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这题能用dp做吗？opt[i][j]=min( opt[i-1][k] )+i*cost[j] ( 1<=k<=n && k!=j && k可以作为j的前驱) 所谓前驱，就是涂完k后，可以直接涂j的点，也就是j的父节点或者是父节点的兄弟节点的孩子节点. 但是预处理要把j的前驱找出来有点麻烦。。。 Followed by:
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