shortest-path

问题 I know this is usually done with breadth first, but we are asked to do it with both, I already accomplished breadth first.... I feel like this is a pretty typical example of using a depth first search, so I was hoping I could get some help here... I am attempting to find the shortest path through a maze using depth first search, but as of right now I cannot wrap my head around how exactly to do it. This is my code so far: void maze::findPathRecursive(graph &g, int position, int goal) { if

问题 I took an Olympiad Exam three days ago. I ran into a nice question as follows. We know the bellman-ford algorithm checks all edges in each step, and for each edge if, d(v)>d(u)+w(u,v) then d(v) being updated such that w(u,v) is the weight of edge (u, v) and d(u) is the length of best finding path for vertex u . if in one step we have no update for vertexes , the algorithms terminates . Supposing this algorithm for finding all shortest path from vertex s in graph G with n vertex after k < n

问题 I took an Olympiad Exam three days ago. I ran into a nice question as follows. We know the bellman-ford algorithm checks all edges in each step, and for each edge if, d(v)>d(u)+w(u,v) then d(v) being updated such that w(u,v) is the weight of edge (u, v) and d(u) is the length of best finding path for vertex u . if in one step we have no update for vertexes , the algorithms terminates . Supposing this algorithm for finding all shortest path from vertex s in graph G with n vertex after k < n

来源： https://stackoverflow.com/questions/59343417/how-to-get-the-shortest-path-in-a-weighted-graph-with-networkx

来源： https://stackoverflow.com/questions/59343417/how-to-get-the-shortest-path-in-a-weighted-graph-with-networkx

问题 I have a networkx graph in Python, with weighted edges. I want to get the weight of the smallest path between two nodes. Currently, I am getting the nodes in the shortest path from the nx.shortest_path implementation, and then iterating through each pair and summing over the weights between each pair of node. shortest_path = nx.shortest_path(G, source, destination, 'distance') #function to iterate over each pair import itertools def pairwise(iterable): a, b = itertools.tee(iterable) next(b,

问题 I would like to know how to calculate the shortest distance between two properties (points) for my code below. There are two shapefile files, one being a points shapefile, the other a roads shapefile. For testing, both shapefiles can be downloaded from the following website: https://github.com/JovaniSouza/JovaniSouza5/blob/master/Example.zip library(sf) roads <- st_read('Roads/Roads.shp') pts <- st_read('Points/Points.shp') %>% st_transform(crs=st_crs(roads)) plot(st_geometry(roads)) plot(st

问题 I would like to know how to calculate the shortest distance between two properties (points) for my code below. There are two shapefile files, one being a points shapefile, the other a roads shapefile. For testing, both shapefiles can be downloaded from the following website: https://github.com/JovaniSouza/JovaniSouza5/blob/master/Example.zip library(sf) roads <- st_read('Roads/Roads.shp') pts <- st_read('Points/Points.shp') %>% st_transform(crs=st_crs(roads)) plot(st_geometry(roads)) plot(st

问题 i am trying to get started with openstreetmap in R and try to replicate the example given in osmar package documentation. I get a bit data for munich. src <- osmsource_api(url = "https://api.openstreetmap.org/api/0.6/") muc_bbox <- center_bbox(11.575278, 48.137222, 1000, 1000) muc <- get_osm(muc_bbox, src) I get a subset of all highways in munich hways_muc <- subset(muc, way_ids = find(muc, way(tags(k == "highway")))) hways <- find(hways_muc, way(tags(k == "name"))) hways <- find_down(muc,

问题 Hi I am struggling with this question. It is the following: Devise an algorithm to find the lowest-weight cycle(i.e. of all cycles in the graph, the one with the smallest sum of edge weights), in a weighted, directed graph G = (V,E). Briefly justify the runtime and space complexity. Assume all edges are non-negative. It should run in O(|V||E|log|V|) time. Hint: Use multiple calls to Dijkstra's algorithm. I have seen solutions that use Floyd-Warshall but I was wondering how we would do this