Stone Game

思路：

state: dp[i][j]代表从第i个石头到第j个石头，能堆的最小cost。
transition: dp[i][j]=min(dp[i][k]+dp[k+1][j]+sum[i][j]), i<=k<j。s[i][j]是拆分后最后一步需要的cost。
initial value: dp[i][i]=0

实现细节：取最小值循环前，应该赋给当前dp最大值。

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