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题目意思是：求最短路径的同时还要保证这些路径的总权值最小
在两条总权值相等的路径中，经过结点最多的那个最好。你想想，如果我能从原来已有的结点上再加一条到达木雕。是不是好过从上一个结点再拉一条呢。
代码只要在原来的dijkstra算法上改一下就好了。新加一个存储到父结点的边权值，在每次更新最短路径的时候，看新的父结点的边值是否小于以前的，是就更改父结点和父结点边的权值。
python代码：
def dij(start, graph): n = len(graph) costs = [99999 for _ in range(n)] parents = [-1 for _ in range(n)] parents_costs = [99999 for _ in range(n)] costs[start] = 0 visited = [False for _ in range(n)] t = [] while len(t) < n: # 每次从costs里找最短花销，把最短花销的结点加进t中 minCost = 99999 minNode = start for i in range(len(costs)): if not visited[i] and costs[i] < minCost: minNode = i minCost = costs[i] t.append(minNode) visited[minNode] = True # 找这点的邻边（子结点不在t里面） # 如果邻边的子结点的路径花销小于原来的花销，就更新花销和父结点 for edge in graph[minNode]: if not visited[edge[0]] and edge[1] + minCost <= costs[edge[0]]: if edge[1] + minCost < costs[edge[0]]: costs[edge[0]] = edge[1] + minCost parents[edge[0]] = minNode parents_costs[edge[0]] = edge[1] elif edge[1] < parents_costs[edge[0]]: parents[edge[0]] = minNode parents_costs[edge[0]] = edge[1] return costs, parents, parents_costs n, m = map(int, input().split()) # 邻接表 graph = [[] for _ in range(n)] for i in range(m): edge = list(map(int, input().split())) graph[edge[0] - 1].append([edge[1] - 1, edge[2]]) graph[edge[1] - 1].append([edge[0] - 1, edge[2]]) costs, parents, parents_costs = dij(0, graph) # 算总费用 total = 0 for i in range(1, len(parents_costs)): total += parents_costs[i] print(parents) print(parents_costs) print(total) 转载请标明出处:CCF-201609-4-交通规划
文章来源: CCF-201609-4-交通规划