BOP 2016复赛题目解析

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复赛题
Microsoft Academic Graph (MAG) is a large heterogeneous graph containing entities such as authors， papers， journals， conferences and relations between them. Microsoft provides Academic Knowledge API for this contest. The Entity attributes are defined here.

Participants are supposed to provide a REST service endpoint that can find all the 1-hop， 2-hop， and 3-hop graph paths connecting a given pair of entity identifiers in MAG. The given pair of entity identifiers could be [Id， Id]， [Id， AA.AuId]， [AA.AuId， Id]， [AA.AuId， AA.AuId]. Each node of a path should be one of the following identifiers: Id， F.Fid， J.JId， C.CId， AA.AuId， AA.AfId. Possible edges (a pair of adjacent nodes) of a path are:

For each test case， the REST service endpoint will receive a JSON array via HTTP with a pair of entity identifiers， where the identifiers are 64-bit integers， e.g. [123， 456]. The service endpoint needs to respond with a JSON array within 300 seconds. The response JSON array consists of a list of graph paths in the form of [path1， path2， …， pathn]， where each path is an array of entity identifiers. For example， if your program finds one 1-hop paths， two 2-hop paths， and one 3-hop paths， the results may look like this: [[123，456]， [123，2，456]， [123，3，456]， [123，4，5，456]]. For a path such as [123，4，5，456]， the integers are the identifiers of the entities on the path. After receiving the response， the evaluator will wait for a random period of time before sending the next requests.

Evaluation Metric
The REST service must be deployed to a Standard_A3 virtual machine for the final test. There are no constraints on the programming language you can use.

The test cases are not available before the final evaluation. When the evaluation starts， the evaluator system sends test cases to the REST endpoint of each team individually. Each team will receive 10 test cases (Q1to Q10). The response time for test case Qi is recorded as Ti(1≤i≤10). The final score is calculated using:

where Ni is the size of the solution (the total number of correct paths) for Qi ， Ki is the total number of paths returned by the REST service， Mi is the number of distinct correct paths returned by the REST service.

思路

题意解析：
为了帮助理解，我把文章实体各个属性含义列在下面，这里只说明比赛中要用到的带id的属性。
其中CC属性让我怨念颇深……比赛的时候完全没注意到，傻傻的用了RId.length，但是排名靠前的队伍基本都用上了，所以还是不够细心啊……心塞

Name	Description	Type	Operations
Id	Entity ID	Int64	Equals
CC	Citation count	Int32	none
AA.AuId	Author ID	Int64	Equals
AA.AfId	Author affiliation ID	Int64	Equals
F.FId	Field of study ID	Int64	Equals
J.Id	Journal ID	Int64	Equals
C.Id	Conference series ID	Int64	Equals
RId	Reference ID	Int64	Equals

从上面规则描述中的hop的定义可以看出，路径的组成只有11种：Id-Id， Id-FId， FId-Id， Id-JId， JId-Id， Id-CId， CId-Id，AuId-AFId， AFId-AuId， AuId-Id， Id-AuId。那么针对不同的Id对儿，可以找出下面的规律。

Id-Id，共计15种
- 1跳，1种直达
- 2跳，5种 Id1-Id-Id2，这种情况单独处理，用RId=Id2的反向查询更快捷。
  Id1-AuId-Id2， Id1-FId-Id2， Id1-JId-Id2， Id1-CId-Id2，
- 3跳，9种 Id1-Id-Id-Id2，这种情况比较麻烦，需要前向和反向查询，url编写复杂度较高
  Id1-AuId-Id-Id2， Id1-FId-Id-Id2， Id1-JId-Id-Id2， Id1-CId-Id-Id2，Id1-Id-AuId-Id2， Id1-Id-FId-Id2，Id1-Id-JId-Id2，Id1-Id-CId-Id2
Id-AuId，共计8种
- 1跳，1种直达
- 2跳，1种 Id-Id-AuId，1次查询就好
- 3跳，6种 Id-Id-Id-AuId， Id-AuId-AfId-AuId，Id-AuId-Id-AuId，Id-FId-Id-AuId，Id-JId-Id-AuId，Id-CId-Id-AuId
AuId-Id，共计8种
- 1跳，1种直达
- 2跳，1种 AuId-Id-Id
- 3跳，6种 AuId-Id-Id-Id， AuId-AfId-AuId-Id， AuId-Id-JId-Id， AuId-Id-CId-Id， AuId-Id-FId-Id，AuId-Id-AuId-Id
AuId-AuId 共计3种
- 1跳，木有
- 2跳，2种 AuId-Id-AuId， AuId-AfId-AuId，
- 3跳，1种 AuId-Id-Id-AuId
看起来这是比较复杂的，需要分别写出34种的情况，但是这些可能性中有不少可以复用的。比如，通过查询一次id1,id2的属性值，就可以写出Id1-AuId-Id2， Id1-FId-Id2， Id1-JId-Id2， Id1-CId-Id2这四种2hop的了。