Problem
Each fence is increased at most 2 times, so dp[i][j] is the min cost to make A[0, i] great with the last fence A[i] increased j times.
The answer is min of dp[n - 1][].
标签：Again,Great,increased,fence,min,Make,times,dp
来源： https://www.cnblogs.com/lz87/p/15816194.html