983. 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 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 <= days.length <= 365
1 <= days[i] <= 365
days is in strictly increasing order.
costs.length == 3
1 <= costs[i] <= 1000

class Solution {
    public int mincostTickets(int[] days, int[] costs) {
        int[] dp = new int[366];
        dp[0] = 0;
        boolean[] travel = new boolean[366];
        for(int i: days) travel[i] = true;
        for(int i = 1; i <= 365; i++) {
            if(!travel[i]) {
                dp[i] = dp[i-1];
                continue;
            }
            for(int j = 0; j < costs.length; j++) {
                if(i >= costs[j]) dp[i] = Math.min(dp[i], dp[i - costs[j]] + costs[j]);
            }
        }
        return dp[365];
    }
}

一开始啊，我这么写，我寻思妹啥毛病啊，结果不对

后来一想，这么写的话会导致买不到票，也就是第一天也买不到票，因为dp[i] = 0。所以要换个方法能让前几天买到票

class Solution {
    public int mincostTickets(int[] days, int[] costs) {
        int[] dp = new int[366];
        dp[0] = 0;
        boolean[] travel = new boolean[366];
        for(int i: days) travel[i] = true;
        for(int i = 1; i <= 365; i++) {
            if(!travel[i]) {
                dp[i] = dp[i-1];
                continue;
            }
            dp[i] = Integer.MAX_VALUE;
            dp[i] = Math.min(dp[i], dp[Math.max(0, i - 1)] + costs[0]);
            dp[i] = Math.min(dp[i], dp[Math.max(0, i - 7)] + costs[1]);
            dp[i] = Math.min(dp[i], dp[Math.max(0, i - 30)] + costs[2]);
        }
        return dp[365];
    }
}

首先，dp【i】表示在第i天玩的时候最小要花费多少来买票，如果当前天数不在travel的days，那就让他等于前一日的cost

上面Math.max(0, i - 1/7/30)就保证了所有天都能买到票。意思是在三种可能的情况下挑一个花费最小的

https://www.youtube.com/watch?v=BgWDlQy5JR0

总结：

1. unbounded knapsack, we calculate in i days the minimum cost of buying tickets in three possible ways

attention the first several days need to check if the index is >= 0,

and if current day is not in our wishlist, we just let it equal to the previous one.