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資料結構與演算法

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國立臺灣大學 2024 · 考試 · 第 10 題

考科演算法

(5 points) Consider a sequence of numbers (v1,…,vn)(v_1, \dots, v_n). Now we remove kk numbers from the sequence so that we have k+1k+1 non-empty segments of numbers. For example, consider (2,3,8,1,4)(2, 3, 8, 1, 4). If we remove 8 then we have two segments (2,3)(2,3) and (1,4)(1,4).

Now we want to minimize the maximum sum of numbers of a segment. Let m(1,n,k)m(1, n, k) be the answer, then what is the correct recursion for mm when 1≤i<j≤n1 \le i < j \le n and 0<k≤(j−i)/20 < k \le (j-i)/2?

  • (A)
    m(i,j,k)=min⁡x=ijmax⁡(∑y=xjvy, m(i,x,k−1))m(i,j,k) = \min_{x=i}^{j} \max\left(\sum_{y=x}^{j} v_y,\ m(i,x,k-1)\right)
  • (B)
    m(i,j,k)=min⁡x=i+1j−1max⁡(∑y=x+1jvy, m(i,x−1,k−1))m(i,j,k) = \min_{x=i+1}^{j-1} \max\left(\sum_{y=x+1}^{j} v_y,\ m(i,x-1,k-1)\right)
  • (C)
    m(i,j,k)=min⁡x=i+1j−1max⁡(m(x+1,j,k−1), ∑y=ix−1vy)m(i,j,k) = \min_{x=i+1}^{j-1} \max\left(m(x+1,j,k-1),\ \sum_{y=i}^{x-1} v_y\right)
  • (D)
    m(i,j,k)=min⁡v+w=k−1, v,w≥0 min⁡x=i+1j−1(max⁡(m(i,x−1,v), m(x+1,j,w)))m(i,j,k) = \min_{v+w=k-1,\ v,w \ge 0}\ \min_{x=i+1}^{j-1}\left(\max\big(m(i,x-1,v),\ m(x+1,j,w)\big)\right)
  • (E)
    m(i,j,k)=min⁡v+w=k−1, v,w≥0 min⁡x=ij(max⁡(m(i,x,v), m(x,j,w)))m(i,j,k) = \min_{v+w=k-1,\ v,w \ge 0}\ \min_{x=i}^{j}\left(\max\big(m(i,x,v),\ m(x,j,w)\big)\right)

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