zoukankan      html  css  js  c++  java
  • Leetcode983 Minimum Cost For Tickets

    Description

    Problem

    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.

    Sample

    Sample1

    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.

    Sample2

    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.

    Analysis

    不方便看出线性关系时一定要考虑全部的不变量,即使往外扩展有多种可能,不变的还是天数之间的差值,以此列出状态转移方程

    Code

    以旅游日入手

    考虑upper_bound

    dp(j)=min(dp(j-1)+costs[0],dp(j-7)+costs[1],dp(j-30)+costs[2])

    class Solution {
    public:
        int mincostTickets(vector<int>& days, vector<int>& costs) {
            int n=days.size();
            auto start=days.begin();
            for(int i=0;i<n;++i){
                int b=(int)(upper_bound(start,start+i,days[i]-7)-start);
                int c=(int)(upper_bound(start,start+i,days[i]-30)-start);
                dp[i+1]=min(dp[i]+costs[0],min(dp[b]+costs[1],dp[c]+costs[2]));
            }
            return dp[n];
        }
    private:
            int dp[366];
    };
    

    以365天入手

    dp(i)=dp(i-1)

    dp(i)=min{cost(j)+dp(i-j)},j∈{1,7,30}

    class Solution {
    public:
        int mincostTickets(vector<int>& days, vector<int>& costs) {
            int n=days.size();
            int end=days[n-1];
            int idx=0;
            int x,y;
            for(int i=days[0];i<=end;++i){
                if(i==days[idx]){
                    x=i-7>=0?dp[i-7]:0;
                    y=i-30>=0?dp[i-30]:0;
                    dp[i]=min(dp[i-1]+costs[0],min(x+costs[1],y+costs[2]));
                    idx++;
                }
                else dp[i]=dp[i-1];
            }
            return dp[end];
        }
    private:
            int dp[366];
    };
    
  • 相关阅读:
    tf_upgrade_v2.exe实验
    tf.random_uniform出错tensorflow2.0出错
    Tensorflow2.0变化
    Anaconda安装PyTorch
    Anaconda是如何进行版本管理的?
    CUDA开发指南
    Tensorflow视频教程&Pytorch视频教程
    Applied Spatiotemporal Data Mining应用时空数据挖掘
    GAN one-shot
    基于深度学习的图像超分辨率重建 一种基于非负矩阵分解的高光谱影像模拟方法
  • 原文地址:https://www.cnblogs.com/chanceYu/p/12839666.html
Copyright © 2011-2022 走看看