is one way to buy passes that lets you travel your travel plan:
On day 1, you bought a 30-day pass for costs[2] = $15 which covered days 1, 2, ..., 30.
On day 31, you bought a 1-day pass for costs[0] = $2 which covered day 31.
In total you spent $17 and covered all the days of your travel.
Note:
1 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000
分析
分组背包，组loop在v内。这里v是天数，注意不在出差天内不管直接dp[i] = dp[i-1]
出差天里3个取min。但是这里value在范围外(负数）可以设置为0
class Solution:
def mincostTickets(self, days: List[int], costs: List[int]) -> int:
n = days[-1]
dp = [float('inf')]*(n+1)
A = [1,7,30]
dp[0] = 0
for d in range(1,n+1):
if d not in days:
dp[d] = dp[d-1] #不在出差天里可以不买票不管。
else:#出差天里必有覆盖
for ind,a in enumerate(A):
dp[d] = min(dp[d],dp[max(0,d-a)] + costs[ind]) #错在max(0,d-a)， 不能判断d-a>=0才进来算Min,如果不够大直接按0天算。表示覆盖率、
return dp[-1]