"Algorithmic Learning Notes" 84. Sequence DP relaxation + cost processing code_vs 1048 Stone Merge

Simple DP, when processing as far as possible with Len from 1 to N), I from 1 to len-n] to traverse.

Note that Len at this time represents a sequence of Len elements after the start of I

 for (int i = n1; I >=1 ;---i)    {for (int j = i+1; j < = N; + +J)        {= INF        ;  for (int k = i; k < J; + +k)            {= min (Dp[i][j], dp[i][k] + dp[k+1][j] + weight (i,j ) );        }    }}

Note First initialize DP[I][J] as inf

K Insert from I to J

The cost of dividing the I.J into [i,k] and [k+1,j] (slack handling) Two segment two is the total weight

The last dp[1][n] is the answer.

"Algorithmic Learning Notes" 84. Sequence DP relaxation + cost processing code_vs 1048 Stone Merge