T2[USACO22DEC]MakingFriendsP考虑删除一个点，会有如下的点相连接：题目要求如果两两个点建立联系，只会建立一次。所以，神奇地，我们取出当前待删的点所连接的最小的点，将它和剩下的点连接。手摸一下会发现这样就巧妙地给每个改建的边都建了一次。所以用一个set启发式合并就做完

P8907[USACO22DEC]MakingFriendsP题解我们考虑维护每个\(i\)，在\(i\)的后面有多少个点和它有朋友关系。初步的想法是每删掉一个人就给集合里所有的点连边。但是我们发现这样太不优了，有很多边会重复连很多次。优化的想法是对于\(i\)，删去之后连的边就成了一个完全图，于是

题目思路我们可以统计出所有点对之间应该有的边的数量然后再减去之前的\(m\)。对于每个点维护一个集合，统计该点应该连接的点。对于每条初始边，我们将其看作从编号小的点连向编号大的点，并在编号小的集合中放入编号大的点。处理删除\(i\)点操作，答案加上目前集合\(i\)的大

ACCT2002CostAnalysisfor Decision MakingTrimester2A,2024SyllabusThisisamanagementaccountingfoundationunitthatsupportsmanagerialplanninganddecisionmakingthroughtheintegrationofethicsandstrategytocostingmodelsandprofitplanni

文章目录引言一、决策的重要性二、常见的决策方式1.理性决策（RationalDecisionMaking）2.有限理性（BoundedRationality）3.直觉决策（IntuitiveDecisionMaking）4.循证管理（Evidence-BasedManagement）5.**众包（Crowdsourcing）**6.增量决策（IncrementalDecisionMaking）7.

1.介绍：Lambda表达式是JDK8开始新增的一种语法形式。2.作用：用于简化匿名内部类的代码写法。3.注意: Lambda表达式只能简化函数式接口的匿名内部类。4.函数式接口：有且仅有一个抽象方法的接口。5.注意：大部分函数式接口上面都可能会有一个@FunctionalInterface的注解，有该注

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1)Tomakeadigitalstory,youmustfirstwriteagoodscriptandstoryboard,sothatyouhaveaclearunderstandingandgraspofthedigitalstoryyouwanttotell.Posteriorly,IfoundanillustrationofthestoryfromtheInternetanduseditasaback

明明看着不难的题目，却意外的卡人。思路考虑两头奶牛可以成为朋友条件是什么。存在一条路径连接这两头奶牛。且除去端点外的路径上的所有点的编号小于两端点的较小值。充分必要性都比较显然。如何维护。我们可以从小到大加入点，维护这些路径。对于每个点维护一个\(\text{se

Management-Decision-{RationalD.M.:Logical,ConsistentandmaximizevalueBoundedRationalD.M.:"GoodEnough"basedonrealityIntuitiveD.M.:onthebasisofExperience,feelingsandaccumulatedjudgement.}DecisionMakingIntuitiveD.M.:So

Problem: YouwanttomakeanHTTPrequesttoawebserver.Solution: Usethenet/httppackagetomakeanHTTPrequest. HTTPisarequest-respondprotocol,andservingrequestsisonlyhalfofthestory.Theotherhalfismakingrequests.Thenet/http

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Whichistheuseofquantitativetechniquestoimprovemanagerialdecisionmaking.Alsoknownasmanagementscience.Bettermanagerialdecisionmakingbyusingquantitativetechniques,suchas:Thequantitativeapproachevolvedfrommathematicalandstati

1VocabularyVocabularyaffiliatedaffiliatedUK/əˈfɪlieɪtɪd/US/əˈfɪlieɪtɪd/adj.[onlybeforenoun]1.closelyconnectedtoorcontrolledbyagrouporanorganizationeg:Allaffiliatedmemberscanvote.agovernment-affiliatedinstituteDialo

最近好像一直懒得写题解，但是感觉还是写一写比较好。首先若干个向量组成一个凸包有经典做法，就是把向量按照极角排序，然后按照极角顺序依次拼接，得到的就是一个凸包，且方案唯一（由于本题限制不存在共线的两个向量）。那么我们实际上只需要知道每个向量最终用了多少就可以了。设第\(i\)

题目大意：给你一个数组A，要求将这个数组变成数组B，使得Sum(abs(A[i]-B[i]))达到最小，且B是单调的解题思路：因为答案要求输出单调非递增或者单调非递减的的任意一个，那就只考虑单调非递增吧，因为两个的思路是相同的如果要变化的话，且变化的值要达到最小的话，那么只能变成和前一个相同或者

FIT5094ITforManagementDecisionMakingSemester1,2023Assignment1–AnalysisofaStrategicDecisionFormat:IndividualReportWeight:25%ofthemarksavailableforFIT5094IndicativeLength:2,500–3,000wordsDuedate:ThursdayApril6,2023@4:30

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