尔雅Advanced Data Structures and Algorithm Analysis课后答案(学习通2023题目答案)

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Lecture 1. AVL Trees,尔雅 Splay Trees, and Amortized Analysis

1.1 AVL Trees (21:47)随堂测验

1、Insert 2,课后 1, 4, 5, 9, 3, 6, 7 into an initially empty AVL tree. Which one of the following statements is FALSE?
A、4 is 答案the root
B、3 and 7 are siblings
C、学习2 and 6 are siblings
D、通题9 is 目答the parent of 7

2、If the depth of an AVL tree is 尔雅6 (the depth of an empty tree is defined to be -1), then the minimum possible number of nodes in this tree is:
A、13
B、课后17
C、答案20
D、学习33

1.2 Splay Trees (13:16)随堂测验

1、通题For the result of accessing the keys 3,目答 9, 1, 5 in order in the splay tree in the following figure, which one of the following statements is FALSE?
A、5 is 尔雅the root
B、1 and 9 are siblings
C、课后6 and 10 are siblings
D、答案3 is the parent of 4

2、Finding the maximum key from a splay tree will result in a tree with its root having no right subtree.

1.3 Amortized Analysis (32:33)随堂测验

1、Consider the following buffer management problem. Initially the buffer size (the number of blocks) is one. Each block can accommodate exactly one item. As soon as a new item arrives, check if there is an available block. If yes, put the item into the block, induced a cost of one. Otherwise, the buffer size is doubled, and then the item is able to put into. Moreover, the old items have to be moved into the new buffer so it costs k+1 to make this insertion, where k is the number of old items. Clearly, if there are N items, the worst-case cost for one insertion can be Ω(N). To show that the average cost is O(1), let us turn to the amortized analysis. To simplify the problem, assume that the buffer is full after all the N items are placed. Which of the following potential functions works?
A、The number of items currently in the buffer
B、The opposite number of items currently in the buffer
C、The number of available blocks currently in the buffer
D、The opposite number of available blocks in the buffer

2、For one operation, if its amortized time bound is O(logN), then its worst-case time bound must be O(logN).

Lecture 2. Red-Black Trees and B+ Trees

2.1 Red-Black Trees: Definition (14:05)随堂测验

1、In a Red-Black tree, the path from the root to the farthest leaf is no more than twice as long as the path from the root to the nearest leaf.

2、For any red node X in a Red-Black tree, if it has two children, then the children's colors must be the same.

2.2 Red-Black Trees: Operations (21:24)随堂测验

1、After inserting { 3, 4, 5, 6, 1, 2, 7 } into an initially empty red-black tree, which of following is False?
A、The resulting tree is a full tree.
B、4 is the root with the black height as 2.
C、3 is the right child of 2, and the color of 3 is red.
D、5 is the left child of 6, and the color of 5 is black.

2.3 B+ Trees (19:29)随堂测验

1、A B+ tree of order 3 with 21 numbers has at most __ nodes of degree 3.
A、1
B、2
C、3
D、4

2、In a B+ tree, leaves and nonleaf nodes have some key values in common.

Lecture 3. Inverted File Index

3.1 Structure (18:06)随堂测验

1、Which of the following is NOT a step in the process of building an inverted file index?
A、Read in strings and parse to get words
B、Use stemming and stop words filter to obtain terms
C、Check dictionary with each term: if it is not in, insert it into the dictionary
D、Get the posting list for each term and calculate the precision

3.2 Modules (8:13)随堂测验

1、While accessing a term stored in a B+ tree in an inverted file index, range searches are expensive.

2、While accessing a term, hashing is faster than search trees.

3、Word stemming is to eliminate the commonly used words from the original documents.

3.3 Topics (16:42)随堂测验

1、For the document-partitioned strategy in distributed indexing, each node contains a subset of all documents that have a specific range of index.

2、In a search engine, thresholding for query retrieves the top k documents according to their weights.

3.4 Measures (13:46)随堂测验

1、Two spam mail detection systems are tested on a dataset with 7981 ordinary mails and 2019 spam mails. System A detects 200 ordinary mails and 1800 spam mails, and system B detects 160 ordinary mails and 1500 spam mails. If our primary concern is to keep the important mails safe, which of the following is correct?
A、Precision is our primary concern and system A is better.
B、Recall is our primary concern and system B is better.
C、Precision is our primary concern and system B is better.
D、Recall is our primary concern and system A is better.

2、When measuring the relevancy of the answer set, if the precision is high but the recall is low, it means that most of the relevant documents are retrieved, but too many irrelevant documents are returned as well.

Lecture 4. Leftist Heaps and Skew Heaps

4.1 Leftist Heap: Definition (15:32)随堂测验

1、A leftist heap with the null path length of the root being must have at least nodes.

4.2 Leftist Heap: Operations (13:22)随堂测验

1、Merge the two leftist heaps in the following figure. Which one of the following statements is FALSE?
A、the null path length of 6 is the same as that of 2
B、1 is the root with 3 being its right child
C、Along the left most path from top down, we have 1, 2, 4, and 5
D、6 is the left child of 2

2、Delete the minimum number from the given leftist heap. Which one of the following statements is TRUE?
A、9 is NOT the root
B、24 is the left child of 18
C、18 is the right child of 11
D、13 is the left child of 12

4.3 Skew Heaps: Definition (09:15)随堂测验

1、With the same operations, the resulting skew heap is always more balanced than the leftist heap.

2、The right path of a skew heap can be arbitrarily long.

4.4 Skew Heaps: Analysis (13:41)随堂测验

1、Which one of the following statements is FALSE about a skew heap?
A、Skew heaps do not need to maintain the null path length of any node
B、Comparing to leftist heaps, skew heaps are always more efficient in space
C、Skew heaps have O(logN) worst-case cost for merging
D、Skew heaps have O(logN) amortized cost per operation

2、The amortized time bound of insertions for a skew heap of size N is O(logN).

Lecture 5. Binomial Queue

5.1 Definition (10:47)随堂测验

1、Which of the following binomial trees can represent a binomial queue of size 42?
A、B0 B1 B2 B3 B4 B5
B、B1 B3 B5
C、B1 B5
D、B2 B4

5.2 Operations (14:37)随堂测验

1、Delete the minimum number from the given binomial queues in the following figure. Which one of the following statements must be FALSE?
A、there are two binomial trees after deletion, which are B1 and B2
B、11 and 15 can be the children of 4
C、29 can never be the root of any resulting binomial tree
D、if 29 is a child of 4, then 15 must be the root of B1

2、Merge the two binomial queues in the following figure. Which one of the following statements must be FALSE?
A、there are two binomial trees after merging, which are B2 and B4
B、13 and 15 are the children of 4
C、if 23 is a child of 2, then 12 must be another child of 2
D、if 4 is a child of 2, then 23 must be another child of 2

3、After inserting number 20 into a binomial queue of 6 numbers { 12, 13, 14, 23, 24, 35 }, which of the followings is impossible?
A、the LeftChild link of the node 20 is NULL
B、the NextSibling link of the node 20 is NULL
C、the NextSibling link of node 14 may point to node 20
D、the LeftChild link of node 12 may point to node 14

4、Inserting a number into a binomial heap with 15 nodes costs less time than inserting a number into a binomial heap with 19 nodes.

5.3 Implementations (21:09)随堂测验

1、To implement a binomial queue, the subtrees of a binomial tree are linked in increasing sizes.

2、To implement a binomial queue, left-child-next-sibling structure is used to represent each binomial tree.

5.4 Analysis (12:35)随堂测验

1、The potential function Q of a binomial queue is the number of the trees. After merging two binomial queues H1 with 12 nodes and H2 with 13 nodes，what is the potential change Q(H1+H2)?(Q(H1)+Q(H2)) ?
A、2
B、0
C、-2
D、-3

2、Making N insertions into an initally empty binomial queue takes Θ(NlogN) time in the worst case.

Lecture 6. Backtracking

6.1 Introduction (05:08)随堂测验

1、It is guaranteed that an exhaustive search can always find the solution in finite time.

6.2 Eight Queens (14:18)随堂测验

1、In the 4-queens problem, (x1, x2, x3, x4) correspond to the 4 queens' column indices. During backtracking, (1, 3, 4, ?) will be checked before (1, 4, 2, ?), and none of them has any solution in their branches.

6.3 Turnpike Reconstruction (18:09)随堂测验

1、In a Turnpike Reconstruction Problem, given distance set D = { 2, 2, 4, 4, 6, 8 }, x1~x4 = ( 0, 2, 4, 8 ) is the only solution provided that x1 = 0.

6.4 Games (13:10)随堂测验

1、Given the following game tree, the red node will be pruned with α-β pruning algorithm if and only if __.
A、6≤x≤13
B、x≥13
C、6≤x≤9
D、x≥9

Lecture 7. Divide and Conquer

7.1 Closest Points (13:29)随堂测验

1、How many of the following sorting methods use(s) Divide and Conquer algorithm? Heap Sort Insertion Sort Merge Sort Quick Sort Selection Sort Shell Sort
A、2
B、3
C、4
D、5

7.2 Substitution and Recursion-tree (22:56)随堂测验

1、3-way-mergesort : Suppose instead of dividing in two halves at each step of the mergesort, we divide into three one thirds, sort each part, and finally combine all of them using a three-way-merge. What is the overall time complexity of this algorithm for sorting elements?
A、
B、
C、
D、

7.3 Master Method (18:24)随堂测验

1、When solving a problem with input size by divide and conquer, if at each stage the problem is divided into 8 sub-problems of equal size , and the conquer step takes to form the solution from the sub-solutions, then the overall time complexity is __.
A、
B、
C、
D、

2、To solve a problem with input size N by divide and conquer algorithm, among the following methods, __ is the worst.
A、divide into 2 sub-problems of equal complexity N/3 and conquer in O(N)
B、divide into 2 sub-problems of equal complexity N/3 and conquer in O(NlogN)
C、divide into 3 sub-problems of equal complexity N/2 and conquer in O(N)
D、divide into 3 sub-problems of equal complexity N/3 and conquer in O(NlogN)

Lecture 8. Dynamic Programming

8.1 Fibonacci Numbers (08:53)随堂测验

1、To solve a problem by dynamic programming instead of recursions, the key approach is to store the results of computations for the subproblems so that we only have to compute each different subproblem once. Those solutions can be stored in an array or a hash table.

8.2 Matrix Multiplications (18:47)随堂测验

1、We can tell that there must be a lot of redundant calculations during the exhaustive search for the matrix multiplication problem, because the search work load is the Catalan number, yet there are only ___ different sub-problems .
A、
B、
C、
D、

8.3 Optimal Binary Search Trees (24:02)随堂测验

1、The root of an optimal binary search tree always contains the key with the highest search probability.

8.4 Floyd Shortest Path Algorithm (08:54)随堂测验

1、Why doesn't Floyd algorithm work if there are negative-cost cycles?
A、Because Floyd didn't like negative numbers.
B、Because Floyd algorithm will terminate after finite steps, yet the shortest distance is negative infinity if there is a negative-cost cycle.
C、Because Floyd algorithm will fall into infinite loops.
D、Because a negative-cost cycle will result in a negative D[i][i], yet Floyd algorithm can only accept positive weights.

8.5 Product Assembly (16:39)随堂测验

1、In dynamic programming, we derive a recurrence relation for the solution to one subproblem in terms of solutions to other subproblems. To turn this relation into a bottom up dynamic programming algorithm, we need an order to fill in the solution cells in a table, such that all needed subproblems are solved before solving a subproblem. Among the following relations, which one is impossible to be computed?
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))

Lecture 9. Greedy Algorithms

9.1 Introduction (05:20)随堂测验

1、Greedy algorithm works only if the local optimum is equal to the global optimum.

2、In a greedy algorithm, a decision made in one stage is not changed in a later stage.

9.2 Activity Selection (19:04)随堂测验

1、Let us consider the following problem: given the set of activities S, we must schedule them all using the minimum number of rooms. Greedy1: Use the optimal algorithm for the Activity Selection Problem to find the max number of activities that can be scheduled in one room. Delete and repeat on the rest, until no activities left. Greedy2: - Sort activities by start time. Open room 1 for . - for i=2 to n if can fit in any open room, schedule it in that room; otherwise open a new room for . Which of the following statements is correct?
A、None of the above two greedy algorithms are optimal.
B、Greedy1 is an optimal algorithm and Greedy2 is not.
C、Greedy2 is an optimal algorithm and Greedy1 is not.
D、Both of the above two greedy algorithms are optimal.

2、Let S be the set of activities in Activity Selection Problem. Then there must be some maximum-size subset of mutually compatible activities of S that includes the earliest finish activity.

9.3 Huffman Codes (22:48)随堂测验

1、Given 4 cases of frequences of four characters. In which case(s) that the total bits taken by Huffman codes are the same as that of the ordinary equal length codes? (1) 4 2 11 6 (2) 6 5 7 12 (3) 3 2 3 4 (4) 8 3 10 7
A、(3) only
B、(1) and (3)
C、(3) and (4)
D、none

2、Given four characters (a, b, c, d) with distinct frequencies in a text. Suppose that a and b are the two characters having the lowest frequencies. Which of the following sets of code is a possible Huffman code for this text?
A、a: 000, b:001, c:01, d:1
B、a: 000, b:001, c:01, d:11
C、a: 000, b:001, c:10, d:1
D、a: 010, b:001, c:01, d:1

Lecture 10. NP-Completeness

10.1 Definition (18:52)随堂测验

1、All decidable problems are NP problems.

2、All NP problems are decidable.

3、All NP-complete problems are NP problems.

4、All NP problems can be solved in polynomial time in a non-deterministic machine.

10.2 HCP to TSP (07:55)随堂测验

1、The first problem that was proven to be NP-complete was Circuit-SAT.

10.3 Formal Language (13:27)随堂测验

1、All the languages can be decided by a non-deterministic machine.

2、Let A and B be decision problems in NP, and assume P NP. If A B and B A, then both A and B are NP-complete.

3、A language L belongs to NP iff there exist a two-input polynomial-time algorithm A that verifies language L in polynomial time.

Lecture 11. Approximation

11.1 Introduction (06:26)随堂测验

1、An approximation scheme that runs in for any fixed is a polynomial-time approximation scheme.

2、An -approximation scheme of time complexity is a PTAS but not an FPTAS.

11.2 Bin Packing (17:28)随堂测验

1、In the bin packing problem, we are asked to pack a list of items L to the minimum number of bins of capacity 1. For the instance L, let FF(L) denote the number of bins used by the algorithm First Fit. The instance L' is derived from L by deleting one item from L. Then FF(L') is at most of FF(L).

11.3 Knapsack Problems (18:47)随堂测验

1、For the 0-1 version of the Knapsack problem, if we are greedy on taking the maximum profit or profit density, then the resulting profit must be bounded below by the optimal solution minus the maximum profit.

11.4 K-center (20:54)随堂测验

1、The K-center problem can be solved optimally in polynomial time if K is a given constant.

Lecture 12. Local Search

12.1 Introduction (06:12)随堂测验

1、In local search, if the optimization function has a constant value in a neighborhood, there will be a problem.

2、Greedy method is a special case of local search.

12.2 Vertex Cover (13:19)随堂测验

1、Random restarts can help a local search algorithm to better find global maxima that are surrounded by local maxima.

2、In Metropolis Algorithm, the probability of jumping up depends on T, the temperature. When the temperature is high, it'll be close to the original gradiant descent method.

12.4 Maximum Cut (19:29)随堂测验

1、In the Maximum Cut problem, let us define S' be the neighbor of S such that S' can be obtained from S by moving one node from A to B, or one from B to A. We only choose a node which, when flipped, increases the cut value by at least w(A,B)/|V|. Then which of the following is true?
A、Upon the termination of the algorithm, the algorithm returns a cut (A,B) so that 2.5w(A,B)≥w(A*,B*), where (A*,B*) is an optimal partition.
B、The algorithm terminates after at most flips, where is the total weight of edges.
C、Upon the termination of the algorithm, the algorithm returns a cut (A,B) so that 2w(A,B)≥w(A*,B*).
D、The algorithm terminates after at most flips.

2、Since finding a locally optimal solution is presumably easier than finding an optimal solution, we can claim that for any local search algorithm, one step of searching in neighborhoods can always be done in polynomial time.

Lecture 13. Randomized Algorithms

13.1 Introduction (04:31)随堂测验

1、If we repeatedly perform independent trials of an experiment, each of which succeeds with probability p>0, then the expected number of trials we need to perform until the first success is:
A、p/(1?p)
B、1/(1?p)
C、1/p
D、None of the above

2、Randomized algorithms are for solving the problems with randomly generated inputs.

13.2 Hiring Problems (21:27)随堂测验

1、Consider the online hiring problem, in which we have total k candidates. First of all, we interview n candidates but reject them all. Then we hire the first candidate who is better than all of the previous candidates you have interviewed. It is true that the probability of the m-th candidate is the best is n/[k(m?1)], where m>n.

13.3 Randomized QuickSort (07:54)随堂测验

1、Given a 3-SAT formula with k clauses, in which each clause has three variables, the MAX-3SAT problem is to find a truth assignment that satisfies as many clauses as possible. A simple randomized algorithm is to flip a coin, and to set each variable true with probability 1/2, independently for each variable. Which of the following statements is FALSE?
A、The expected number of clauses satisfied by this random assignment is 7k/8.
B、For every instance of 3-SAT, there is a truth assignment that satisfies at least a 7/8 fraction of all clauses.
C、If we repeatedly generate random truth assignments until one of them satisfies ≥7k/8 clauses, then this algorithm is a 8/7-approximation algorithm.
D、The probability that a random assignment satisfies at least 7k/8 clauses is at most 1/(8k).

2、A randomized Quicksort algorithm has an O(NlogN) expected running time, only if all the input permutations are equally likely.

3、The worst-case running time is equal to the expected running time within constant factors for any randomized algorithm.

Lecture 14. Parallel Algorithms

14.1 Introduction (18:21)随堂测验

1、EREW does not allow simultaneous access by more than one processor to the same memory location for read or write purposes.

2、CRCW allows concurrent access for both reads and writes.

3、CREW allows concurrent access for reads but not for writes.

4、In Work-Depth presentation, each time unit consists of a sequence of instructions to be performed concurrently; the sequence of instructions may include any number.

14.4 Maximum Finding (15:08)随堂测验

1、If we translate a serial algorithm into a reasonably efficient parallel algorithm, the work load and the worst-case running time are usually reduced.

Lecture 15. External Sorting

15.1 Introduction (09:00)随堂测验

1、Suppose we only have 2 tapes, Ta and Tb, to do external sorting. Suppose that the data which has N records is initially on Ta. Suppose further that the internal memory can hold (and sort) M records at a time. A simple algorithm works as the following: Step 1: read M records at a time from Ta, sort the records internally, and then write the sorted records to Tb. Step 2: read M records at a time from Ta, sort the records internally, and merge them with sorted records from Tb, and write them (2M records) to Ta. Step 3: read M records from Ta, sort them internally, and merge them with sorted 2M records from Ta, and write them (3M records) to Tb. Repeat steps 2 and 3 until all the records are sorted. This algorithm will require __ passes.
A、
B、
C、
D、

15.2 Pass Reduction (12:17)随堂测验

1、For a k-way merge in external sorting, the primary reason for k not assuming a large value is that:
A、the I\O time would increase
B、during merging, the number of comparisons would increase
C、k is bounded above by the number of runs
D、k has to be a finite integer

2、Polyphase merge is a method for speeding up k-way merge in external sorting.

15.3 Buffer Handling (05:34)随堂测验

1、The bottleneck of external sorting is to merge the records from input buffers to the output buffers.

2、For the purpose of parallel operations, we need 2k input buffers and 2 output buffers for a k-way merge.

15.4 Run Generation and Merge (08:48)随堂测验

1、Suppose that the replacement selection is applied to generate longer runs with a priority queue of size 5. Given the sequence of numbers { 17, 2, 6, 57, 51, 86, 5, 94, 43, 54, 39, 87, 29}, the longest run contains ____ numbers.
A、5
B、6
C、7
D、8

2、Replacement selection is a method for generating longer runs in external sorting.

学习通Advanced Data Structures and Algorithm Analysis

Advanced Data Structures and Algorithm Analysis是一门深入学习数据结构和算法的课程。本课程主要讲授复杂数据结构和算法的实现和分析，对于提高程序运行效率和解决实际问题有很大的帮助。

1.大纲

• 数据结构的高级应用
• 算法分析
• 高级算法设计
• 高级数据结构设计
• 高级问题求解

2.课程内容

2.1 数据结构的高级应用

本部分主要讲授高级数据结构的实现和使用，包括：

• 堆和优先队列
• 平衡树
• 哈希表
• 图和图算法

2.1.1 堆和优先队列

堆和优先队列是一种有序的数据结构，能够快速定位最大或最小值，常用于排序、任务调度等场景。堆可以分为最大堆和最小堆，优先队列可以根据实际需求选择是返回最大值还是最小值。

堆和优先队列的实现可以使用数组或树结构。在数组实现中，可以使用完全二叉树的性质，将堆或优先队列存储在数组中。在树结构实现中，可以使用二叉堆、斐波那契堆等高级数据结构。

2.1.2 平衡树

平衡树是一种保证树的高度比传统二叉树更小的树结构。平衡树常用于数据库、文件系统等需要快速插入、删除、查找数据的场景，可以保证数据操作的时间复杂度为O(log n)。

常见的平衡树有：

• AVL树
• 红黑树
• Splay树

2.1.3 哈希表

哈希表是一种利用哈希函数将数据映射到桶中的数据结构。哈希表常用于实现字典、缓存等场景，可以快速查找、插入、删除数据。

哈希表的实现可以使用开放寻址法或链表法。开放寻址法将冲突的数据存储在其他空桶中，链表法将冲突的数据存储在同一个桶的链表中。

2.1.4 图和图算法

图是由节点和边构成的一种数据结构，常用于模拟现实中的网络、社交、交通等场景。图算法常用于最短路径、最小生成树、拓扑排序等场景。

图的实现可以使用邻接矩阵或邻接表。邻接矩阵使用二维数组表示节点之间的连通关系，邻接表使用链表表示每个节点的邻居。

2.2 算法分析

本部分主要讲授时间复杂度分析和空间复杂度分析，能够帮助学生对算法进行正确的评估和选择。

2.2.1 时间复杂度分析

时间复杂度是衡量算法执行时间的指标，可以用大O符号表示。常见的时间复杂度有：

• O(1)：常数复杂度
• O(log n)：对数复杂度
• O(n)：线性复杂度
• O(n log n)：线性对数复杂度
• O(n^2)：平方复杂度
• O(2^n)：指数复杂度

2.2.2 空间复杂度分析

空间复杂度是衡量算法内存占用的指标，可以用大O符号表示。常见的空间复杂度有：

• O(1)：常数空间
• O(n)：线性空间
• O(n^2)：平方空间

2.3 高级算法设计

本部分主要讲授高级算法的设计思路和实现。

2.3.1 分治算法

分治算法是一种递归的算法思想，将问题分解成子问题，逐一解决。常用于求解最近点对、矩阵乘法等问题。

2.3.2 动态规划算法

动态规划算法是一种将问题分解成子问题进行求解的算法思想，并将子问题的解保存下来，避免重复计算。常用于求解最长公共子序列、背包问题、最短路径等问题。

2.3.3 贪心算法

贪心算法是一种在每一步都选择当前最优解的算法思想，但不一定能够得到全局最优解。常用于求解霍夫曼编码、最小生成树等问题。

2.3.4 回溯算法

回溯算法是一种试错的算法思想，尝试所有可能的解，并找出最优解。常用于求解N皇后、数独等问题。

2.4 高级数据结构设计

本部分主要讲授高级数据结构的设计思路和实现。

2.4.1 线段树

线段树是一种用于求解区间最值的数据结构。线段树将区间划分成若干个子区间，每个子区间对应线段树中的一个节点。线段树可以支持区间查询、区间修改等操作，时间复杂度为O(log n)。

2.4.2 字典树

字典树是一种用于快速查询和匹配字符串的数据结构。字典树将字符串的每个字符存储在树的节点中，可以支持前缀查询、全文检索等操作。

2.4.3 双端队列

双端队列是一种能够在队列两端进行插入、删除操作的数据结构。双端队列可以用于实现滑动窗口等算法。

2.5 高级问题求解

本部分主要讲授如何利用高级算法和数据结构求解实际问题。

2.5.1 最短路径问题

最短路径问题常用于求解两点之间的最短距离。常用的算法有Dijkstra算法、Bellman-Ford算法、Floyd算法等。

2.5.2 最小生成树问题

最小生成树问题常用于求解图的最小连通子图。常用的算法有Prim算法、Kruskal算法等。

2.5.3 图匹配问题

图匹配问题常用于求解两个图之间的相似度。常用的算法有最大流算法、Hopcroft-Karp算法等。

3.总结

Advanced Data Structures and Algorithm Analysis是一门重要的课程，对于提高编程能力和解决实际问题都有很大的帮助。通过学习本课程，学生可以掌握高级数据结构和算法的实现和分析，为之后的学习和工作打下坚实的基础。