depth-first-search

Stop boost::depth_first_search along a particular depth if certain criteria is met

一笑奈何 提交于 2019-12-05 06:42:22
I'm using BGL to store my DAG. Vertices have states. Given a change in state in one of the vertices i want to update dependent vertices. This i'm able to do using boost::depth_first_search and a custom visitor. Now the logic is that i dont want to update a searched vertex and its dependent if the vertex is in a particular state. Basically i want to control over en-queuing of vertices in either dfs or bfs. What is the best way to achieve this in BGL. Thanks. It seems that boost::depth_first_search does not support this, but the underlying boost::depth_first_visit does, through its 2nd overload

Difference between BFS and DFS

落花浮王杯 提交于 2019-12-05 02:22:49
问题 I am reading about DFS in Introduction to Algorithms by Cormen. Following is text snippet. Unlike BFS, whose predecessor subgraph forms a tree, the predecessor subgrpah produced by DFS may be composed of several trees, because search may be repeated from multiple sources. In addition to above notes, following is mentioned. It may seem arbitrary that BFS is limited to only one source where as DFS may search from multiple sources. Although conceptually, BFS could proceed from mutilple sources

How can I store data in a table as a trie? (SQL Server)

為{幸葍}努か 提交于 2019-12-05 02:15:31
问题 To make things easier, the table contains all the words in the English dictionary. What I would like to do is be able to store the data as a trie. This way I can traverse the different branches of the trie and return the most relevant result. First, how do I store the data in the table as a trie? Second, how do I traverse the tree? If it helps at all, the suggestion in this previous question is where this question was sparked from. Please make sure it's SQL we're talking about. I understood

Puzzle Game Android DFS algorithm

非 Y 不嫁゛ 提交于 2019-12-04 23:14:20
I have a android application called Islands and bridges also known as Hashiwokakero The application uses A 2 Dimensional array that spawns the Islands randomly everytime the user restarts the game It form a Matrix with number from 0 to 4 where 0=null and 1-4 = Island There can be 2 bridges comming out of one Island to connect the other , The map at the moment is not solvable. To solve the game the user needs to connect the islands using bridges so if an island = 4 it needs 4 connection to it if an island = 2 it needs 2 connection and so on.. in my research i found out that the best algorithm

algorithm to enumerate all possible paths

三世轮回 提交于 2019-12-04 20:59:10
问题 Consider the following graph: I'm trying to find a way to enumerate all possible paths from a source node to a target node. For example, from A to E, we have the following possible paths: A B C D E A B C E A C D E A C E Note that for A C D E, there are actually 2 paths, since one of the paths uses edge F3 and the other uses edge F5. Also, since there's a cycle between A and C, you could end up with an infinite number of paths, but for the purposes of this I'm only interested in paths in which

how find fields of all member variables contained in java bean

帅比萌擦擦* 提交于 2019-12-04 19:49:26
I want to make a GUI using Java in which a user can select a bean, edit its fields, and then add an instance of the created bean to a queue. My question though is about accessing the fields. I have a class MyCompositeObject that inherits from MyParentObject. The MyParentObject is composed of multiple beans, each being composed of more beans. The class MyCompositeObject is also composed of beans. I want to find all accessible fields from MyCompositeObject. Class MyParentObject { MyObjectOne fieldOne; MyObjectTwo fieldTwo; String name; ... } Class MyCompositeObject extends MyParentObject {

Complexity of finding all simple paths using depth first search?

不问归期 提交于 2019-12-04 12:34:24
Thanks to everyone replying with ideas and alternate solutions. More efficient ways of solving problems are always welcome, as well as reminders to question my assumptions. That said, I'd like you to ignore for a moment what problem I'm trying to solve with the algorithm, and just help me analyze the big-Oh complexity of my algorithm as I've written it -- an all simple paths in a graph using depth-limited search as described here , and implemented here . Thanks! Edit: This is homework, but I've already submitted this assignment and I'd just like to know if my answer correct. I get a little

Is the runtime of BFS and DFS on a binary tree O(N)?

♀尐吖头ヾ 提交于 2019-12-04 10:27:16
问题 I realize that runtime of BFS and DFS on a generic graph is O(n+m), where n is number of nodes and m is number of edges, and this is because for each node its adjacency list must be considered. However, what is the runtime of BFS and DFS when it is executed on a binary tree? I believe it should be O(n) because the possible number of edges that can go out of a node is constant (i.e., 2). Please confirm if this is the correct understanding. If not, then please explain what is the correct time

Which Procedure we can use for Maze exploration BFS or DFS

自古美人都是妖i 提交于 2019-12-04 10:00:56
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 They have similar running time , but either may greatly outperform the other on any given problem simply due to the order in which the

Dividing a graph in three parts such the maximum of the sum of weights of the three parts is minimized

孤人 提交于 2019-12-04 09:32:34
I want to divide a graph with N weighted-vertices and N-1 edges into three parts such that the maximum of the sum of weights of all the vertices in each of the parts is minimized. This is the actual problem i am trying to solve, http://www.iarcs.org.in/inoi/contests/jan2006/Advanced-1.php I considered the following method /*Edges are stored in an array E, and also in an adjacency matrix for depth first search. Every edge in E has two attributes a and b which are the nodes of the edge*/ min-max = infinity for i -> 0 to length(E): for j -> i+1 to length(E): /*Call depth first search on the nodes