Algorithm 一个有趣的图形任务
有一棵树有n个顶点。我们被要求计算一个多集的最小大小,例如,对于树中的每条边(u,v),至少有以下一个保持:Algorithm 一个有趣的图形任务,algorithm,dynamic-programming,graph-theory,graph-algorithm,Algorithm,Dynamic Programming,Graph Theory,Graph Algorithm,有一棵树有n个顶点。我们被要求计算一个多集的最小大小,例如,对于树中的每条边(u,v),至少有以下一个保持: u∈ v∈ S中至少有两个顶点,每个顶点都与u或v相邻 由于S是一个多重集,一个顶点可能在S中多次出现 我的直觉如下。首先,我们考虑以下事实:在最优解中,每个顶点最多在S中两次。所以我们可以按后序遍历树,计算三种情况下的结果,其中一个顶点不在最优S中,它在一次和两次中 不幸的是,我不能将子问题之间的关系联系起来,我不确定这个想法是否正确 欢迎提供提示或参考。非常感谢。代表
- u∈
- v∈
- S中至少有两个顶点,每个顶点都与u或v相邻
1
/|\
2 3 4
|
5
是
这里有一个自下而上的验证函数
> type ZeroOneOrTwo = Int
> data Result = Failure
> | Success { numDist0Provided :: ZeroOneOrTwo
> , numDist1Provided :: ZeroOneOrTwo
> , numDist1Required :: ZeroOneOrTwo
> }
>
> plus, minus :: ZeroOneOrTwo -> ZeroOneOrTwo -> ZeroOneOrTwo
> x `plus` y = min 2 (x + y)
> x `minus` y = max 0 (x - y)
>
> evaluate :: Tree ZeroOneOrTwo -> Bool
> evaluate t = case evaluate' t of
> Failure -> False
> s -> 0 == numDist1Required s
>
> evaluate' :: Tree ZeroOneOrTwo -> Result
> evaluate' (Node x) = Success { numDist0Provided = x
> , numDist1Provided = 0
> , numDist1Required = 0
> }
> evaluate' (Cons t1 t2) = case (evaluate' t1, evaluate' t2) of
> (Failure, _) -> Failure
> (_, Failure) -> Failure
> (s1, s2) ->
> if numDist0Provided s2 < numDist1Required s1 then Failure
> else Success { numDist0Provided = numDist0Provided s2
> , numDist1Provided = numDist1Provided s2 `plus` numDist0Provided s1
> , numDist1Required = max (numDist1Required s2 `minus` numDist0Provided s1)
> (if 0 < numDist0Provided s1 || 0 < numDist0Provided s2 then 0
> else 2 `minus` (numDist1Provided s1 `plus` numDist1Provided s2))
> }
>键入ZeroOneOrTwo=Int
>数据结果=失败
>|成功{numDist0Provided::ZeroOneOrTwo
>,numDist1Provided::ZeroOneOrTwo
>,numDist1Required::ZeroOneOrTwo
> }
>
>加,减::零一或二->零一或二->零一或二
>x`plus`y=min2(x+y)
>x`减去`y=max 0(x-y)
>
>评估::树zeroneortwo->Bool
>评估t=案例评估t
>失败->错误
>s->0==numdist1所需的s
>
>评估“::树zeroneortwo->结果
>evaluate'(节点x)=成功{numDist0Provided=x
>,numDist1Provided=0
>,numDist1Required=0
> }
>评估’(Cons t1 t2)=案例(评估’t1,评估’t2)
>(失败,824;)->失败
>(_,故障)->故障
>(s1,s2)->
>如果NUMDIST0提供s2else Success{numDist0Provided=numDist0Provided s2
>,numDist1Provided=numDist1Provided s2`plus`numDist0Provided s1
>,numDist1Required=max(numDist1Required s2`减去'numdist0s1)
>(如果0else 2`减去'(NUMDist1提供的s1`加上'NUMDist1提供的s2))
> }
我将保留相应的线性时间DP作为练习。听起来像是的相对值。谢谢。我以类似的方式表示一棵树,但我不知道如何实现DP算法。问题是,有很多案例。例如,位于S中的root的两个子项覆盖从root传出的其他边。所以也许贪婪的方法存在并且更容易编码…@SteveMadden不是
树zeroneortwo->Result
,而是树()->[(Result,Int)]
(()
在Haskell中意味着无效),而且,您可以扔掉成本更高的选项,以保持输出列表的大小最多为27个元素。
(Cons (Node 2)
(Cons (Cons (Node 5) (Node 3))
(Cons (Node 4) (Node 1)))) :: Tree Int .
> type ZeroOneOrTwo = Int
> data Result = Failure
> | Success { numDist0Provided :: ZeroOneOrTwo
> , numDist1Provided :: ZeroOneOrTwo
> , numDist1Required :: ZeroOneOrTwo
> }
>
> plus, minus :: ZeroOneOrTwo -> ZeroOneOrTwo -> ZeroOneOrTwo
> x `plus` y = min 2 (x + y)
> x `minus` y = max 0 (x - y)
>
> evaluate :: Tree ZeroOneOrTwo -> Bool
> evaluate t = case evaluate' t of
> Failure -> False
> s -> 0 == numDist1Required s
>
> evaluate' :: Tree ZeroOneOrTwo -> Result
> evaluate' (Node x) = Success { numDist0Provided = x
> , numDist1Provided = 0
> , numDist1Required = 0
> }
> evaluate' (Cons t1 t2) = case (evaluate' t1, evaluate' t2) of
> (Failure, _) -> Failure
> (_, Failure) -> Failure
> (s1, s2) ->
> if numDist0Provided s2 < numDist1Required s1 then Failure
> else Success { numDist0Provided = numDist0Provided s2
> , numDist1Provided = numDist1Provided s2 `plus` numDist0Provided s1
> , numDist1Required = max (numDist1Required s2 `minus` numDist0Provided s1)
> (if 0 < numDist0Provided s1 || 0 < numDist0Provided s2 then 0
> else 2 `minus` (numDist1Provided s1 `plus` numDist1Provided s2))
> }