【笔记】第6章 图

A. 概述

一、邻接 + 关联

1. 基本术语
   
2. 邻接(adjacency)：v-v，同一条边的两个顶点彼此邻接
3. 关联(incidence)：v-e，顶点与其所属的边彼此关联
4. 度(degree/valency):与同一顶点关联的边数
5. 自环(self-loop)：同一顶点自我邻接
6. 简单图(simple graph)：不含自环及重边；非简单图(non-simple graph)

二、无向图 + 有向图

1. 无向边(undirected edge)：若邻接顶点u和v的次序无所谓，则(u,v)为无向边
2. 无向图(undigraph)：所有边均无方向的图
3. 有向图(digraph)：所有边均为有向边(directed edge)，u,v分别称作边(u,v)的尾(tail)、头(head)
4. 混合图(mixed graph)：无向边、有向边并存的图

三、路径 + 环路

1. 路径π：顶点按一定的序列进行组织0, v₁, …, v_k>
2. 长度|π|：k
3. 简单路径：V₀ ≠ V_k，除非 i = j
4. 环/环路：V₀ = V_k
5. 有向无环图(DAG)
6. 欧拉环路：|π| = |E|，各边恰好出现一次
7. 哈密尔顿环路：|π| = |V|，各顶点恰好出现一次

四、支撑树 + 带权网络 + 最小支撑树*

1. 图G = (V;；E) 的子图T = (V；F)若是树，即为其支撑树(spanning tree)，通常并不唯一
2. 各边e均有对应的权值wt(e)，则为带权网络(weighted network)
3. 同一网络的支撑树中，总权重最小者为最小支撑树(MST)

五、作业

1. The relationship between two vertices connected by a edge is called:两个通过一条边连起来的顶点之间的关系称为：adjacent 邻接
2. The one-touch-drawing question is to find out: 一笔画问题即要找出：Euler path 欧拉路径

B. 邻接矩阵

构思

1. 算法
   
2. 邻接矩阵(adjacency matrix)：记录顶点之间的邻接关系；空间复杂度Θ(n²)，与图中实际的边数无关
3. 关联矩阵(incidence matrix)：记录顶点与边之间的关联关系；空间复杂度Θ(n*e) = O（n³）；空间利用率2e/ne = 2/n
   

二、模板实现

1. 顶点与边
2. 图矩阵
   

三、静态操作

1. 顶点的读写
   
2. 边的读写
   
3. 邻点的枚举
   

四、动态操作

• 点的操作
  
• 边的操作
  

五、性能分析

• 优点
  1. 直观，易于理解和实现
  2. 适用广泛，eg：digraph、network、cyclic、尤其是稠密图(dense graph)
  3. 判断两点之间是否存在联边：O(1)
  4. 获取顶点的(出/入)度数：O(1)
  5. 添加、删除边后更新度数：O(1)
• 缺点
  1. 空间复杂的Θ(n²)，与边数无关，且e << n²
  2. 欧拉定理(Euler’s formula)： v - e +f - c =1，（v顶点数目，e一维元素:边，f二维元素:压缩面片的总数，c连通域总数）
  3. 平面图(planar graph)：可嵌入于平面的图。e ≤ 3n - 6 = O(n) << n²，空间利用率 ≈ 1 / n → 0
  4. 稀疏图(sparse graph)

六、作业

1. A graph with an (undirected) edge between any two vertices is called a complete graph, and a complete graph containing n vertices is represented by K_n. Which of the following figures must not be a plan? 任何两个顶点间都有一条（无向）边的图称为完全图，包含n个顶点的完全图用 K_n表示。下列哪个图一定不是平面图？K₅ 【Obviously the other three are all plans. K_5 is not a plan that can be proved by Euler’s formula and the fact that a face in a plan corresponds to a maximum of 3/2 edges.显然其余三个都是平面图。K_5不是平面图可以用欧拉公式以及平面图中的一个面最多平均对应3/2条边这一事实来证明。】
2. The adjacency matrix of the above digraph is (in order of A, B, C, D) 以上有向图的邻接矩阵为（以A、B、C、D为顺序）
   
3. In the graph implemented with adjacency matrices with n vertices, the vertex v has m neighbors, and the time complexity of traversing all m neighbors is:在包含n个顶点的用邻接矩阵实现的图中，顶点v有m个邻居，遍历所有m个邻居的时间复杂度为：O(n) 【Need to access a row in the adjacency list 需要访问邻接表中的一行】
4. The graph G contains n vertices (n>0), implemented with an adjacency matrix. How many items have been added to the adjacency matrix after adding a new vertex?图G包含n个顶点(n>0)，用邻接矩阵实现。在其中加入一个新的顶点后邻接矩阵增加了多少项？2n+1 【Adjacency matrix adds one row and one column邻接矩阵增加了一行一列】

D. 广度优先搜索BFS(Breadth-First Search)

一、算法

1. 化繁为简：traversal使图变为非线性结构
2. 策略
   
3. 算法实现
   

二、实例-无向图

三、推广

四、性质及应用*

五、作业

1. Traversing graphs in a sense is to translate the graph into: 对图进行遍历某种意义上是将图转化为：Tree树
2. The breadth-first search of a graph is similar to that of a binary tree: 图的广度优先搜索访问各顶点的模式类似于二叉树的：Level-order traversal 层次遍历
3. The BFS is performed on the above undigraph with the vertex s as the starting point. The neighbors of the same vertex are in the order of a~z. When the vertex c is just out of the queue, the vertices in the queue from the head of the queue to the end of the queue are: 以顶点s为起点对以上无向图进行BFS，同一顶点的邻居之间以a-z为顺序，顶点c刚出队时队列中顶点从队头到队尾为：d,e
   
4. For graphs with n vertices and e edges implemented with adjacency list, the time complexity of BFS is:对于用邻接表实现的包含n个顶点e条边的图，BFS的时间复杂度为：O(n + e)

E. 深度优先搜索DFS(Depth-First Search)

一、算法

1. 策略：
   
2. 算法实现：
   

二、实例：无向图

三、实例：有向图

四、性质

1. 从顶点s出发的DFS：
   · 在无向图中将访问与s连通的所有顶点(connectivity)
   · 在有向图中将访问由s可达的所有顶点(reachability)
2. 经DFS确定的树边，不会构成回路
3. DFS树及森林由parent指针描述(只不过所有边去反向)
4. 括号引理
   
5. 边分类
   

五、作业

1. DFS on the above undigraph with A as the starting point, and the neighbors of the same vertex are in order of a~z. The order in which the vertices are accessed is: 以A为起点对以上无向图进行DFS，同一顶点的邻居以a-z为序，各顶点被访问的顺序为：a, b, c, f, g, d, i, h, j, e
   
2. u and v are two vertices in the graph. After performing DFS on the graph, dTime(u) < dTime(v) < fTime(v) < fTime(u), then the relationship between u and v in the DFS forest is: u 和 v 为图中两个顶点，对图进行 DFS 后，dTime(u) < dTime(v) < fTime(v) < fTime(u)，则 u 和 v 在 DFS 森林中的关系是：u is the ancestor of v u 为 v 的祖先
3. Run breadth-first search (BFS) and depth-first search (DFS) respectively on the same undirected graph, the numbers of TREE edges satisfy: 对同一个无向图分别运行广度优先算法和深度优先算法，得到的树边数量：they result in same number of TREE edges两种算法得到的树边一样多 【The number of TREE edges is always equal to the number of vertices minus the number of connected components.TREE 边的数量总是等于顶点数减去连通分量的数量】
4. DFS on a graph, which situation means that the graph contains a loop 对图进行DFS，一下哪种情况意味着该图包含环：There has a BACKWARD edge 有BACKWARD边

遍历算法应用举例

F1. 拓扑排序之零入度算法

一、拓扑排序

1. 有向无环图(Directed Acyclic Graph)
2. 任给有向图G(不一定是DAG)，尝试将所有顶点排成一个线性序列，使其次序须与原图相容(每一个顶点都不会通过边指向前驱顶点)
3. 接口要求：若原图存在回路(即并非DAG)，检查并报告；否则给出一个相容的线性序列

二、算法

从零入度的点开始顺序输出

三、实例

F2. 拓扑排序之零初度算法

一、算法

1. 基于DFS，借住栈s，逆序输出零出度顶点
2. 对图做DFS：每当有顶点被标记为VISITED，则将其压入s；一旦发现有后向边，则报告"NOT_A_DAG"并退出；DFS结束后，顺序弹出s中的各个顶点
3. 实现
   

二、实例

无向图的边数 = 各顶点度数的一半

测验

1. In a simple undigraph with 20 vertices, the maximum number of edges is:在含20个顶点的简单无向图中，边的数量最多为：190
   The degree of the vertex with the smallest degree at this time is:此时度最小的顶点的度为：19
2. A total of 7 people took part in the banquet and a friendly handshake took place among the participants. The number of hands-on handshakes known to each of them is:3, 1, 2, 2, 3, 1, 2。How many handshakes have occurred at the banquet?某宴会一共有7个人参加，与会者之间进行了亲切的握手。已知他们中的每个人进行握手的次数分别为：3, 1, 2, 2, 3, 1, 2。请问宴会上总共发生了多少次握手？7
3. （接上题）In the long history of humanity, everyone may have to shake hands with other people. If someone makes an odd number of handshakes in his life, he is called a Class A person, otherwise he is called a Class B person. The number of people of type A since ancient times is: (assuming that humans can only shake hands with humans)在人类的历史长河中，每个人都可能要与其他人握手。如果某人在他的一生中进行握手的次数为奇数，则称他为A类人，否则称为B类人。试问从古至今A类人的个数是：(假设人类只能和人类握手) Even number偶数 【无向图的边数等于各顶点度数之和的一半】
4. The adjacency matrix of the above digraph is (in order of A, B, C, D) 以上有向图的邻接矩阵为（图中顶点以A、B、C、D为顺序）
   
5. For a simple undigraph with n vertices and e edges, which of the following argument for its adjacency matrix A is wrong: 对于包含n个顶点e条边的简单无向图，以下关于它的邻接矩阵A的说法中错误的是：A has n rows and e columns, where the elements take a value of {0, 1} A有n行e列，其中元素取值于{0, 1} 【n rows and n columns n行n列】
6. G is a simple undigraph, A is an adjacency matrix of G, M is an associative matrix of G, and D is a diagonal matrix of the degree of the i-th element of the vertex on the diagonal. Their relationship is: G是简单无向图，A为G的邻接矩阵，M为G的关联矩阵，D是对角线上第i个元素为顶点i的度的对角矩阵,它们的关系是：A + D = MM^T 【Use the example to verify, or see the analysis of the textbook matching exercises 6-1 用例子验证，或见教材配套习题解析6-1】
7. Using the adjacency matrix to implement a graph with n vertices and e edges: 用邻接矩阵实现含n个顶点e条边的图：Space complexity: 空间复杂度：O(n²)
8. Time complexity of deleting edge(i,j): 删除边(i, j)的时间复杂度：O(1)
9. Time complexity of traversing all the neighbors of vertex v: 遍历顶点v的所有邻居的时间复杂度：O(n)
10. Time complexity of accessing data stored in vertex v: 访问顶点v中存储的数据的时间复杂度：O(1)
11. G is a directed acyclic graph, and (u, v) is an edge in G that points from u to v. The result of DFS on G is: G是有向无环图，(u, v)是G中的一条由u指向v的边。对G进行DFS的结果是：fTime(u) > fTime(v) 【G does not contain a loop, (u, v) cannot be BACKWARD, and access to v must have ended when access to u ends G不含环路，(u, v)不可能是BACKWARD，对u的访问结束时对v的访问必然已经结束】
12. The following is the dTime and fTime of each vertex after performing a DFS on a simple undigraph: 下面是对一个简单无向图进行DFS后得到各顶点的dTime和fTime:The DFS tree is: 得到的DFS树为：
    
13. Starting from s, perform BFS on the above undigraph, order a-z between the neighbors of the same vertex, and find the dTime of the vertex 从s开始，对以上无向图进行BFS，同一顶点的邻居之间以a~z为序，求顶点的dTime，dTime of s =1，
    
    dtime of a = 2，
    dtime of b = 6，
    dtime of e = 5，
    dtime of f = 7
14. Starting from s, perform DFS on the above undigraph, order a~z between the neighbors of the same vertex, and find the dTime and fTime of each vertex
    从s开始，对以上无向图进行DFS，同一顶点的邻居之间以a~z为序，求各顶点的dTime和fTime。dTime of s = 1，fTime of s = 16
    
    dTime of c = 3
    fTime of c = 14
    dTime of g = 7
    fTime of g = 12