※ 引述《Leon (Achilles)》之铭言:
: 嗯, 刚刚想了一下.
: 这个题目, 是 KMP 的变形.
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: 难的地方在于, 你怎么想到正确的方向
: (听起来有点废话, 哈哈..)
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: 这里我采用之前的定义.
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: For a sentance, we are given the occurance index.
: Take above example, say, only 3 words {a, b, c}.
: The sentance is [b b a c b a].
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: The the occurance index will be
: a = {3, 6}.
: b = {1, 2, 5}.
: c = {4 }.
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: 在开始之前, 我们先看几个非常简单的作法.
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: 1. Naive: Arbitary choose 2 position from [1, N],
: then check the condition.
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: 2. Choose one position as the left boundary, then,
: use the max-min of the rest set the decide the right boundary.
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: For example, if I choose 2 as my left boundary,
: then I need to choose the right boundary by the max of
: d(2,3) and d(2,4) which cover set {a} and set {c}
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: The result window is [2, 4].
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: And the complexity should be O(N* K).
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: 接下来, 就是有趣的地方了.
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: 3. Follow the previous algorithm, inspired by KMP.
: When we start from 3, actually we don't need to compare all K groups.
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: Because from previous step, we know [2] covers {b} for sure.
: And [3,4] covers {a,c} for sure.
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: In this case, we only need to check the condition for set {b} !
: which is d(3,5).
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: Then the computational complexity is O(N).
: I'll leave the detail for you.
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