Dijkstra

 Dijkstra（迪杰斯特拉算法）

#Dijkstra(迪杰斯特拉算法) 带权无向图中寻求最短路径
#动态演示过程： https://www.bilibili.com/video/BV1q4411M7r9?from=search&seid=5585063956337033177
#实现的话用优先队列实现（出队的都是确认是到该点的最短路径）
import heapq
import math
graph = {
    "A": {"B": 5, "C": 1},
    "B": {"A": 5, "C": 2, "D": 1},
    "C": {"A": 1, "B": 2, "D": 4, "E": 8},
    "D": {"B": 1, "C": 4, "E": 3, "F": 6},
    "E": {"C": 8, "D": 3},
    "F": {"D": 6}
}
def init_distance(graph,s):
    distance = {s:0}
    nodes = graph.keys()
    for node in nodes:
        if node !=s:
            distance[node] = math.inf
    return distance

def dijkstra(graph,s):
    #用heapq（小根堆）实现优先队列
    pqueue = []
    heapq.heappush(pqueue,(0,s))
    #此处的seen和BFS，DFS中的seen不一样，此处的seen都是出队的，已经确认最短路径的节点
    seen = set()
    #          A  B  C  D  E  F
    #distance(初始化位正无穷)
    #parent（初始化为-1）
    distance = init_distance(graph,s)
    parent = {s:None}
    #出队的都是已经确认是最短距离的
    while len(pqueue) > 0:
        pair = heapq.heappop(pqueue)
        dist = pair[0]
        cur = pair[1]
        seen.add(cur)
        #与cur相邻的节点
        nodes = graph[cur].keys()
        for n in nodes:
            if n not in seen:
                #满足下面关系式才更新
                if dist + graph[cur][n] < distance[n]:
                    heapq.heappush(pqueue,(dist + graph[cur][n],n))
                    parent[n] = cur
                    distance[n] = dist + graph[cur][n]
            
    return parent, distance