983. Minimum Cost For Tickets 最低票价

@TOC

# 题目描述

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 array days. Each day is an integer from 1 to 365.

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 of days.

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. `1 <= days.length <= 365`
2. `1 <= days[i] <= 365`
3. days is in strictly increasing order.
4. `costs.length == 3`
5. `1 <= costs[i] <= 1000`

# 解题方法

# 动态规划

1. dp[i] = dp[i - 1] 当第i天不用旅行
2. dp[i] = min(dp[i - 1] + costs[0], dp[i - 7] + costs[1], dp[i - 30] + costs[2]) 当第i天需要旅行

C++代码如下：

``````class Solution {
public:
int mincostTickets(vector<int>& days, vector<int>& costs) {
// dp[i] means min cost for day i
vector<int> dp(366, INT_MAX);
for (int day : days)
dp[day] = 0;
dp[0] = 0;
for (int i = 1; i < 366; ++i) {
if (dp[i] == INT_MAX)
dp[i] = dp[i - 1];
else {
int cur = dp[i - 1] + costs[0];
cur = min(cur, costs[1] + dp[max(0, i - 7)]);
cur = min(cur, costs[2] + dp[max(0, i - 30)]);
dp[i] = cur;
}
}
return dp[days[days.size() - 1]];
}
};
``````

Python代码如下：

``````class Solution(object):
def mincostTickets(self, days, costs):
"""
:type days: List[int]
:type costs: List[int]
:rtype: int
"""
dp = [float("inf")] * 366
for day in days:
dp[day] = 0
dp[0] = 0
for i in range(1, 366):
if dp[i] == float("inf"):
dp[i] = dp[i - 1]
else:
cur = dp[i - 1] + costs[0]
cur = min(dp[max(0, i - 7)] + costs[1], cur)
cur = min(dp[max(0, i - 30)] + costs[2], cur)
dp[i] = cur
return dp[days[-1]]
``````

# 日期

2019 年 1 月 29 日 —— 小年好 2020 年 5 月 6 日 —— 今天的每日一题