基于python的数学建模---图论模型（Dijkstra）

from collections import defaultdict
from heapq import *  # 堆--先进后出

inf = 99999  # 不连通值
mtx_graph = [[0, 1, inf, 3, inf, inf, inf, inf, inf],
             [1, 0, 5, inf, 2, inf, inf, inf, inf],
             [inf, inf, 0, 1, inf, 6, inf, inf, inf],
             [inf, inf, inf, 0, inf, 7, inf, 9, inf],
             [inf, 2, 3, inf, 0, 4, 2, inf, 8],
             [inf, inf, 6, 7, inf, 0, inf, 2, inf],
             [inf, inf, inf, inf, inf, 1, 0, inf, 3],
             [inf, inf, inf, inf, inf, inf, 1, 0, 2],
             [inf, inf, inf, inf, 8, inf, inf, 2, 0]]
m_n = len(mtx_graph)  # 带权连接矩阵的阶数
edges = []  # 保存连通的两个点之间的距离(点A、点B、距离)
for i in range(m_n):
    for j in range(m_n):
        # 去除inf和0
        if i != j and mtx_graph[i][j] != inf:
            edges.append((i, j, mtx_graph[i][j]))

# edges 就变成了带有数字的元素
def dijkstra(edges, from_node, to_node):
    global v1, path, cost
    go_path = []
    # range 从0开始 记得后面+1
    to_node = to_node - 1
    g = defaultdict(list)
    for l, r, c in edges:  # 对应dijkstra方法中的元素
        # l:点A r:点B c:距离
        # 字典： [点A -> (距离，点B)]
        g[l].append((c, r))
    q, seen = [(0, from_node - 1, ())], set()  # set集合-->去重
    while q:
        (cost, v1, path) = heappop(q)  # 堆弹出当前路径最小成本
        # seen指的是已经到达的地方
        if v1 not in seen:
            seen.add(v1)
            path = (v1, path)
            if v1 == to_node:
                break
            for c, v2 in g.get(v1, ()):
                if v2 not in seen:
                    heappush(q, (cost + c, v2, path))

    if v1 != to_node:  # 无法到达
        return float('inf'), []
    # 下面是逆序进行
    if len(path) > 0:
        left = path[0]
        go_path.append(left)
        right = path[1]
        while len(right) > 0:
            left = right[0]
            go_path.append(left)
            right = right[1]

        go_path.reverse()  # 逆序变换
        for i in range(len(go_path)):  # 标号加1
            go_path[i] = go_path[i] + 1
    return cost, go_path

leght, path = dijkstra(edges, 1, 9)
print('最短距离为：' + str(leght))
print('前进路径为：' + str(path)

最短距离为：8
前进路径为：[1, 2, 5, 7, 9]