狄克斯特拉算法用于在加權(quán)圖中查找最短路徑（權(quán)重不能為負(fù)） '''實(shí)現(xiàn)狄克斯特拉算法''' graph = {} graph['start'] = {}#在散列表graph中再建一個散列表為start，start即是里面的散列表的名字，也是graph中的key graph['start']['a'] = 6 graph['start']['b'] = 2 #print(graph['start'].keys()) 調(diào)用key函數(shù)獲取關(guān)鍵字 #print(graph['start'])單獨(dú)打印start散列表 ---{'a': 6, 'b': 2} #print(graph) 打印整個散列表 ---{'start': {'a': 6, 'b': 2}} graph['a'] = {} graph['a']['fin'] = 1graph['b'] = {} #b節(jié)點(diǎn)有兩個鄰點(diǎn) graph['b']['a'] = 3 graph['b']['fin'] = 5 graph['fin'] = {} #print(graph) ---{'start': {'a': 6, 'b': 2}, 'a': {'fin': 1}, 'b': {'a': 3, 'fin': 5}} infinity = float('inf')#無窮大 costs = {} #存儲現(xiàn)有的到每個節(jié)點(diǎn)的花銷；后面就會更新 costs['a'] = 6 costs['b'] = 2 costs['fin'] = infinity parents = {} parents['a'] = 'start' parents['b'] = 'start' parents['fin'] = None processed = []#定義一個函數(shù)來找cost最小的節(jié)點(diǎn) #參數(shù)兩個：1.該節(jié)點(diǎn)的cost；2.該節(jié)點(diǎn)對應(yīng)的key #循環(huán)所有的節(jié)點(diǎn)，將每個節(jié)點(diǎn)的costs與最小比較，如果滿足節(jié)點(diǎn)未經(jīng)處理且比最小還小，就進(jìn)行賦值；否則就找下一個節(jié)點(diǎn) def find_lowest_node(costs):lowest_cost = infinitylowest_cost_node = Nonefor node in costs:if costs[node] < lowest_cost and node not in processed:lowest_cost = costs[node]lowest_cost_node = nodereturn lowest_cost_nodenode = find_lowest_node(costs) while node is not None:cost = costs[node]#獲取開銷最小的節(jié)點(diǎn)的costsneighbors = graph[node]#獲取這個節(jié)點(diǎn)的鄰點(diǎn)，應(yīng)該是一個散列表#print(neighbors)for i in neighbors.keys():new_cost = cost + neighbors[i] #應(yīng)該要把neighbors[n]的values拿去相加if costs[i] > new_cost:costs[i] = new_costparents[i] = nodeprocessed.append(node)node = find_lowest_node(costs) print(graph) print(costs['fin']) '''實(shí)現(xiàn)狄克斯特拉算法''' # ------------整個圖的散列表（字典）-------- graph = {} #起點(diǎn) graph['start'] = {} #起點(diǎn)是一個散列表 graph['start']['B'] = 2 #存儲權(quán)重start---B graph['start']['c'] = 5 #存儲權(quán)重start---C #存儲其他節(jié)點(diǎn) graph['B'] = {} #節(jié)點(diǎn)B所有的鄰節(jié)點(diǎn) graph['B']['C'] = 8 #B---C graph['B']['E'] = 7 #B---E graph['C'] = {} graph['C']['D'] = 4 graph['C']['E'] = 2 graph['D'] = {} graph['D']['E'] = 6 graph['D']['fin'] = 3 graph['E'] = {} graph['E']['fin'] = 1 #終點(diǎn) graph['fin'] = {} # ---------實(shí)時計算消耗權(quán)重的散列表（字典）------------ infinity = float("inf") # python中表示無窮大，因?yàn)榇丝痰拈_銷還不知搭配 costs = {} costs['B'] = 2 costs["C"] = 5 # costs["destination"] = infinity # 表示路徑還沒選擇，此時到達(dá)終點(diǎn)所消耗權(quán)重未知 # -----------存儲父節(jié)點(diǎn)的散列表---------------- parents = {} parents["B"] = "start" parents["C"] = "start" parents["destination"] = None # ----------記錄存儲過的節(jié)點(diǎn)----------------- processed = [] print(graph) def find_lowest_node(costs):lowest_cost = infinitylowest_cost_node = Nonefor node in costs:if costs[node] < lowest_cost and node not in processed:lowest_cost = costs[node]lowest_cost_node = nodereturn lowest_cost_nodenode = find_lowest_node(costs) while node is not None:cost = costs[node]#獲取開銷最小的節(jié)點(diǎn)的costsneighbors = graph[node]#獲取這個節(jié)點(diǎn)的鄰點(diǎn)，應(yīng)該是一個散列表#print(neighbors)for i in neighbors.keys():new_cost = cost + neighbors[i] #應(yīng)該要把neighbors[n]的values拿去相加if costs[i] > new_cost:costs[i] = new_costparents[i] = nodeprocessed.append(node)node = find_lowest_node(costs)print(costs['fin'])

第二個程序報錯：KeyError: 'E'（未解決）

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