2016微软编程之美复赛
复赛采用团队赛的形式，基于Azure云计算平台在实际大数据的基础上展开前沿课题的较量。
在编程之美挑战赛复赛中，选手需要通过组队共同完成复赛题，考查选手编程能力的同时，也考验选手的团队协作能力。选手点此进入组队。
三人组队完成后，需要提供申请Azure账号的必要信息。然后我们会以邮件形式发送给选手Azure账号以及微软学术搜索API订阅key。
复赛题是基于微软Oxford 学术搜索API的系统编程题。Oxford 学术搜索是一个能够快速检索海量论文的搜索引擎。Oxford 学术搜索API的作用是从最新的Microsoft Academic Graph (MAG)中提取丰富的学术信息。复赛最终必须通过Azure进行提交，提交系统将在大赛官网主页公布。
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 Qiis 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.
思路总结：
**思路：**
判断给定的参数id是Id还是AuId：
发请求expr=Composite(AA.AuId=id)&attributes=Id,AA.AuId,AA.AfId，如果返回的entities长度为0，则为Id，否则为AuId。
而且当为AuId时，该请求所得结果可以直接用于后续计算，避免后面重复发送同样的请求
***
**[id1,id2]**
2-hop：[id1-other-id2]和[id1-id3-id2]
请求一：expr=Or(Id=id1,Id=id2)&attributes=RId,J.JId,C.CId,F.FId,AA.AuId
=>id1.RId, id1.Other 和 id2.Other
请求二：expr=RId=Id2&attributes=Id
=>{id3 | id3.RId==id2}
然后对 id1.Other 和 id2.Other 求交集（hashset），对id1.RId和 {id3| id3.RId==id2} 求交集
1-hop：从id1.RId中找里面是否有id2
3-hop：
1）可以有环 && 1-hop存在：id1 -> id2 -> id2.Other -> id2
2)
a) [id1-id4-id3-id2]和[id1-id4-other-id2] :
把id1.RId中的所有Id（除了Id2以外），指定几个为一组，使用
expr=Or(Or(Id=a,Id=b),Or(Id=c,Id=d))&attributes=Id,RId,J.JId,C.CId,F.FId,AA.AuId
的形式，求出id1.RId[index].RId和id1.RId[index].Other，然后求每个id1.RId[index].RId与{id3|id3.RId==id2}交集，以及每个id1.RId[index].Other与id2.Other的交集；
b) [id1-other-id3-id2] :
把{id3}中的所有Id，指定几个为一组，使用
expr=Or(Or(Id=a,Id=b),Or(Id=c,Id=d))&attributes=Id,J.JId,C.CId,F.FId,AA.AuId
的形式，求出{id3}[index].Other，
然后分别与id1.Other 求交集
***
**[id, AuId]**
expr=Id=id&attributes=RId,J.JId,C.CId,F.FId,AA.AuId
=> id.RId, id.Other, id.AuId
expr=Composite(AA.AuId=AuId)&attributes=Id,AA.AuId,AA.AfId
=> {id2 | id2<->AuId}, AuId.AfId
1-hop：从{id2}中找里面是否有id
2-hop：[id-id2-AuId]
求id.RId和{id2}的交集即可
3-hop：
a)
[id-other-id2-AuId]：把{id2}中的每一个元素id2[index]按照expr=Or(Or(Id=a,Id=b),Or(Id=c,Id=d))&attributes=Id,J.JId,C.CId,F.FId,AA.AuId
的形式，求出id2[index].Other，然后分别与id.Other求交集
b)
[id-id3-id2-AuId]：把id.RId中的每个元素id.RId[index]，按照expr=Or(Or(Id=a,Id=b),Or(Id=c,Id=d))&attributes=Id,RId
的形式，求出id.RId[index].RId，然后分别与{id2}求交集
c)
[id-AuId2-AfId-AuId]：(多线程or拼起来or分着发请求)把id.AuId中的每个元素id.AuId[index]（除去AuId，如果有等于AuId的那个，单独算）按照expr=Composite(AA.AuId=a)&attributes=AA.AuId,AA.AfId的形式，找出id.Auid[index].AfId，然后分别与AuId.AfId求交集
***
**[AuId, Id]**
expr=Id=id&attributes=J.JId,C.CId,F.FId,AA.AuId
=> Id.Other, Id.AuId
expr=RId=Id&attributes=Id
=> {id3 | id3.RId==Id}
expr=Composite(AA.AuId=AuId)&attributes=Id,AA.AuId,AA.AfId
=> {id2 | id2<->AuId}, AuId.AfId
1-hop：从{id2}中找里面是否有Id
2-hop：[AuId-id2-Id]
求{id2}与{id3}的交集即可
3-hop：
a) [AuId-id2-other-Id]和[AuId-id2-id3-Id]：
把{id2}中的每一个元素id2[index]按照expr=Or(Or(Id=a,Id=b),Or(Id=c,Id=d))&attributes=Id,RId,J.JId,C.CId,F.FId,AA.AuId
的形式，求出id2[index].Other和id2[index].RId，然后分别与Id.Other和{id3}求交集
b)[AuId-AfId-AuId2-Id]：
如果Id.AuId包含AuId，则单独算出一类3-hop路径—AuId-AuId.AfId-AuId-Id，然后对于剩余的每个Id.AuId[index]，按照expr=Composite(AA.AuId=a)&attributes=AA.AuId,AA.AfId的形式，找出id.Auid[index].AfId，然后分别与AuId.AfId求交集
***
**[AuId1, AuId2]**
expr=Composite(AA.AuId=AuId1)&attributes=Id,AA.AuId,AA.AfId
=> {id1| id1<->AuId1},
AuId1.AfId
expr=Composite(AA.AuId=AuId2)&attributes=Id,AA.AuId,AA.AfId
=> {id2| id2<->AuId2},
AuId2.AfId
没有1-hop
2-hop：
a)
[AuId1-id-AuId2]：求{id1}与{id2}的交集
b)
[AuId1-AfId-AuId2]：求AuId1.AfId与AuId2.AfId的交集
3-hop：[AuId1-id1-id2-AuId2]
对于{id1}中的每一个元素{id1}[index]，按照expr=Or(Or(Id=a,Id=b),Or(Id=c,Id=d))&attributes=Id,RId
的形式，
求出{id1}[index].RId，然后分别与{id2}求交集
***
**注**
get请求最大长度为2048，因此expr表达式长度设为1800
Id的最大长度为基本为10（但目前是按照最大19考虑的）
Or()操作拼64个Id
query最大条数count目前只设为10000，对于各别论文，这个值是不够的，可以多获取一个CC（Citation count）字段，根据此字段值的大小决定count设为多少
***
**RESTful参考资料**
[how-to-build-restful-service-with-java-using-jax-rs-and-jersey](http://crunchify.com/how-to-build-restful-service-with-java-using-jax-rs-and-jersey/)
[restful](http://wiki.jikexueyuan.com/project/restful/)
[jersey-25-maven-easy-rest-web-services](http://poor-developer.blogspot.com/2014/02/jersey-25-maven-easy-rest-web-services.html)
[maven-dynamic-web-project](http://poor-developer.blogspot.com/2014/02/maven-dynamic-web-project-ouch-how-do-i.html)
[tomcat](http://www.cnblogs.com/pannysp/archive/2012/03/07/2383364.html)
