dijkstra最短路代码模板更新

本文参考了C++ ACM的dijkstra模板：

先说之前自己写错了的：

from collections import defaultdict
from heapq import heappush, heappop
 
 
def dijkstra(edges, start_node, end_node):
    graph = defaultdict(dict)
    for src, dst, distance in edges:
        graph[src][dst] = distance
 
    q = [(0, start_node, None)]
    found_min_dist_nodes = set()
    distances = {start_node: 0}
    back_paths = {}
 
    while q:
        cost, min_dist_node, src_node = heappop(q)
 
        #  下面这行代码非常关键，是为了去除优先级队列q里冗余的push，见后注释说明重复节点push问题
        #if min_dist_node in found_min_dist_nodes:
        #    continue
 
        found_min_dist_nodes.add(min_dist_node)
        back_paths[min_dist_node] = src_node
 
        if min_dist_node == end_node:
            return cost, back_paths
 
        for neibor_node, distance in graph[min_dist_node].items():
            if neibor_node in found_min_dist_nodes:
                continue
 
            prev_dist = distances.get(neibor_node, float('inf'))
            new_dist = cost + distance
            if new_dist < prev_dist:
                distances[neibor_node] = new_dist
                # 下面这个代码，正常情况下，应该是更新优先级队列里neibor_node的priority value
                # 但因为priority queue无原生更新api支持，所以下面代码是在没有remove neibor_node的情况直接push，会导致重复节点push
                heappush(q, (new_dist, neibor_node, min_dist_node))
 
    return float("inf"), back_paths
 
 
def find_path(back_paths, start_node, end_node):
    ans = [end_node]
    while end_node != start_node:
        end_node = back_paths[end_node]
        ans.append(end_node)
 
    return ans[::-1]
 
 
if __name__ == "__main__":
    edges = [
        ("A", "B", 5),
        ("A", "C", 10),
        ("C", "D", 10),
        ("B", "C", 2)]
 
    dist, backpaths = dijkstra(edges, "A", "D")
    print("dist: ", dist)
    print(find_path(backpaths, "A", "D"))

上面案例中的图示例：

A--------(5)--------->B

| /

(10) / (2)

| /

|(10)

肉眼看，A->D的最短路距离是17，路径是ABCD。

如果没有下面的代码：

1 2	`if` `min_dist_node` `in` `found_min_dist_nodes:` `continue<br><br>输出A到D的最短距离和路径为：`

dist: 17
['A', 'C', 'D']

路径这个答案是错的！！！路径计算错了！但是距离计算是ok的！

为啥呢？？？因此C节点会重复push，debug下就可以看出来了：

因此，我们加上上述if判定进行去重，因为之前已经pop过了，再pop重复节点已经没有意义：

from collections import defaultdict
from heapq import heappush, heappop
 
 
def dijkstra(edges, start_node, end_node):
    graph = defaultdict(dict)
    for src, dst, distance in edges:
        graph[src][dst] = distance
 
    q = [(0, start_node, None)]
    found_min_dist_nodes = set()
    distances = {start_node: 0}
    back_paths = {}
 
    while q:
        cost, min_dist_node, src_node = heappop(q)
 
        #  下面这行代码非常关键，是为了去除优先级队列q里冗余的push，见后注释说明重复节点push问题
        if min_dist_node in found_min_dist_nodes:
            continue
 
        found_min_dist_nodes.add(min_dist_node)
        back_paths[min_dist_node] = src_node
 
        if min_dist_node == end_node:
            return cost, back_paths
 
        for neibor_node, distance in graph[min_dist_node].items():
            if neibor_node in found_min_dist_nodes:
                continue
 
            prev_dist = distances.get(neibor_node, float('inf'))
            new_dist = cost + distance
            if new_dist < prev_dist:
                distances[neibor_node] = new_dist
                # 下面这个代码，正常情况下，应该是更新优先级队列里neibor_node的priority value
                # 但因为priority queue无原生更新api支持，所以下面代码是在没有remove neibor_node的情况直接push，会导致重复节点push
                heappush(q, (new_dist, neibor_node, min_dist_node))
 
    return float("inf"), back_paths
 
 
def find_path(back_paths, start_node, end_node):
    ans = [end_node]
    while end_node != start_node:
        end_node = back_paths[end_node]
        ans.append(end_node)
 
    return ans[::-1]
 
 
if __name__ == "__main__":
    edges = [
        ("A", "B", 5),
        ("A", "C", 10),
        ("C", "D", 10),
        ("B", "C", 2)]
 
    dist, backpaths = dijkstra(edges, "A", "D")
    print("dist: ", dist)
    print(find_path(backpaths, "A", "D"))