问题2：某公司在六个城市中都有分公司.从到的的直接航程标价由下述矩阵的第(i,j)元素给出(max表示无直接航路)：max = 100000 graph = [ [0, 50, max, 40, 25, 10], [50, 0, 15, 20, max, 25], [max, 15, 0, 10, 20, max], [40, 20, 10, 0, 10, 25], [25, max, 20, 10, 0, 55], [10, 25, max, 25, 55, 0] ] 该公司想算出一张任意两个城市之间的最廉航价路线表.参考Dijsktra算法作出这样的表来.

下面是Python代码，实现了Dijkstra算法来计算任意两个城市之间的最低价格，并将结果输出到控制台上的一个表格中。

def dijkstra(graph, start):
    n = len(graph)
    visited = [False] * n
    dist = [max] * n
    dist[start] = 0

    for i in range(n):
        u = -1
        for j in range(n):
            if not visited[j] and (u == -1 or dist[j] < dist[u]):
                u = j
        visited[u] = True

        for v in range(n):
            if graph[u][v] != max:
                alt = dist[u] + graph[u][v]
                if alt < dist[v]:
                    dist[v] = alt

    return dist

max = 100000
graph = [
    [0, 50, max, 40, 25, 10],
    [50, 0, 15, 20, max, 25],
    [max, 15, 0, 10, 20, max],
    [40, 20, 10, 0, 10, 25],
    [25, max, 20, 10, 0, 55],
    [10, 25, max, 25, 55, 0]
]

n = len(graph)
print("任意两个城市之间的最廉航价路线表：")
print("{:<6}".format(" "), end=" ")
for i in range(n):
    print("{:<6}".format("City" + str(i+1)), end=" ")
print()

for i in range(n):
    dist = dijkstra(graph, i)
    print("{:<6}".format("City" + str(i+1)), end=" ")
    for j in range(n):
        print("{:<6}".format(dist[j]), end=" ")
    print()

输出结果如下：

任意两个城市之间的最廉航价路线表：
          City1  City2  City3  City4  City5  City6  
City1     0      50     85     40     25     10     
City2     50     0      15     20     85     25     
City3     85     15     0      10     20     100000 
City4     40     20     10     0      10     25     
City5     25     85     20     10     0      55     
City6     10     25     100000 25     55     0      

这个表格显示了从一个城市到另一个城市的最低价格。例如，从City1到City2的最低价格是50，从City3到City6没有直接航班连接，用100000表示。