题意：a->b c :攻克b城堡前要先攻克a城堡攻克b城堡将获得c

共有n个城堡（编号为-n）最多攻克m个城堡能获得的最大收益

把作为树的根把节点连成一棵树

p[root][i]表示在以root为根的子树中选取i个点能获得的最大收益

p[root][i]=max(p[root][i],p[root][i-j]+p[v[e]][j])

当root!=0时每棵子树根必选所以p[root][i]=p[root][i-1]+val[root]

结果为p[0][m]

-Meaning of the questions: a-> bc: overcome b Castle before the first capture of a castle overcome b Castle will get c

The total n Castle (number-n) up to overcome the m castles can get maximum benefit

The root of the tree as a node connected into a tree

p [root] [i] i points to get selected in the sub-tree root-root maximum benefit

p [root] [i] = max (p [root] [i], p [root] [ij]+ p [v [e]] [j])

When the root! = 0 Meike subtree root Required so p [root] [i] = p [root] [i-1]+ val [root]

The result is p [0] [m]