判断题
1.如果一个问题可以用动态规划算法解决,则总是可以在多项式时间内解决的。
T | F |
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2.最优二叉搜索树的根结点一定存放的是搜索概率最高的那个关键字。
T | F |
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3.用动态规划而非递归的方法去解决问题时,关键是将子问题的计算结果保存起来,使得每个不同的子问题只需要被计算一次。子问题的解可以被保存在数组或哈希散列表中。
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4.For finding an optimal binary search tree, we can use the same greedy algorithm as the one for building a Huffman tree.
T | F |
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5.The weighted Activity Selection problem with weights in the set {1, 2} can be solved optimally by the same greedy algorithm used for the unweighted case.
T | F |
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选择题
1.切原木问题:给定一根长度为N米的原木;另有一个分段价格表,给出长度L=1,2,⋯,M对应的价格PL 。要求你找出适当切割原木分段出售所能获得的最大收益RN 。例如,根据下面给出的价格表,若要出售一段8米长的原木,最优解是将其切割为2米和6米的两段,这样可以获得最大收益R8 =P2 +P6=5+17=22。而若要出售一段3米长的原木,最优解是根本不要切割,直接售出。
下列哪句陈述是错的?
选项 | |
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A | 此问题可以用动态规划求解 |
B | 若N≤M,则有RN =max{PN ,max1≤i<N {Ri +RN-i}} |
C | 若N>M,则有RN=max1≤i<N {Ri +RN−M } |
D | 算法的时间复杂度是O(N2) |
2.在求解最优二叉搜索树问题时,我们用到递推式 cij=mini≤l≤j {wij +ci,l−1 +cl+1,j}。要通过迭代求解此式,必须用以下哪种方式填表:
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A | for i= 1 to n-1 do; for j= i to n do; for l= i to j do |
B | for j= 1 to n-1 do; for i= 1 to j do; for l= i to j do |
C | for k= 1 to n-1 do; for i= 1 to n-k do; set j = i+k; for l= i to j do |
D | for k= 1 to n-1 do; for i= 1 to n do; set j = i+k; for l= i to j do |
3.在动态规划中,我们要推导出一个子问题的解与其他子问题解的递推关系。要将这种关系转换为自底向上的动态规划算法,我们需要以正确的顺序填写子问题解的表格,使得在解任一子问题时,所有它需要的子问题都已经被解决了。在下列关系式中,哪一个是不可能被计算的?
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A | A(i,j)=min(A(i−1,j),A(i,j−1),A(i−1,j−1)) |
B | A(i,j)=F(A(min{i,j}−1,min{i,j}−1),A(max{i,j}−1,max{i,j}−1)) |
C | A(i,j)=F(A(i,j−1),A(i−1,j−1),A(i−1,j+1)) |
D | A(i,j)=F(A(i−2,j−2),A(i+2,j+2)) |
4.给定递推方程 fi,j,k=fi,j+1,k +min0≤l≤k {fi−1,j,l +wj,l }。要通过循环解此方程,我们一定不能用下列哪种方法填表?
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A | for k in 0 to n: for i in 0 to n: for j in n to 0 |
B | for i in 0 to n: for j in 0 to n: for k in 0 to n |
C | for i in 0 to n: for j in n to 0: for k in n to 0 |
D | for i in 0 to n: for j in n to 0: for k in 0 to n |
5.Which one of the following problems can be best solved by dynamic programming?
选项 | |
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A | Mergesort |
B | Closest pair of points problem |
C | Quicksort |
D | Longest common subsequence problem |