from queue import Queue
# 邻接矩阵存储
class Graph:
def __init__(self, mat, unconn=0):
# mat = [[*, *, *, *], [*, *, *, *], [*, *, *, *], [*, *, *, *]]
vnum = len(mat)
for x in mat:
if len(x) != vnum:
# 检查是否为方阵
raise ValueError("mat不是邻接矩阵的格式")
self._mat = [mat[i][:] for i in range(vnum)] # 矩阵的拷贝
self._unconn = unconn
self._vnum = vnum
def vertex_num(self):
# 计算顶点个数
return self._vnum
def add_edge(self, vi, vj, val=1):
# 加边
self._mat[vi][vj] = val
def get_edge(self, vi, vj):
# 获取某两个点之间的边
if self._mat[vi][vj] == 0:
return 100 # 如果不存在边, 存一个很大的数, 因为我底下求最短路的边长最长也就8
return self._mat[vi][vj]
# 一个点到其他点之间的边
@staticmethod
def _out_edges(row, unconn):
edges = []
for i in range(len(row)):
if row[i] != unconn:
edges.append((i, row[i]))
return edges # 返回当前一个点到其他点的边
def out_edges(self, vi):
return self._out_edges(self._mat[vi], self._unconn)
# 邻接表存储
class GraphAL(Graph):
# 直接将邻接矩阵转换为邻接表
def __init__(self, mat=[], unconn=0):
vnum = len(mat)
for x in mat:
if len(x) != vnum: # 检查矩阵是否为方阵
raise ValueError("输错了")
self._mat = [Graph._out_edges(mat[i], unconn)
for i in range(vnum)] # 查每个定点到其他点的边,构成邻接表
self._vnum = vnum
self._unconn = unconn
# [
# [(2, 1), (3, 1)] # 代表0能直接到2长度1, 到3长度1
# [(0, 1)] # 代表1能直接到0长度1
# [(1, 1), (3, 1)] # 代表2能直接到1长度1, 到3长度1
# [(1, 1)] # 代表3能直接到1长度1
# ]
def add_vertex(self):
# 增加节点
self._mat.append([])
self._vnum += 1
return self._vnum - 1
def add_edge(self, vi, vj, val=1):
row = self._mat[vi]
i = 0
while i < len(row):
if row[i][0] == vj: # 修改mat[vi][vj]值
self._mat[vi][i] = (vj, val)
return
if row[i][0] > vj: # 原来没有到vj的边, 退出循环加边
break
i += 1
self._mat[vi].insert(i, (vj, val))
def get_edge(self, vi, vj):
for i, val in self._mat[vi]:
if i == vj:
return val
return self._unconn
def out_edges(self, vi):
return self._mat[vi]
# 深度优先遍历 这个是数据结构与算法给出的实例代码 我没怎么懂,自己写了一个DFS(下一个函数)
# def DFS1(graph, v0):
# vnum = graph.vertex_num() # 获取节点数
# visited = [0]*vnum # visited=0 代表都没有被访问
# visited[v0] = 1 # 代表初始节点已经被访问
# dfs_list = [v0] # 遍历的序列放这里面
# stack = []
# stack.append((0, graph.out_edges(v0))) # 获取与v0点连接的
# while stack:
# i, edges = stack.pop()
# if i < len(edges):
# v, e = edges[i]
# stack.append((i+1, edges))
# if visited[v] == 0:
# dfs_list.append(v)
# visited[v] = 1
# stack.append((0, graph.out_edges(v)))
# return dfs_list
def DFS(graph, v0):
# 自己实现的深度优先遍历
vnum = graph.vertex_num() # 获取当前的节点数
visited = [0]*vnum # visited=0 代表都没有被访问
visited[v0] = 1
dfs_list = [v0]
stack = []
member1 = graph.out_edges(v0)
member1.reverse()
for i in member1:
stack.append(i)
while stack:
v, e = stack.pop()
if visited[v] == 0:
dfs_list.append(v)
visited[v] = 1
member = graph.out_edges(v)
member.reverse()
for e in member:
stack.append(e)
return dfs_list
def BFS(graph, v0):
# 广度优先遍历
vnum = graph.vertex_num() # 获取当前的节点数
visited = [0] * vnum # visited=0 代表都没有被访问
bfs_list = []
que = Queue()
que.put(v0)
visited[v0[0]] = 1 # 当前出队的那个元素标记为访问
while que.empty() == False:
temp, v = que.get(block=False) # 非阻塞获取
bfs_list.append(temp) # 将当前元素放在咱们等会输出的那个队列
member = graph.out_edges(temp) # 获取与当前点直接相连的点
member.reverse() # 逆转一下,这一步自己手动分析一下
for (i, v) in member:
if visited[i] == 0:
que.put((i, v))
visited[i] = 1
return bfs_list
# dijkstra求最短路的算法
path = [0, 0, 0, 0, 0, 0, 0] # 存每个节点当前路径的前驱
dist = [0, 0, 0, 0, 0, 0, 0] # 存走到当前每个点的最短路径
visited = [0, 0, 0, 0, 0, 0, 0] # 标志当前点是否被访问
def dijkstra(graph1, v0):
vnum = graph1.vertex_num()
# print(vnum)
for i in range(vnum):
dist[i] = graph1.get_edge(v0, i)
visited[i] = 0 # 代表未被访问
if graph1.get_edge(v0, i) < inf:
path[i] = v0
else:
path[i] = -1
visited[v0] = 1
path[v0] = -1
for i in range(vnum-1):
min = inf
# 这个循环每次从剩余顶点中选出一个顶点,通过这个顶点的路径是通往所有剩余顶点的路径中长度最短的
for j in range(vnum):
if visited[j] == 0 and dist[j] < min:
u = j
min = dist[j]
visited[u] = 1 # 将选取的顶点加入到已确定的集合里
for j in range(vnum):
if visited[j] == 0 and dist[u] + graph1.get_edge(u, j) < dist[j]:
dist[j] = dist[u] + graph1.get_edge(u, j)
path[j] = u
# dist[]数组存放着v点到其余顶点的最短路径, path[]数组存放v点到其余顶点的最短路径
def print_path(path, a):
# 想看一下0 到 a 的最短路径 打印路径
short_path = []
while path[a] != -1:
short_path.append(a)
a = path[a]
short_path.append(a)
short_path.reverse()
return short_path
# Floyd算法
def floyd(graph1, path):
vnum = graph1.vertex_num()
A = [[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]] # 等会存任意两点之间的最短的路径
for i in range(vnum):
for j in range(vnum):
A[i][j] = graph1.get_edge(i, j)
path[i][j] = -1 # 双重for循环是为了初始化
for k in range(vnum):
for i in range(vnum):
for j in range(vnum):
if A[i][j] > A[i][k] + A[k][j]:
A[i][j] = A[i][k] + A[k][j]
path[i][j] = k
# A[0][4] 代表的是0到4的最短路径 A[i][j]代表i到j的最短路径
return A, path
short_floyd_path = []
def print_floyd_path(u, v, path):
if path[u][v] == -1:
return
short_floyd_path.append(path[u][v])
# 这个路径打印出来不包括u, v, 我们最后输出的时候把u v记得***来
mid = path[u][v]
print_floyd_path(mid, v, path) # 处理mid的后半段
print_floyd_path(u, mid, path) # 处理mid的前半段
if __name__ == '__main__':
# 输入的邻接矩阵 无向图
a = [[0, 1, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 0],
[0, 0, 0, 1, 0, 1],
[1, 1, 0, 0, 1, 0],
[0, 0, 0, 0, 0, 1],
[0, 0, 0, 0, 0, 0]]
g = [[0, 0, 6, 3, 0, 0, 0],
[11, 0, 4, 0, 0, 7, 0],
[0, 3, 0, 0, 5, 0, 0],
[0, 0, 0, 0, 5, 0, 0],
[0, 0, 0, 0, 0, 0, 9],
[0, 0, 0, 0, 0, 0, 10],
[0, 0, 0, 0, 0, 0, 0]]
# c邻接矩阵是等会求最短路径的图
inf = float('inf') # inf表示最大，代表没有边
c = [[0, 4, 6, 6, inf, inf, inf],
[inf, 0, 1, inf, 7, inf, inf],
[inf, inf, 0, inf, 6, 4, inf],
[inf, inf, 2, 0, inf, 5, inf],
[inf, inf, inf, inf, 0, inf, 6],
[inf, inf, inf, inf, 1, 0, 8],
[inf, inf, inf, inf, inf, inf, 0]]
# 建立大家执行代码之前，先根据邻接矩阵画出对应的图.
graph = GraphAL(g)
# dfs_list = DFS1(graph, 2)
# print("深度优先遍历:", dfs_list)
dfs_list1 = DFS(graph, 2)
print("深度优先遍历:", dfs_list1)
bfs_list1 = BFS(graph, (2, 0))
print("广度优先遍历:", bfs_list1)
graph1 = Graph(c)
# dijkstra算法求最短路算法
dijkstra(graph1, 0) # 调用dijkstra算法 算出0到任意点的最短路径
short_path = print_path(path, 4) # 从0点开始,到6终结
print("Dijlstra算法求得最短路径:", short_path)
# floyd算法求最短路
path = [[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0],
[0, 0, 0, 0, 0, 0, 0]]
u = 0; v = 4
A, path = floyd(graph1, path)
print_floyd_path(u, v, path)
short_floyd_path.reverse()
short_floyd_path.append(v)
short_floyd_path.insert(u, 0)
print("Floyd算法求得最短路径:", short_floyd_path)
# 输出结果:
# 深度优先遍历: [2, 1, 0, 3, 4, 6, 5]
# 广度优先遍历: [2, 1, 4, 0, 5, 6, 3]
# Dijlstra算法求得最短路径: [0, 1, 2, 5, 4]
# Floyd算法求得最短路径: [0, 1, 2, 5, 4]
京公网安备 11010502036488号