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  • 利用hadoop来解决“共同好友”的问题

    假设A有好友B C D;B有好友A C D E;C有好友A B D E;D有好友A B C E;E有好友B C D。
    A -> B C D
    B -> A C D E
    C -> A B D E
    D -> A B C E
    E -> B C D
    对于A来说,A -> B C D
    (A B) -> B C D
    (A C) -> B C D
    (A D) -> B C D
    对于B来说,B -> A C D E
    (A B) -> A C D E
    (B C) -> A C D E
    (B D) -> A C D E
    (B E) -> A C D E
    对于C来说,C -> A B D E
    (A C) -> A B D E
    (B C) -> A B D E
    (C D) -> A B D E
    (C E) -> A B D E
    对于D来说,D -> A B C E
    (A D) -> A B C E
    (B D) -> A B C E
    (C D) -> A B C E
    (D E) -> A B C E
    对于E来说,E -> B C D
    (B E) -> B C D
    (C E) -> B C D
    (D E) -> B C D
    汇总得到
    (A B) -> (A C D E) (B C D)
    (A C) -> (A B D E) (B C D)
    (A D) -> (A B C E) (B C D)
    (B C) -> (A B D E) (A C D E)
    (B D) -> (A B C E) (A C D E)
    (B E) -> (A C D E) (B C D)
    (C D) -> (A B C E) (A B D E)
    (C E) -> (A B D E) (B C D)
    (D E) -> (A B C E) (B C D)
    则共同好友的关系是
    (A B) -> (C D)
    (A C) -> (B D)
    (A D) -> (B C)
    (B C) -> (A D E)
    (B D) -> (A C E)
    (B E) -> (C D)
    (C D) -> (A B E)
    (C E) -> (B D)
    (D E) -> (B C)
     
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  • 原文地址:https://www.cnblogs.com/tommyli/p/3338798.html
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