breadth-first-search

问题 Problem Statement: Given a matrix of people(denoted by small alphabets) and bikes(denoted by capital alphabets), find the nearest bike for a given person. How will you change your solution if you have to find bikes for a set of people? (assuming multiple bikes can be at the same distance from 1 person)? I know Dijkstra's algorithm and Bellman Ford's algorithm, but curious about the implementation here. Or can it be solved by BFS(Breadth First Search)? 回答1: To assign a single bike to a single

问题 Problem Statement: Given a matrix of people(denoted by small alphabets) and bikes(denoted by capital alphabets), find the nearest bike for a given person. How will you change your solution if you have to find bikes for a set of people? (assuming multiple bikes can be at the same distance from 1 person)? I know Dijkstra's algorithm and Bellman Ford's algorithm, but curious about the implementation here. Or can it be solved by BFS(Breadth First Search)? 回答1: To assign a single bike to a single

问题 I've seen quite a few posts (viz. post1, post2, post3) on this topic but none of the posts provides an algorithm to back up respective queries. Consequently I'm not sure to accept the answers to those posts. Here I present a BFS based shortest-path (single source) algorithm that works for non-negative weighted graph. Can anyone help me understand why BFS (in light of below BFS based algorithm) are not used for such problems (involving weighted graph)! The Algorithm: SingleSourceShortestPath

问题 I've seen quite a few posts (viz. post1, post2, post3) on this topic but none of the posts provides an algorithm to back up respective queries. Consequently I'm not sure to accept the answers to those posts. Here I present a BFS based shortest-path (single source) algorithm that works for non-negative weighted graph. Can anyone help me understand why BFS (in light of below BFS based algorithm) are not used for such problems (involving weighted graph)! The Algorithm: SingleSourceShortestPath

问题 Some pseudocode here (disregard my style) Starting from v1(enqueued): function BFS(queue Q) v2 = dequeue Q enqueue all unvisited connected nodes of v2 into Q BFS(Q) end // maybe minor problems here Since there are V vertices in the graph, and these V vertices are connected to E edges, and visiting getting connected nodes (equivalent to visiting connected edges) is in the inner loop (the outer loop is the recursion itself), it seems to me that the complexity should be O(V*E) rather than O(V+E)

问题 I understand there are 3 common ways to represent graphs: Adjacency Matrix Adjacency List Edge list That said, problems I’ve solved on LeetCode often use matrices and the solution requires DFS or BFS. For example, given the matrix below, find if a target string exists when you go left, right, up, and down (but not diagonal). [ [‘a’,‘p’,’p’], [‘e’,’a’,’l’], [‘r’,’t’,’e’] ] This required a DFS approach. Is this because this matrix represents a graph or does DFS and BFS apply to matrices too and

问题 Given a binary matrix of size N x M. The task is to find the distance of nearest 1 in the matrix for each cell. The distance is calculated as |i1 – i2| + |j1 – j2|, where i1, j1 are the row number and column number of the current cell and i2, j2 are the row number and column number of the nearest cell having value 1. Input: The first line of input is an integer T denoting the number of test cases. Then T test cases follow. Each test case consists of 2 lines . The first line of each test case

问题 The problem i'm trying to solve is this: Given a graph G = (V,E) such that every edge is colored in one of 10 colors, and two vertices: s, t. I need to find an algorithm that produces a (shortest) path from s to t, that goes over a minimal amount of colors. My idea was to duplicate the graph 10 times: The first duplicate will include only edges of one color The second will include only edges of two colors... and so on. Also, I connect an outer node: s' to every "s" node in every duplicate.

问题 The problem is as follow: Given an oriented weighted graph, a start node, an end node and a number k, verify if exist a path from the start node to the end node with at least length k. This is the code i wrote and it's correct but only in specific graph. For example g1 with weights1 is as follows: let weights1 = [(2,1,1);(2,1,3);(2,1,4);(1,1,5);(5,1,2);(5,1,6);(3,1,6);(6,1,7);(4,1,3)];; let f1 = function 1 -> [5] | 2 -> [1;3;4] | 3 -> [6] | 4 -> [3] | 5 -> [2;6] | 6 -> [7] | _ -> [];; type 'a

问题 The problem is as follow: Given an oriented weighted graph, a start node, an end node and a number k, verify if exist a path from the start node to the end node with at least length k. This is the code i wrote and it's correct but only in specific graph. For example g1 with weights1 is as follows: let weights1 = [(2,1,1);(2,1,3);(2,1,4);(1,1,5);(5,1,2);(5,1,6);(3,1,6);(6,1,7);(4,1,3)];; let f1 = function 1 -> [5] | 2 -> [1;3;4] | 3 -> [6] | 4 -> [3] | 5 -> [2;6] | 6 -> [7] | _ -> [];; type 'a