Dijkstra算法的Python实现

最新推荐文章于 2024-04-02 03:24:24 发布

Wayne_Mai

最新推荐文章于 2024-04-02 03:24:24 发布

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本文介绍了Dijkstra算法，并提供了详细的Python代码实现，通过实例展示了算法的运行过程和输出结果。

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太白发句，谓之开门见山。
代码一一如下。

# This is a program to illustrate the detail of dijkstra algorithm.
# 这里为了引入无穷大这个概念
import sys

size = 6
adjMatrix = [[0, 50, sys.maxsize, 40, 25, 10], [50, 0, 15, 20, sys.maxsize, 25], [sys.maxsize, 15, 0, 10, 20, sys.maxsize], [40, 20, 10, 0, 10, 25],
             [25, sys.maxsize, 20, 10, 0, 55], [10, 25, sys.maxsize, 25, 55, 0]]

"""
这一部分是用作手动输入的代码
for i in range(size):
    tempRow = []
    print("Please input Row " + str(i) + " of the matrix.")
    for j in range(size):
        temp = int(input("Input the weight.(Use sys.maxsize to represent infinity)"))
        if temp == sys.maxsize:
            temp = sys.maxsize
        tempRow.append(temp)
    adjMatrix.append(tempRow)
"""
visited = [0 for i in range(size)]    #已经访问过的节点列表，0与1表示了有与无的概念
visited[0] = 1
cityDis = adjMatrix[0]    #表示与起始点的距离
routine = [[0] for i in range(size)]   #存储了到各个点的最短路径
for i in range(1, size):
    print(cityDis)
    minDisPos = int()
    minDis = sys.maxsize
    for j in range(1, size):
        if visited[j] == 1:
            continue
        if cityDis[j] < minDis:
            minDisPos = j
            minDis = cityDis[j]
    visited[minDisPos] = 1
    for j in range(1, size):
        if visited[j] == 0:
            if cityDis[j] > cityDis[minDisPos] + adjMatrix[minDisPos][j]:  #如果新加入的点使得距离更短则刷新
                cityDis[j] = cityDis[minDisPos] + adjMatrix[minDisPos][j]
                routine[j].append(minDisPos)

print("Final result " + str(cityDis) + "\n" + str(routine))

输出一一如下。

[0, 50, 9223372036854775807, 40, 25, 10]
[0, 35, 9223372036854775807, 35, 25, 10]
[0, 35, 45, 35, 25, 10]
[0, 35, 45, 35, 25, 10]
[0, 35, 45, 35, 25, 10]
Final result [0, 35, 45, 35, 25, 10]
[[0], [0, 5], [0, 4], [0, 5], [0], [0]]

Process finished with exit code 0