我是一名毒贩，附有送货路线图，每张都值得一定数量。但我不能覆盖整个区域，我只需要打破我的一点点草皮。但是，我可以选择哪一块草皮可以解决。
此图表的顶点是街角，所有街道长度相同和1路。有些街道比其他街道值得多。我想找到行走的最大节拍，这样我必须经过的街角数量最少。所有路径均从42街地铁站和罗斯福岛电车的终点站开始。
我已经快速浏览了一下网，但似乎并不像我记得在任何地方看过的那样。
使用动态编程在DAG 上轻松计算最大权重/流量/成本路径。
但是如果我们还需要找到一个尽可能短的最大路径怎么办？ 好吧，我们可以迭代所有长度为k的最大路径，以获得接近某个球场值的k值。但是，如何获得长度为k的最大路径？
有什么想法可以找到指向这个的指针吗？
编辑1
好的..我有一种感觉Floyd-Warshall可以在特定长度做最大路径。这可能足够好，我可以在所有可能的长度上运行它。但是......任何更好的想法？
编辑2
修改为仅提及DAG。
编辑3
Floyd-Warshall可以执行特定长度的最大路径，k：意义，将DAG最大路径与使用Floyd Warshall相结合，以获得任何顶点和最终停靠点之间的最短路径（沿边数）。因此，在最大路径算法中的任何顶点处，当当前边数（当前最大路径数）+（Floyd-Warshall确定的）最小边数与该顶点之间的边数之和时，杀死该顶点处的搜索分支最后一站将超过我们的特定长度，k。
编辑4
我想我有一个答案。
要查找此图的最大长度k，我们假设对于图中的任何节点，我们知道该节点与终点之间的最小边数。
所以我们只运行这样的最大路径算法，确保我们检查给定分支可能的最小边数是否低于我们的排除边界
在此代码中，图形是由其起始顶点索引的边的字典，每个边是元组（destination_edge，weight）。此算法最大化权重，但确保其返回的路径不超过参数k：
def maxpathoflength(k,start_vertex,graph_out_edges,memory,frame_size):
# first of all we leverage dynamic programming to save any results we have checked before
if start_vertex in memory:
return memory[start_vertex]
# otherwise we calculate it
# first we set the current maximum gold at this vertex
maxgold = 0
# corresponding to the following maximum path
maxpath = [start_vertex]
#then we check to see if we have broken our k-length path constraint
if k < 0:
# we failed, yet somehow we should never make it to here anyway
return { 'gold' : maxgold, 'path' : maxpath }
# if on the other hand k were equal to zero and this was the end point
# we would have succeeded
# we save this size since we can use it to calculate the lower bound
# on the number of edges required to get to the finish from this vertex
graph_size = len(graph_out_edges)
# we get the edges adjacent to start_vertex
# they are all out_edges, the graph has the DAG property
# the graph also has this property : there are no out edges from the last vertex :
if start_vertex > graph_size - 1:
return { 'gold' : maxgold, 'path' : maxpath }
out_edges = graph_out_edges[start_vertex]
print 'Out edges: ' + str(out_edges)
for edge in out_edges:
# each edge is stored as a 2-tuple (to_vertex,edge_weight)
arrival_vertex = edge[0]
edge_weight = edge[1]
# the following property holds for this graph, hence we can know the lower bound on edges of any path from this vertex
min_edges_from_arrival_vertex_to_end = ceil((1.0*graph_size - int(arrival_vertex))/frame_size)
# and so we can determine if we want should search it or not
if min_edges_from_arrival_vertex_to_end > k-1:
print 'Skipping this edge'
continue
# or if this branch is ok
# we search down this branch and take 1 off the edges
# remaining
# we search it and find the result of searching it is...
result = maxpathoflength(k-1,arrival_vertex,graph_out_edges,memory,frame_size)
gold = edge_weight + result.get('gold')
path = result.get('path')
# if that result is greater than our current best
if gold > maxgold:
# we make that result our new personal best
maxgold = gold
maxpath = [start_vertex] + path
# our return value is what we found the maximum to be from this vertex
returnvalue = { 'gold' : maxgold, 'path' : maxpath }
# we save this so we never ever have to do this again
memory[start_vertex] = returnvalue
# and send it back to up to the calling function
return returnvalue