合肥生活安徽新聞合肥交通合肥房產生活服務合肥教育合肥招聘合肥旅游文化藝術合肥美食合肥地圖合肥社保合肥醫院企業服務合肥法律

        代寫 CS 336、代做 java/c++設計程序
        代寫 CS 336、代做 java/c++設計程序

        時間:2024-11-10  來源:合肥網hfw.cc  作者:hfw.cc 我要糾錯



        CS 336: Algorithms Problem Set 5 Date: Thursday, October 31, 2024 Due: Thursday, November 7, 2024
        Submit your solution on Gradescope.
        Please, solve all problems on your own. Do not collaborate with other students.
        Problem 1. The page limit for Problem 1 is 2 pages.
        Similarly to HW2, you want to travel from city A to city B located on a straight line (A is
        located in position 0 and B is located in position M ≥ 0), and you can travel at most distance D ≥ 0 miles per day, and you can only move to the right. Similarly, you have hotels between A and B with locations a1, . . . , an, where you can stay for a night.
        You are a person who likes to optimize all aspects of your life. In particular, if you didn’t fully use all D miles per day, it causes you great distress. Namely, if on some day you traveled distance d miles (out of possible D miles), the amount of distress is 2D−d.
        You start at city A. Your goal is to reach city B while suffering the least total amount of distress. Example: Assume that D = 4 and city B is located in position 6. You have two hotels in locations
        2 and 3. The following routes have the following distress:
        • 0→2→6: 24−(2−0) +24−(6−2) =4+1=5
        • 0→2→3→6: 24−(2−0) +24−(3−2) +24−(6−3) =4+8+2=14 • 0→2→6: 24−(3−0) +24−(6−3) =2+2=4
        The last route is optimal.
        Please do the following:
        • Formulate the subproblem. Please state it as precisely as possible. • Design a dynamic programming algorithm for solving this problem:
        – State the base case.
        – State the recurrence relation.
        – Explain why the recurrence relation is correct (from your explanation, one should un- derstand how to get your the recurrence relation).
        – Please provide the pseudocode. Please use the bottom-up approach.
        – Explain:
        ∗ What is the running time of your algorithm (all arithmetic operations take constant time).
        ∗ How to recover the maximum reward.
        ∗ How to recover the optimal route. You don’t need to write a pseudocode.
        ∗ How your algorithm correctly handles the case when an optimal solution doesn’t
        exist.
         1

        Problem 2. There is a new series in your streaming platform, Panopto. The series contains n episodes in total. Episodes need to be watched in order; that is, you cannot watch episode j before episode i if i < j. Since you’re busy, you decide to skip some subset of episodes (potentially empty). Your goal is to minimize the total amount of energy needed for this series, computed as follows:
        • You figure out that if you skip episode i, you would have to spend pi energy at the end of the year to figure out the missed content.
        • In addition, each episode has excitement value ei. You don’t want to dramatically change your emotions as well. So, for any consecutive episode i and j you watch, you need to spend |ei − ej | energy to adjust your mood as well.
        For example, if there are 5 episodes:
        • If you decide to watch episodes 1, 3, and 4, you need to spend p2 +p5 +|e1 −e3|+|e3 −e4| units of energy.
        • If you only decide to watch episode 3, you need to spend p1 + p2 + p4 + p5 units of energy.
        • If you decide to watch none of the episodes, you need to spend p1 +p2 +p3 +p4 +p5 units of
        energy.
        Implement the following function, which returns the list of episodes you decided to watch in the sorted order (the episodes are **indexed). For example, if you decide to watch first, third, and fourth episodes, your function must return a vector with items 1,3,4, in exactly this order. The input arrays are e and p respectively. It is guaranteed that for all test cases, the optimal answer is unique.
            vector<int> Episodes(const vector<int>& excitement, const vector<int>& penalty)
        Time limit The instructions are similar to the previous programming assignments. Your program should pass each tests in no more than 1 second. You can assume that 1 ≤ n ≤ 104 and all numbers are between 1 and 109.



        請加QQ:99515681  郵箱:99515681@qq.com   WX:codinghelp

        掃一掃在手機打開當前頁
      1. 上一篇:代做CMPT 401、代寫 c++設計程序
      2. 下一篇:代寫 CP3405、代做 Python/C++語言編程
      3. 無相關信息
        合肥生活資訊

        合肥圖文信息
        出評 開團工具
        出評 開團工具
        挖掘機濾芯提升發動機性能
        挖掘機濾芯提升發動機性能
        戴納斯帝壁掛爐全國售后服務電話24小時官網400(全國服務熱線)
        戴納斯帝壁掛爐全國售后服務電話24小時官網
        菲斯曼壁掛爐全國統一400售后維修服務電話24小時服務熱線
        菲斯曼壁掛爐全國統一400售后維修服務電話2
        美的熱水器售后服務技術咨詢電話全國24小時客服熱線
        美的熱水器售后服務技術咨詢電話全國24小時
        海信羅馬假日洗衣機亮相AWE  復古美學與現代科技完美結合
        海信羅馬假日洗衣機亮相AWE 復古美學與現代
        合肥機場巴士4號線
        合肥機場巴士4號線
        合肥機場巴士3號線
        合肥機場巴士3號線
      4. 上海廠房出租 短信驗證碼 酒店vi設計

        主站蜘蛛池模板: 国产免费一区二区三区免费视频 | 亚洲一区二区精品视频| 国产在线精品一区二区在线观看 | 久久99国产一区二区三区| 激情啪啪精品一区二区| 亚洲视频一区在线| 国产亚洲情侣一区二区无| 国产亚洲福利一区二区免费看 | 亚洲国产精品一区二区第一页免| 亚洲一区二区三区高清| 国产免费播放一区二区| 国产精品一区二区AV麻豆| 无码中文字幕乱码一区| 老熟妇仑乱一区二区视頻| 中文字幕精品一区二区精品 | 国产一区二区草草影院| 伊人久久精品一区二区三区| 精品无码日韩一区二区三区不卡| 亚洲乱码一区二区三区在线观看 | 久久久久久免费一区二区三区| 无码日韩人妻AV一区免费l| 国产在线一区观看| 精品免费国产一区二区| 亚洲一区无码精品色| 国产美女口爆吞精一区二区| 国产小仙女视频一区二区三区| 3d动漫精品啪啪一区二区中| 国产精品视频第一区二区三区| 中文字幕一区二区日产乱码| 国产综合精品一区二区| 精品国产一区二区三区www| 国产人妖视频一区二区破除| 国产福利一区视频| 在线观看午夜亚洲一区| 久久久无码精品国产一区| 午夜AV内射一区二区三区红桃视| 日本在线视频一区二区| 久久久99精品一区二区| 久久国产午夜一区二区福利| 无码人妻精品一区二区三区在线| 亚洲AV无码一区二区三区牛牛|