python数据结构实现（六）：图及相关LeetCode题

1. 图

1.1 python实现图的邻接矩阵表示方法

1.1.1 python实现无向无权图

1. 通过传入一个二维数组确立邻接矩阵，无向无权图的邻接矩阵元素为0和1；
2. adjMatrix[ i ][ j ] = 0 说明顶点i和j不连接，若为1则说明连接；
3. 无向图的邻接矩阵总是为对称矩阵

import numpy as np  

class MatrixGraph:
    '''
    无向图的邻接矩阵表示(为对称矩阵)
    用矩阵存储顶点间的关系：顶点a与顶点b相连，则adjMatrix[vex2num[a],vex2num[b]]
    的值为1，否则为0
    vex2num为字典，将顶点映射为矩阵的编号
    '''
    def __init__(self,vertex=[]):
        '''
        根据传入的顶点信息表建造邻接矩阵和顶点字典
        :param vertex：无向图的所有顶点组成的列表
        '''
        self.vexNum = len(vertex)
        self.adjMatrix = np.zeros((self.vexNum,self.vexNum))
        self.vex2num = {}
        for index, vertex in enumerate(vertex):
            self.vex2num[vertex] = index
        
    def createGraph(self, maxtrix):
        '''
        传入一个矩阵确立顶点间的关系
        '''
        if maxtrix.shape == self.adjMatrix.shape:
            self.adjMatrix = maxtrix
        else:
            raise Exception('wrong matrix shape')
    
    def isLinked(self, vex1, vex2):
    	# 判断图中两个顶点是否连接
        return self.adjMatrix[self.vex2num[vex1], self.vex2num[vex2]] == 1
    
    def linkVex(self, vex1, vex2):
        # 连接无向图的两个顶点
        if vex1 != vex2:
            self.adjMatrix[self.vex2num[vex1], self.vex2num[vex2]] = 1
            self.adjMatrix[self.vex2num[vex2], self.vex2num[vex1]] = 1

    def splitVex(self, vex1, vex2):
        # 断开无向图的两个顶点
        self.adjMatrix[self.vex2num[vex1], self.vex2num[vex2]] = 0
        self.adjMatrix[self.vex2num[vex2], self.vex2num[vex1]] = 0

生成无向无权图：

vexs = ['a','b','c','d','e']
Mgraph = MatrixGraph(vexs)
matrix = np.array([[0,1,1,0,0],[1,0,1,1,0],\
					[1,1,0,1,0],[0,1,1,0,1],[0,0,0,1,0]])
Mgraph.createGraph(matrix)      
print(Mgraph.adjMatrix)
Mgraph.isLinked('a','d')           
=====================================
[[0 1 1 0 0]
 [1 0 1 1 0]
 [1 1 0 1 0]
 [0 1 1 0 1]
 [0 0 0 1 0]]
 False

1.1.2 python实现无向有权图

1. 无向有权图与无向无权图的区别仅在与邻接矩阵中的元素
2. adjMatrix[i][j] 为路权大小则说明顶点i和j相连；若为inf则说明两个顶点间没有直接路径，即不相连

import numpy as np

inf = float('inf')
class MatrixGraph:
    '''
    无向有权图的邻接矩阵表示(为对称矩阵)
    用矩阵存储顶点间的关系：顶点a与顶点b相连，则adjMatrix[vex2num[a],vex2num[b]]的值为路权，否则为inf
    vex2num为字典，将顶点映射为矩阵的编号
    '''
    def __init__(self,vertex=[]):
        '''
        根据传入的顶点信息表建造邻接矩阵和顶点字典
        :param vertex：无向图的所有顶点组成的列表
        '''
        self.vex2num = {}
        self.vexNum = len(vertex)
        self.adjMatrix = np.array([[inf]*self.vexNum]*self.vexNum)
        
        for index, vertex in enumerate(vertex):
            self.vex2num[vertex] = index
        
    def createGraph(self, maxtrix):
        '''
        传入一个带权矩阵确立顶点间的关系
        '''
        if maxtrix.shape == self.adjMatrix.shape:
            self.adjMatrix = maxtrix
        else:
            raise Exception('wrong matrix shape')
    
    def isLinked(self, vex1, vex2):
        return not self.adjMatrix[self.vex2num[vex1], self.vex2num[vex2]] == inf
    
    def linkVex(self, vex1, vex2, weight):
        # 连接无向图的两个顶点
        if vex1 != vex2:
            self.adjMatrix[self.vex2num[vex1], self.vex2num[vex2]] = weight
            self.adjMatrix[self.vex2num[vex2], self.vex2num[vex1]] = weight

    def splitVex(self, vex1, vex2):
        # 断开无向图的两个顶点
        self.adjMatrix[self.vex2num[vex1], self.vex2num[vex2]] = inf
        self.adjMatrix[self.vex2num[vex2], self.vex2num[vex1]] = inf

生成无向有权图：

vexs = ['a','b','c','d','e']
Mgraph = MatrixGraph(vexs)
matrix = np.array([[inf,9,12,inf,9],[9,inf,7,10,inf],[12,7,inf,3,inf],\
					[inf,10,3,inf,14],[9,inf,inf,14,inf]])
Mgraph.createGraph(matrix)
print(Mgraph.adjMatrix)
print(Mgraph.isLinked('a','d'))
Mgraph.linkVex('a','d', 11)
print(Mgraph.adjMatrix)
print(Mgraph.isLinked('a','d'))
=================================
[[inf  9. 12. inf  9.]
 [ 9. inf  7. 10. inf]
 [12.  7. inf  3. inf]
 [inf 10.  3. inf 14.]
 [ 9. inf inf 14. inf]]
False
[[inf  9. 12. 11.  9.]
 [ 9. inf  7. 10. inf]
 [12.  7. inf  3. inf]
 [11. 10.  3. inf 14.]
 [ 9. inf inf 14. inf]]
True