문제
Matrix i has size dimensions[i] × dimensions[i+1]. Multiplying a p×q matrix by a q×r matrix costs p·q·r scalar multiplications. Choose where to put parentheses so multiplying the whole chain costs the least.
Implement min_multiplications(dimensions).
Input
dimensions: n + 1 sizes describing n matrices.
Output
Return the minimum total cost.
제약 조건
2 ≤ length of dimensions ≤ 200; 1 ≤ each size ≤ 500
예시
입력:
dimensions = [10, 30, 5, 60]
예상 출력:
expected = 4500입력:
std::vector<long long> dimensions = {10, 30, 5, 60};
예상 출력:
long long expected = 4500;입력:
long[] dimensions = new long[] {10, 30, 5, 60};
예상 출력:
long expected = 4500;설명
(A×B)×C costs 10·30·5 + 10·5·60 = 4500, cheaper than A×(B×C) = 27000.
입력:
dimensions = [5, 10]
예상 출력:
expected = 0입력:
std::vector<long long> dimensions = {5, 10};
예상 출력:
long long expected = 0;입력:
long[] dimensions = new long[] {5, 10};
예상 출력:
long expected = 0;설명
A single matrix needs no multiplication: 0.
접근 방법
먼저 스스로 풀어 본 뒤 힌트를 하나씩 열어 보세요.
힌트 1
The order of multiplication changes the total cost even though the final matrix shape stays the same.
힌트 2
Try every possible split point for each interval and reuse smaller interval results.
힌트 3
Use the dimensions array carefully: each subproblem depends on the endpoints around the matrices, not the matrix count alone.
예상 복잡도
O(n^3) time, O(n^2) extra space
참고 자료
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