poj 1703 Find them, Catch them（种类并查集）

Findthem,Catchthem

TimeLimit:1000MS		MemoryLimit:10000K
TotalSubmissions:31532		Accepted:9713

DescriptionThepoliceofficeinTaduCitydecidestosayendstothechaos,aslaunchactionstorootuptheTWOgangsinthecity,GangDragonandGangSnake.However,thepolicefirstneedstoidentifywhichgangacriminalbelongsto.Thepresentquestionis,giventwocriminals;dotheybelongtoasameclan?Youmustgiveyourjudgmentbasedonincompleteinformation.(Sincethegangstersarealwaysactingsecretly.)AssumeN(N<=10^5)criminalsarecurrentlyinTaduCity,numberedfrom1toN.Andofcourse,atleastoneofthembelongstoGangDragon,andthesameforGangSnake.YouwillbegivenM(M<=10^5)messagesinsequence,whichareinthefollowingtwokinds:1.D[a][b]where[a]and[b]arethenumbersoftwocriminals,andtheybelongtodifferentgangs.2.A[a][b]where[a]and[b]arethenumbersoftwocriminals.Thisrequiresyoutodecidewhetheraandbbelongtoasamegang.InputThefirstlineoftheinputcontainsasingleintegerT(1<=T<=20),thenumberoftestcases.ThenTcasesfollow.EachtestcasebeginswithalinewithtwointegersNandM,followedbyMlineseachcontainingonemessageasdescribedabove.OutputForeachmessage"A[a][b]"ineachcase,yourprogramshouldgivethejudgmentbasedontheinformationgotbefore.Theanswersmightbeoneof"Inthesamegang.","Indifferentgangs."and"Notsureyet."SampleInput

1
55
A12
D12
A12
D24
A14

SampleOutput

Notsureyet.
Indifferentgangs.
Inthesamegang.

题意：警察要肃清两个帮派，该城市有龙帮和蛇帮两个帮派，给出两个罪犯，让我们判断是否属于一个帮派。

当前有N（1-N）个罪犯在这个城市，至少他们其中一个属于龙帮或者蛇帮，给出M行信息，信息分两种：

1.D[a][b]：a和b罪犯不属于同一个帮派。

2.A[a][b]:判断a和b是不是在一个帮派。是就输出Inthesamegang.不是就输出Indifferentgangs.不能判断就输出Notsureyet.

分析：种类并查集；

两个数组，一个为储存根节点的数组，另外一个为储存该点与根节点之间的关系（0,或者1）在寻找根节点的时候需要更新该点与根节点之间的关系；x与y合并的时候，若x1为y1的根节点，那么需要求ans[y1]；

ans[y1]=(ans[x]-ans[y]+1)%2;如下图所示可得出