LEETCODE ALGORITHM:983. Minimum Cost For Tickets

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题目

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:

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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:

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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

题解

从后向前动态规划

转移方程： \[ dp(i)=min\{cost(j)+dp(i+j)\},j∈{1,7,30} \] i为日期，如果当天不是出行日，则\(dp[i]=dp[i+1]\),若该天为出行日，转移方程如上

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class Solution {
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        vector<int> dp(367,0);
        int j=days.size()-1;
        for(int i=365;i>0;i--)
        {
            if(i==days[j]){
                    if(i+30<=365) dp[i]=min(costs[0]+dp[i+1],min(costs[1]+dp[i+7],costs[2]+dp[i+30]));
                    else if(i+7<=365) dp[i]=min(costs[0]+dp[i+1],costs[1]+dp[i+7]);
                    else dp[i]=costs[0]+dp[i+1];
                    j--;
            }else{
                dp[i]=dp[i+1];
            }
            if(j==-1) return dp[i];
        }
        return dp[1];
    }
};

从前往后,原理完全相同，不过只是直接找下一个出行日，直接跳过了不出行的日子，由于使用了递归反而速度更慢了。

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class Solution {
private:
    vector<int> m,days,costs,dur;
    int dp(int i)
    {
        if(i>=days.size()) return 0;
        if(m[i]!=-1) return m[i];
        m[i]=INT_MAX;
        for(int j=0;j<3;j++)
        {
            auto k = upper_bound(days.begin() + i, days.end(), days[i] + dur[j] - 1);
            if (k == days.end()) m[i] = min(m[i], costs[j]);
            else m[i] = min(m[i], costs[j] + dp(distance(days.begin(),k)));
        }
        return m[i];
    }
public:
    int mincostTickets(vector<int>& days, vector<int>& costs) {
        this->days=days;
        this->costs=costs;
        m.assign(days.size(),-1);
        dur=vector<int>{1,7,30};
        return dp(0);
    }
};