breadth-first-search

问题 I'm trying to write the codes for breadth-first search in binary tree. I've stored all the data in a queue, but I can't figure out how to travel to all nodes and consume all their children. Here's my code in C: void breadthFirstSearch (btree *bt, queue **q) { if (bt != NULL) { //store the data to queue if there is if (bt->left != NULL) enqueue (q, bt->left->data); if (bt->right != NULL) enqueue (q, bt->right->data); //recursive if (bt->left != NULL) breadthFirstSearch (bt->left, q); if (bt-

问题 I was wondering what is the time complexity of BFS, if I use: an adjacency matrix adjacency list edge list Is it same as their space complexity? 回答1: The complexity of BFS implemented using an Adjacency Matrix will be O(|V| 2 ). And that when implemented by an Adjacency List is O(|V| + |E|) . Why is it more in the case of Adjacency Matrix? This is mainly because every time we want to find what are the edges adjacent to a given vertex 'U', we would have to traverse the whole array

问题 The site http://web.eecs.utk.edu/~huangj/CS302S04/notes/graph-searching.html describes that when an adjacency list is used then, DFS and BFS have complexity O(V+E), and if an adjacency matrix is used, the complexity is O(V 2 ). Why is this? 回答1: In both cases, the runtime depends on how long it takes to iterate across the outgoing edges of a given node. With an adjacency list, the runtime is directly proportional to the number of outgoing edges. Since each node is visited once, the cost is

问题 Time complexity to go over each adjacent edges of a vertex is say O(N) , where N is number of adjacent edges. So for V number of vertices time complexity becomes O(V*N) = O(E) , where E is the total number of edges in the graph. Since removing and adding a vertex from/to Queue is O(1) , why it is added to the overall time complexity of BFS as O(V+E) . 回答1: Considering the following Graph we see how the time complexity is O(|V|+|E|) but not O(V*E). Adjacency List V E v0:{v1,v2} v1:{v3} v2:{v3}

问题 Time complexity to go over each adjacent edges of a vertex is say O(N) , where N is number of adjacent edges. So for V number of vertices time complexity becomes O(V*N) = O(E) , where E is the total number of edges in the graph. Since removing and adding a vertex from/to Queue is O(1) , why it is added to the overall time complexity of BFS as O(V+E) . 回答1: Considering the following Graph we see how the time complexity is O(|V|+|E|) but not O(V*E). Adjacency List V E v0:{v1,v2} v1:{v3} v2:{v3}

问题 The results below are measured using perf on a compute server with 32 cores. I know my implementation is unoptimized but purposely as I want to make comparisons. I understand that graph algorithms tend to have low locality which researchers try to address. I'm unclear of the results, though. The time elapsed is misleading. My implementation runs through a graph with about 4mm nodes in about 10 seconds and the rest of the time pre processing. The optimized version uses the same input and

问题 The results below are measured using perf on a compute server with 32 cores. I know my implementation is unoptimized but purposely as I want to make comparisons. I understand that graph algorithms tend to have low locality which researchers try to address. I'm unclear of the results, though. The time elapsed is misleading. My implementation runs through a graph with about 4mm nodes in about 10 seconds and the rest of the time pre processing. The optimized version uses the same input and

问题 I have this line of code that tests wether or not there is a path to be found in a labyrinth represented by a matrix. How do I print the path at the end after I've determined wether or not there is a path. I've tried doing a stack but I'm not sure how to proceed. from queue import Queue maze=open(input()) matrix=maze.readlines() matrix=[i.strip() for i in matrix] matrix=[i.split() for i in matrix] numrows, numcols = len(matrix), len(matrix[0]) q=Queue() row=0 col=0 q.put((row,col)) while not

问题 The following question was found in Sedgewick and Wayne book about algorithms in java: 4.2.19 Topological sort and BFS. Explain why the following algorithm does not necessarily produce a topological order: Run BFS, and label the vertices by increasing distance to their respective source. I was trying to prove it finding a counter example. But, everytime I try, I get a topological order. I mean, I don't understand why this not work: If the source of a vertex comes before it, why don't we have

问题 void traverse(Node* root) { queue<Node*> q; Node* temp_node= root; while(temp_node) { cout<<temp_node->value<<endl; if(temp_node->left) q.push(temp_node->left); if(temp_node->right) q.push(temp_node->right); if(!q.empty()) { temp_node = q.front(); q.pop(); } else temp_node = NULL; } } The above posted code is my level order traversal code. This code works fine for me but One thing I dont like is I am explicitly initializing temp_node = NULL or I use break. But it does not seem to be a good