This time we'll put some effort to make an English editorial (at least brief full-score solution explanation of problems except problem I).

I solved it with DP.

Now we consider array a[1...n] and we know the position p such that a[p] = max(a[1...n]).

We can:

Not erase position p:

In this case, every queries have l(i) <= p <= r(i) will pick p, so the total cost = "number of queries such that l(i) <= p <= r(i)" * a[p].

After that, we can skip queries that have l(i) <= p <= r(i) so we can split array [1...n] to array [1...p-1] and [p+1...n].

With that observation we can think about DP with f(l, r, ...).

Erase position p:

We can store variable lim so that next time we only consider position i such that a(i) < lim.

We can compress array so lim <= n.

After merging 2 cases, we can know which states we need to store.

So DP will look like f(l, r, number of elements we can erase, limit).

DP transtion tooks O(N). So total complexity is O(N ^ 5).

My code: https://pastebin.com/iczqhRcZ.