python - 一个recursive的深度优先算法问题

Question

我自己写了一些代码，想知道哪里出了问题，更改的代码已经写在里面了
用recursion来计算两点之间最短距离
例如有一个file

a,b,5
a,c,8
b,d,6
c,d,2
d,e,12
d,f,2
e,g,3
f,g,7

通过代码能变成如图的map

读取文件和生成map的代码已经写出来了
剩下的就是如何计算a到g的最短距离，有点懵逼，没啥思路，已写代码如下哈

"""
Graph search to find shortest path between two cities. 

A road map is represented as a directed graph.
Edges represent road segments and are labeled with distances.
Nodes represent cities. 
Wne use depth first search to find the shortest distance
between two cities. During the depth-first search from
start to end, we compute a 'shortest known distance to here from start'
attribute for each node.  After completion of depth first search,
destination city should be labeled with the shortest distance from
the start city.  
"""

import argparse 


def read_distances(map_file):
    """Read a distance table from a named file into a dict
       mapping sources to lists of (destination, distance) pairs. 
    Args:
       map_file: A readable text file in which each line either 
           begins with #  (indicating a comment line) or is of the form
           from location, to location, distance or time, for example
              Minis Tirith,Cair Andros,5
           indicating one can travel from Minis Tirith to Cair Andros in 5.0 hours.
    Returns:
        Dict in which each entry d[from] is a list [(to, cost), (to, cost), ... ]
        where from is a location name, to is a location name, and cost is time
        or distance
        as a floating point number.  If x,y,z appears in the input file, then 
        we should have both d[x] contains (y,z) and d[y] contains (x,z), i.e., 
        we assume that roads are bi-directional. 
    """
    connections = dict() 
    for line in map_file:
        line = line.strip()  
        if line.startswith("#") or line=="" :
            # print("Skipping comment: ", line)
            continue  ## Discard and go on to next
        # print("Processing line: ", line)
        fields = line.split(",")
        place_from = fields[0]
        place_to = fields[1]
        cost = float(fields[2])
        if not (place_from in connections): 
            connections[place_from] = [ ]
        connections[place_from].append( (place_to, cost) )
        if not (place_to in connections): 
            connections[place_to] = [ ]
        connections[place_to].append( (place_from, cost) ) 
    return connections
    
def show_roads(roads):
    """Print roads from dict (for debugging).
       Args: 
         roads:  A dict in which
                roads[Chicago] == [("Atlanta", 30.0), ("Austin", 25.0)]
              means that there is a 30.0 mile road from Chicago to Atlanta and a 
              25.0 mile road from Chicago to Austin.
       Returns: 
            nothing
       Effects: 
            Prints a textual representation of the road connections
    """
    for from_place in roads:
        print(from_place + ":")
        for hop in roads[from_place]:
            to_place, dist = hop
            print("\t->" + to_place + " (" + str(dist) + ")")
            

def dfs(place, dist_so_far, roads, distances):
    """Depth-first search, which may continue from from_place if dist_so_far    
        is the shortest distance at which it has yet been reached.  
       Args: 
          place: Currently searching from here
          dist_so_far:  Distance at which from_place has been reached 
              this time (which may not be the shortest path to from_place)
          roads:  dict mapping places to lists of hops of the form (place, hop-distance)
          distances: dict mapping places to the shortest distance at which they 
               have been reached so far (up to this time). 
    """
    distances = []
    dist_so_far = distances[place]
    if place not in distances:
        distances[place] = dist_so_far
        for city,dist in road[place]:
            for i in city:
                dfs(i,dist_so_far,roads,distances)
    else:
        if distances[place]<=dist_so_far:
            distances[place] = dist_so_far
    #FIXME
    #   Consider cases:
    #      - We've never been at place before (so it's not in distances)
    #      - We've been at place before, on a path as short as this one (in distances)
    #      - We've been here before, but this way is shorter (dist_so_far)
    #    Consider which are base cases, and which require recursion.
    #    For the cases that require recursion, what is the progress step?
    
    
            

    
def main():
    """
    Main program gets city pair and map file name,
    reports distance or reports lack of connectivity.
    """
    parser = argparse.ArgumentParser(
        description="Find shortest route in road network")
    parser.add_argument('from_place', 
                help = "Starting place (quoted if it contains blanks)")
    parser.add_argument('to_place', 
                help = "Destination place (quoted if it contains blanks)")
    parser.add_argument('map_file', type=argparse.FileType('r'),
                help = "Name of file containing road connections and distances")
    args = parser.parse_args()
    start_place  = args.from_place
    destination = args.to_place
    roads = read_distances(args.map_file)

    if not start_place in roads: 
        print("Start place ", start_place, " is not on the map")
        exit(1)
    if not destination in roads: 
        print("Destination ", destination, " is not on the map")
        exit(1)

    distances = { }

    dfs(start_place, 0.0, roads, distances )

    if destination in distances :
        print("Distance from {} to {} is {}".format(
            start_place,destination, distances[destination]))
    else:
        print("You can't get from {} to {}".format(start_place, destination))

if __name__ == "__main__":
    main()

求大神啊~

Answer 1

지난주 오피스아워를 통해 GTF에 문의한 후 실행 가능한 코드 작성에 성공했습니다

으아악

결국 거의 만점을 받았습니다. 단점은 두 개의 for 루프를 하나로 바꿀 수 있다는 것입니다. 표준답안교수님은 아직 업로드가 안되어 업로드 후 업데이트 하겠습니다.

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표준적인 답변은 다음과 같습니다

으아악

Answer 2

코드는 깊이 우선 검색 알고리즘을 사용하라는 메시지를 표시합니다.
각 단계에서 가보지 않은 지점을 선택하여 끝까지 가세요.