What is the fastest algorithm for this kind of RMQ?

Question

I am doing a task using the following algorithm (pseudo code)

int A = []
int C = { ... } // N non-negative integers 
int R = { ... } // N non-negative integers
for(i = 0 to N){
    // Let j in range [i-R[i], i-1]
    A[i] = Minimum of ( A[j] + C[j] ) where j in [i-R[i], i-1]
}

What the loop trying to do is, use a range of previous computed A[j] to compute current A[i] .

To me it is like doing N dynamic RMQs (Range minimum query).

It is dynamic because we do not know the value of A beforehand, we compute it online using the previously computed values.

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For example,

C = {10,1,5,3}
R = {2,4,2,3}

A[0] = C[0] // Assume for i = 0, this is always true as base case
A[1] = Min(A[j] + C[j]) where -3 <= j <= 0, only j = 0 is valid option
     = Min(A[0] + C[0]) = 20
A[2] = Min(A[j] + C[j]) where 1 <= j <= 1
     = 21
A[3] = Min(A[j] + C[j]) where -1 <= j <= 2
     = Min(A[0]+C[0], A[1]+C[1], A[2]+C[2])
     = Min(20, 21, 24)
     = 20

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So my question is, is there any algorithm which is faster than O(N^2) to achieve this? I feel like there is some method / data structure to speed up the N RMQs, but I do not know how.

Please tell me to elaborate more if the example is not clear.

Answer 1

动态RMQ问题是段树的基本应用，每个查询的复杂度为O(log n) 。