breadth-first-search

问题 I'm trying to implement the BFS in order to find the shortest path from a source to a target in a maze. The problem that I'm having is that I'm unable to print the path, it is printed with '*' in the maze, but how can I extract the path from the predecessors of the BFS without printing all the visited nodes? Here's my code for you to compile: #include <stdio.h> #include <stdlib.h> #include <string.h> struct coord{ //This is a struct I'll use to store coordinates int row; int col; }; //-------

问题 I know we can use DFS for maze exploration. But I think we can also use BFS for maze exploration. I'm little bit confused here because most of the books and articles that I've read had used DFS for this problem. What I think is that the Best Case time complexity of DFS will be better as compared to BFS. But Average and Worst Case time complexity will be same for both BFS & DFS and thats why we prefer DFS over BFS. Am I right or I'm having some misconception 回答1: They have similar running time

问题 It's difficult to tell what is being asked here. This question is ambiguous, vague, incomplete, overly broad, or rhetorical and cannot be reasonably answered in its current form. For help clarifying this question so that it can be reopened, visit the help center. Closed 6 years ago . From Skiena's Book, Design a linear-time algorithm to eliminate each vertex v of degree 2 from a graph by replacing edges (u,v) and (v,w) by an edge (u,w). We also seek to eliminate multiple copies of edges by

问题 I'm trying to solve this problem and i'm fairly new to graphs. I tried BFS to figure this out but i'm not getting the right answer. What am i doing wrong? Also, is there a better way of doing this, other than the approach i am using. public static boolean isThereARoute(int[][] graph ,gNode n1 , gNode n2 ) { // where can we move? - anywhere where the value of x and y is 1 - else can't move // Start with node 1 and then traverse either BFS or DFS to see if the n2 is in the path anywhere //

问题 First of all I want to note that I posted this same question before and did not found a right answer, so sorry for repeating question. Note that I am required to use recursion here. I am aware that shortest path is usually found using BFS, which is not recursive, but I need to know how can this be done recursively. I am working on a rouge game and one of my monsters behaves like this. In a maze, if monster can reach the player in 15 or less steps, it makes the most optimal move possible. In

问题 I have been working on algorithm for a week at last I have successfully made an algorithm to create paths that captures opponent pieces, and then find a Maximum captures Path on 8x8 board. I have created an array of the following move captures. var moves = [[37 , 69] , [69 , 101] , [ 101 , 99] , [69 ,71], [ 69 , 67] , [67 , 99] , [99 , 101] , [71 ,103] ] Note: Can't capture downwards respective to black. Here I don't know the end point so that's why I can use Search Algorithms such as BFS,

问题 My question is mostly algorithm-related and not specific to a specific programming language Assuming we have a graph represented by a list of lists, where each internal list represents two nodes and a numbered edge, Is is possible to implement a recursive BFS (breadth-first search) function using ONLY the following 12 functions? Our bfs recursive function is supposed to take 5 arguments: the graph to be searched the element to find the search queue list of visited nodes current element we're

问题 I have written the code finding the optimal path for a Weighted Graph: SimpleDirectedWeightedGraph<String, DefaultWeightedEdge> graph = new SimpleDirectedWeightedGraph<String, DefaultWeightedEdge>(DefaultWeightedEdge.class); graph.addVertex("1"); graph.addVertex("2"); graph.addVertex("3"); graph.addVertex("4"); graph.addVertex("5"); DefaultWeightedEdge e1 = graph.addEdge("1", "2"); graph.setEdgeWeight(e1, 5); DefaultWeightedEdge e2 = graph.addEdge("2", "3"); graph.setEdgeWeight(e2, 10);

问题 I am working on an algorithm problem. For Simplification, the problem could be reduced to: Given an m * n matrix, with a bunch of special points, find an optimal path from point A to point B in this matrix. The optimal path is a path which passes fewest special points. If there are multiple paths with fewest special points, choose the shortest one. If there are still multiple paths, select one randomly among them. This problem could definitely be solved by BFS. To hold a queue, record

问题 So I'm currently working on a project of a word ladder problem and I have already built the graph for storing all dictionary words in it and added the edges in it, I did this using boost graph library. But what is confusing me is that breadth_first_search() function, seems like the parameters only take the starting vertex but no ending vertex. I checked the documentations and noticed that I can define the BFS visitor for that search function, but since I'm a newbie to the boost library I