bellman-ford

问题 Find the shortest path from source to destination in a directed graph with positive and negative edges, such that at no point in the path the sum of edges coming before it is negative. If no such path exists report that too. I tried to use modified Bellman Ford, but could not find the correct solution. I would like to clarify a few points : yes there can be negative weight cycles. n is the number of edges. Assume that a O(n) length path exists if the problem has a solution. +1/-1 edge weights

问题 After a lot of Googling, I've found that most sources say that the Dijkstra algorithm is "more efficient" than the Bellman-Ford algorithm. But under what circumstances is the Bellman-Ford algorithm better than the Dijkstra algorithm? I know "better" is a broad statement, so specifically I mean in terms of speed and also space if that applies. Surely there is some situation in which the Bellman-Ford approach is better than the Dijkstra approach. 回答1: Bellman-Ford algorithm is a single-source

问题 I have just started learning new algorithms but I got stuck when I read bellman ford algorithms on geeks for geeks :- http://www.geeksforgeeks.org/dynamic-programming-set-23-bellman-ford-algorithm/ There It is written that- the algorithm calculate shortest paths in bottom-up manner. It first calculates the shortest distances for the shortest paths which have at-most one edge in the path. Then, it calculates shortest paths with at-most 2 edges, and so on. After the ith iteration of outer loop,

问题 I have a question regarding the bellman ford algorithm. I created this program that when given a graph will output the shortest distance between a source node and all other nodes. That part is working fantastic so I have outputs like this: The cost table is: Destination: 0 1 2 Cost: 0 4 6 So for instance the shortest distance between my source and node 2 is 6,which is great. But now I would like to get the actual routes instead of just their costs. Like instead of having only the cost on the

问题 I'm trying to code a program for a class that simulates a router and so far I have the basics set up ("router" can send and receive packets through an emulated server to other "routers" connected to the server). Each packet contains only the distance vector for that router. When a router receives a packet it is supposed to update it's own distance vector accordingly using the Bellman-Ford algorithm. The problem I'm having is that I am finding myself unable to implement the actual algorithm

问题 Bellman Ford is used to detect negative weighted cycles in a graph. I would like to know how I can use it to detect cycles which exceeds a certain threshold instead. Example: ---------> ^ |1 0.5 | <------v 1 -----------> 2 ^ | |4 |1 | 2 v 4<------------3 This graph has 2 cycles. One with the product = 1 and the other with the product = 4. If the threshold = 1, the algorithm should output true, since there is 1 cycle with a product > 1. 回答1: I assume you want to detect a simple cycle with

问题 So I came to this beautiful problem that asks you to write a program that finds whether a negative infinity shortest path exists in a directed graph. (Also can be thought of as finding whether a "negative cycle" exists in the graph). Here's a link for the problem: http://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=499 I successfully solved the problem by running Bellman Ford Algorithm twice by starting with any source in the graph. The second time I

问题 I use a matrix d to present a graph. d.(i).(j) means the distance between i and j ; v denotes the number of nodes in the graph. It is possible that there is negative cycle in this graph. I would like to check if a negative cycle exists. I have written something as follows from a variation of Floyd-Warshall: let dr = Matrix.copy d in (* part 1 *) for i = 0 to v - 1 do dr.(i).(i) <- 0 done; (* part 2 *) try for k = 0 to v - 1 do for i = 0 to v - 1 do for j = 0 to v - 1 do let improvement = dr.

问题 Let G = (V;E) be a directed graph whose edges all have non-negative weights. Let s,t be 2 vertices in V, and let e be an edge in E. Describe an algorithm that decides whether all shortest paths from s to t contain the edge e. Well, this is how you can achieve Dijsktra's time complexity: Simply run Dijkstra from s and calculate delta(s,t) (the weight of the shortest path from s to t). Remove the edge e, and run Djikstra again from s in the new graph. If delta(s,t) in the new graph has

问题 I learned that the Bellman-Ford Algorithm has a running time of O(|E|*|V|), in which the E is the number of edges and V the number of vertices. Assume the graph does not have any negative weighted cycles. My first question is that how do we prove that within (|V|-1) iterations (every iteration checks every edge in E), it updates the shortest path to every possible node, given a particular start node? Is it possible that we have iterated (|V|-1) times but still not ending up with shortest