It's a standard knapsack problem, the only difference being that the order you want to eat the food in is increasing by A (you want the longest food you'll eat to be at the end)

Let $$$dp[i][j]$$$ be maximal happiness you can achieve taking some of first i dishes during j minutes. Let $$$backdp[i][j]$$$ be maximal happiness you can achieve taking some of last dishes (with indices from i to n) during j minutes. Now let's go through the dish that we will eat last ($$$i$$$) and time we spent on eating first $$$i - 1$$$ dishes ($$$t'$$$). Try to update answer as $$$min(answer, dp[i - 1][t'] + backdp[i + 1][(t - 1) - t'] + b[i]$$$. It's easy to calc $$$dp$$$ : $$$dp[i][j] = max(dp[i - 1][j], dp[i][j - 1], dp[i - 1][j - a[i]] + b[i]$$$ (if such exist)$$$)$$$. Same with $$$backdp$$$.

If you print the matrix you notice that every third diagonal is part of pascals triangle (so just implement combinations and get the right row and column in the triangle)

best[i][j][k] -> best cost if you only consider the first i columns, the i-th column has value j and you've done k changes already. (j is really big, but you notice that there's no use in changing in something that's not already in the input, or 0). (Note, since you only look 1 in the past with i, you can only store 0/1 for it)

The big observation to make this fit in time is that if the previous one has height X, then if you change the current one, you don't want to make it shorter, since that won't decrease the cost, but might increase it in the future.