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`1`	`1`	`### 3553.Minimum-Weighted-Subgraph-With-the-Required-Paths-II`
`2`	`2`
`3`	`3`	本题的第一个知识点是:在一棵树里,联通u,v,w三个节点的最小子树的权重和,就是`[dist(u,v)+dist(u,w)+dist(v,w)]/2`.
	`4`	`+`
	`5`	+本体的第二个知识点是:在一棵树里,联通x,y两点的路径长度,等于`dist(r,x)+dist(r,y)-2*dist(r,c)`,其中r是整棵树的根节点,c是x和y的LCA(lowest common ancester)。
	`6`	`+`
	`7`	`+任意一点到距离根节点的距离dist(r,x)可以通过DFS得到。于是本题的关键点就是求任意两点的LCA,于是就是一个binary list经典题。`
	`8`	`+`
	`9`	+我们需要处理得到一个数组up[v][k],表示节点v往上(朝根节点方向)走2^k步能够得到的位置。转移方程就是`up[v][k] = up[up[v][k-1]][k-1]`. 边界条件就是对于一对父子节点a->b,有`up[b][0]=a`.

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