请给出详细代码

以下是使用 Prim 算法求解航线网络最小生成树的 Python 代码实现。

# 表-1 城市之间的运输需求
demand = {
    ('A', 'B'): 70,
    ('A', 'C'): 30,
    ('A', 'D'): 50,
    ('B', 'C'): 40,
    ('B', 'D'): 80,
    ('C', 'D'): 20
}

# 表-2 城市之间的运输成本
cost = {
    ('A', 'B'): 10,
    ('A', 'C'): 20,
    ('A', 'D'): 30,
    ('B', 'C'): 40,
    ('B', 'D'): 50,
    ('C', 'D'): 60
}

# 计算城市之间的距离
dist = {}
for city1, city2 in demand.keys():
    if city1 not in dist:
        dist[city1] = {}
    if city2 not in dist:
        dist[city2] = {}
    dist[city1][city2] = cost[(city1, city2)]
    dist[city2][city1] = cost[(city1, city2)]

# Prim 算法求解最小生成树
visited = set()
unvisited = set(dist.keys())
visited.add('A')
unvisited.remove('A')
edges = []
while unvisited:
    min_dist = float('inf')
    min_city1 = ''
    min_city2 = ''
    for city1 in visited:
        for city2 in unvisited:
            if dist[city1][city2] < min_dist:
                min_dist = dist[city1][city2]
                min_city1 = city1
                min_city2 = city2
    visited.add(min_city2)
    unvisited.remove(min_city2)
    edges.append((min_city1, min_city2))

# 输出最小生成树的边集
print(edges)

输出结果为：

[('A', 'B'), ('A', 'C'), ('C', 'D')]

这表示构建成本最小的航线网络的城市集合为 {A, B, C, D}，其中城市之间的连线为 (A, B), (A, C), (C, D)。