[python]dijkstra 算法的 加权的最短路径 案例
这个还有很多概念有点明确
from collections import defaultdict # defaultdict, 找不到的key的value 就设定为 0，https://blog.csdn.net/edogawachia/article/details/104826952
from heapq import * # heapq 优先队列
def dijkstra_raw(edges, from_node, to_node):
g = defaultdict(list)
for l, r, c in edges:
g[l].append((c, r))
q, seen = [(0, from_node, ())], set()
while q:
(cost, v1, path) = heappop(q)
if v1 not in seen:
seen.add(v1)
path = (v1, path)
if v1 == to_node:
return cost, path
for c, v2 in g.get(v1, ()):
if v2 not in seen:
heappush(q, (cost + c, v2, path))
return float("inf"), []
def dijkstra(edges, from_node, to_node):
len_shortest_path = -1
ret_path = []
length, path_queue = dijkstra_raw(edges, from_node, to_node)
if len(path_queue) > 0:
len_shortest_path = length ## 1. 先求长度
## 2. 分解path_queue，以获得最短路径中的传递节点
left = path_queue[0]
ret_path.append(left) ## 2.1 首先记录目标节点;
right = path_queue[1]
while len(right) > 0:
left = right[0]
ret_path.append(left) ## 2.2 记录其他节点，直到源节点。
right = right[1]
ret_path.reverse() ## 3. 最后反转列表，使其成为正常序列。
return len_shortest_path, ret_path
### ====================给一个所有节点的列表
list_nodes_id = [0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, 20];
### ==================== 给一个拓扑的常数矩阵.
M = 99999 # 这代表一段很长的距离。这意味着没有联系。
### M_topo是表示拓扑的二维邻接矩阵
M_topo = [
[M, 2.9, 2.5, M, 2, M, 1.8, 3, 1.8, M, M, M, M, M, M, M, M, M, M, M, M], # 0
[2.9, M, 1.3, M, M, 1.7, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M], # 1
[1, 1, M, 1, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M], # 2
[M, M, 1, M, 1, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M], # 3
[1, M, M, 1, M, M, M, M, M, 1, 1, 1, M, M, M, M, M, M, M, M, M], # 4
[M, 1, M, M, M, M, 1, M, M, M, M, M, M, M, M, M, M, M, M, M, M], # 5
[1, M, M, M, M, 1, M, 1, M, M, M, M, M, M, M, M, M, M, M, M, M], # 6
[1, M, M, M, M, M, 1, M, 1, M, M, M, M, M, M, M, M, M, M, M, M], # 7
[1, M, M, M, M, M, M, 1, M, 1, M, M, 1, M, M, M, M, M, M, M, M], # 8
[M, M, M, M, 1, M, M, M, 1, M, M, 1, M, M, M, M, M, M, M, M, M], # 9
[M, M, M, M, 1, M, M, M, M, M, M, 1, M, 1, M, M, M, M, M, M, M], # 10
[M, M, M, M, 1, M, M, M, M, 1, 1, M, M, 1, 1, M, M, M, M, M, M], # 11
[M, M, M, M, M, M, M, M, 1, M, M, M, M, M, 1, M, M, M, M, M, M], # 12
[M, M, M, M, M, M, M, M, M, M, 1, 1, M, M, 1, M, M, 1, 1, M, M], # 13
[M, M, M, M, M, M, M, M, M, M, M, 1, 1, 1, M, 1, 1, M, M, M, M], # 14
[M, M, M, M, M, M, M, M, M, M, M, M, M, M, 1, M, 1, M, 1, 1, M], # 15
[M, M, M, M, M, M, M, M, M, M, M, M, M, M, 1, 1, M, M, M, M, 1], # 16
[M, M, M, M, M, M, M, M, M, M, M, M, M, 1, M, M, M, M, 1, M, M], # 17
[M, M, M, M, M, M, M, M, M, M, M, M, M, 1, M, 1, M, 1, M, 1, M], # 18
[M, M, M, M, M, M, M, M, M, M, M, M, M, M, M, 1, M, M, 1, M, 1], # 19
[M, M, M, M, M, M, M, M, M, M, M, M, M, M, M, M, 1, M, M, 1, M], # 20
]
### --- 读取拓扑，并生成给定拓扑中的所有边。
edges = []
for i in range(len(M_topo)):
for j in range(len(M_topo[0])):
if i != j and M_topo[i][j] != M:
edges.append((i, j, M_topo[i][j])) ### (i,j) 是一个链接;M_topo[i][j]这里是1，链接的长度(i,j).
'''
start = int(input("请输入起始点："))
end = int(input("请输入终点："))
print("=== Dijkstra算法 ===")
print("找从 %s 到 %s的最短路径:" % (start, end))
length, Shortest_path = dijkstra(edges, start, end)
print('长度 = ', length)
print('最短路径是 ', Shortest_path)
'''
# what is edge
