Dijkstra Algorithm

与BFS不同的是每条路径多了权重

1.步骤:

  1. 找到最便宜的节点,即可在最短时间内前往的节点
  2. 对于该节点的邻居,检查是否有前往它们的更短路径,如果有,就更新其开销。
  3. 重复这个过程,直到对图中的每个节点都这样做了
  4. 计算最终路径。

2.注意

  • 只适用于有向无环图(directed acyclic graph, DAG)
  • 不适用于包含负权重边的图(Bellman-Ford algorithm)

3.实现

Dijkstra Algorithm

3.1 三个散列表和一个数组

根据有向图,需要三个散列表
Dijkstra Algorithm

# the graph
graph = {}
graph["start"] = {}
# 散列表又包含散列表
graph["start"]["a"] = 6
graph["start"]["b"] = 2

graph["a"] = {}
graph["a"]["fin"] = 1

graph["b"] = {}
graph["b"]["a"] = 3
graph["b"]["fin"] = 5

graph["fin"] = {} # 终点没有邻居

Dijkstra Algorithm

# the costs table
infinity = float("inf")
costs = {}
costs["a"] = 6
costs["b"] = 2
costs["fin"] = infinity

Dijkstra Algorithm

# the parents table
parents = {}
parents["a"] = "start"
parents["b"] = "start"
parents["fin"] = None

Dijkstra Algorithm
最后还需要一个数组记录处理过的节点

processed = []
3.2算法

Dijkstra Algorithm

node = find_lowest_cost_node(costs) # 在未处理的节点中找出开销最小的节点 
while node is not None:             # 在所有节点都被处理过后结束 
    cost = costs[node]
    # Go through all the neighbors of this node.
    neighbors = graph[node]
    for n in neighbors.keys():
        new_cost = cost + neighbors[n]
        # If it's cheaper to get to this neighbor by going through this node...
        if costs[n] > new_cost:
            # ... update the cost for this node.
            costs[n] = new_cost
            # This node becomes the new parent for this neighbor.
            parents[n] = node
    # Mark the node as processed.
    processed.append(node)
    # Find the next node to process, and loop.
    node = find_lowest_cost_node(costs)
3.3 找开销最低的节点
def find_lowest_cost_node(costs):
    lowest_cost = float("inf")
    lowest_cost_node = None
    # Go through each node.
    for node in costs:
        cost = costs[node]
        # If it's the lowest cost so far and hasn't been processed yet...
        if cost < lowest_cost and node not in processed:
            # ... set it as the new lowest-cost node.
            lowest_cost = cost
            lowest_cost_node = node
    return lowest_cost_node