Pick The Optimization

The Aliens trick on $i$, one binary search per penalty. Expanding the square makes each $k$ a line in $pre[j]$ with slope $-2,pre[k]$ and intercept $dp[i-1][k] + pre[k]^2$. A penalty per layer drops the $i$ dimension entirely, so one hull is $O(n)$ per probe and $O(n \log V) \approx 6 \cdot 10^6$ overall.

Convex hull trick, one hull per layer. Expanding the square makes each $k$ a line in $pre[j]$ with slope $-2,pre[k]$ and intercept $dp[i-1][k] + pre[k]^2$. Non-negative values make both the slopes and the query points monotone, so a deque hull is $O(n)$ per layer and $O(mn) \approx 2 \cdot 10^7$ overall.

Divide and conquer DP, one recursion per layer. Expanding the square makes each $k$ a line in $pre[j]$ with slope $-2,pre[k]$ and intercept $dp[i-1][k] + pre[k]^2$. It is the only option here, since a squared cost is not Monge, so the recursion is $O(mn \log n) \approx 3 \cdot 10^8$ overall.

Knuth optimization, one interval table per layer. Expanding the square makes each $k$ a line in $pre[j]$ with slope $-2,pre[k]$ and intercept $dp[i-1][k] + pre[k]^2$. Quadrangle monotonicity collapses the split range, so the whole table is $O(n^2) \approx 10^{10}$ overall and fits the limit.