2024-10-05 2024-10-05 随手记 几秒读完 (大约66个字)区间 DPdp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]+cost[i][j])dp[i][j]=max(dp[i][j], dp[i][k]+dp[k+1][j]+cost[i][j])dp[i][j]=max(dp[i][j],dp[i][k]+dp[k+1][j]+cost[i][j]) 从小到大枚举区间的长度 确定 i,j,枚举中间的点,由于先枚举的长度,i和j之间更小的区间都已经被计算过。 Dynamic Programming, Competitive Programming
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