Minimum Cost For Tickets
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 from1
to365
.
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]
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