I am new to PCL and I would like to get all indices from a branch node in pcl::octree.
So, the root node for instance should provide every single index and each sub-branch node the indices of the points of all leaf nodes within this branch node.
What is the best way of achieving this?
I am using a
pcl::octree::OctreePointCloudSearch<pcl::PointXYZ, pcl::octree::OctreeContainerPointIndices, pcl::octree::OctreeContainerPointIndices>
and hoped with the 3rd template parameter for the BranchContainerT also set to pcl::octree::OctreeContainerPointIndices to be able to call something like
std::vector<int> indices;
auto it = m_octree->breadth_begin();
it.getBranchContainer().getPointIndices(indices)
However, the indices vector is empty.
Of course, I can manually iterate over all the nodes, get the leafs and insert the indices, but maybe I am missing something here .. ?
Ok, what I was missing is that the template parameter for BranchContainerT set to pcl::octree::OctreeContainerPointIndices only provides you the option to store the indices in this kind of container. However adding these indices is something you have to do manually.
Given that it is just a matter of checking each node for its type and collect the indices from the leafs and add them to the branch nodes.
Related
I saw the following implementation of topological sort using DFS on Leetcode https://leetcode.com/problems/course-schedule/discuss/58509/18-22-lines-C++-BFSDFS-Solutions
Now the part of this that is confusing me is the representation of the directed graph which is used for the top sort. The graph is created as follows:
vector<unordered_set<int>> make_graph(int numCourses, vector<pair<int, int>>& prerequisites) {
vector<unordered_set<int>> graph(numCourses);
for (auto pre : prerequisites)
graph[pre.second].insert(pre.first);
return graph;
}
What is confusing me is this line:
graph[pre.second].insert(pre.first);
Presumably the graph is being represented as an adjacency list; if so why is each node being represented by the incoming edges and not the outgoing edges? Interestingly, if I flip pre.second and pre.first like this:
graph[pre.first].insert(pre.second);
The top sort still works. However, it seems most implementations of the same problem use the former method. Does this generalize to all directed graphs? I was taught in my undergraduate degree that a directed graph's adjacency list should contain a list of each nodes outgoing nodes. Is the choice of incoming vs outgoing node arbitrary for the representation of the adjacency list?
To the specific problem which only requires answering true or false, it doesn't matter if you flip every edge. That's because a graph is topological sortable if and only if it has no loops. But if you want an order of taking, it doesn't work as you can see in the different results of [[0, 1]] and [[1, 0]].
Which way to save the graph depends on how you solve the problem. In this given case, we need to know the indegrees of every node (course) and also to update it every time we delete a node from the graph (take the course), so that we know if we can delete a node (we can do it when the indegree is 0). When updating, we minus 1 to each node that the deleted node direct to. If you apply this method (as most do), it's clear how you should save the graph
I am trying to find all the possible paths from one node in my graph that will visit all other nodes in the graph. I want the function to produce all possibilities of paths in my n*m graph. Each node in my graph has a vector of all neighbors nodes and a Boolean that check if the node is visited or not.
example:
a b
c d
will produce:
abcd
abdc
acbd
...
I tried the solution in this answer, but only return one path. How can I produce all possible paths?
It seems like in some situations by your description you could have infinite paths and a path of infinite length because you didn't specify that nodes couldn't be revisited.
You should implement depth first search and pass a reference to an array of marked (visited) nodes in your recursive DFS method assuming that you have a count of the number of nodes in your graph. After you visit each node, before you leave that node make sure you set it to false again so that it can be reaccessed via another node.
The implementation of this algorithm is really going to depend on how you implemented your graph structure and without the details all I can do is speculate that you have a linked structure with an adjacency list representing the different nodes. I also have no idea how the different nodes map to characters so that is another detail I have to speculate, but say that the nodes are represented by integers.
You need to pass into a DFS method the following: array of marked nodes, a linked list which contains the path information, starting node, (i.e, current node) and final node
void printAllPaths(LinkedList<Integer> currentPath, boolean[] marked, int current, int last){
for( all nodes adjacent to current, node ){
if(node == last){
currentPath.addLast(last);
System.out.println(currentPath);
currentPath.removeLast();
}else if(!marked[node]){
currentPath.addLast(node);
marked[node] = true;
printAllPaths(currentPath, marked, node, final);
marked[node] = false;
currentPath.removeLast();
}
}
}
This will be the basic idea of the code. I apologize if it doesn't compile in advance, but this should print out all of the paths.
I want to create a graph with nodes and edges, where each node will contain n number of values. We would be given with the n values of the starting node, from which we need to generate other nodes where each value in each node would be of the form either:
t_n=t_(n-1)+2
or
t_n=t_(n-1)-1
When such a node is generated, it should create an edge from the old node to the new node.
I know this might be very trivial job, but I have very limited programming knowledge. I have been suggested to use classes in C++ or structure to represent the nodes. Please help me in creating the graph with nodes that would have multiple values and further the next nodes would be generated from the parent node following the above rule. Some C++ code would be very helpful.
Thanks in Advance.
here you have some code but I don't really fully understand your task.
- graph with nodes and edges
- each node has n number of values
- we are given n values of the starting point
- need to generate other nodes where each value in each node would be either
- t_n=t_(n-1)+2
- t_n=t_(n-1)-1
- when such node is generated, it creates an edge from the old node to the new node.
this starting point: do we have to generate a graph from it? what is with the creation of the edge from the old node and the new node? is old node here the starting point?
does n number of values means to where the point is connected to (as a chain of the other edges to which this edge is connected to)? example we are provided a node with a chain of numbers (6, 4, 5) where this means we need to generate extra edges which would be connected x times (first one linked to our starting point would be linked to 6 edges, one of them being the starting point)
will edit my answer when I have more information. could you please draw an example in paint and upload it online and provide the link? it would be easier to imagine.
Given a tree(By tree , I mean N Nodes , N-1 edges and it is connected) the root of tree is changed to lets say r. Now given another node lets say t , you have to find the sum of all nodes in the subtree rooted at t.
I was trying to implement it in c++ .
std::map<int, std::vector< pair<int,int> > > map;
if I iterate over the vector map[t] , I have to ensure that it does not go to a path which leads to r . How would I ensure that ?
Also is there a better way to store a tree in c++ , given the conditions that the root of the tree might change ? I think there will be because the map does not convey anything about a root . :)
The problem is that you have your tree stored as a general graph. Hence for a given vertex you don't know which is the arc leading towards the root (i.e. the parent).
The solution very much depends on the context. A simple solution would be a depth first search starting from r and looking for t. As soon as you find t you have found also the parent of t, and you can easily identify the subtree starting from t.
Alternatively you can start from t and look for r. When you find r you have found the parent of t and you can traverse all other arcs to find the subtree of t.
About an alternative representation of the graph, usually it is better to keep a list of vertices and for each vertex keep a list of neighbour vertices as in:
std::map<int, std::list<int> >
I have this data that is hierarchical and so I store it in a tree. I want to provide a search function to it. Do I have to create a binary tree for that? I don't want to have thousands of nodes twice. Isn't there a kind of tree that allows me to both store the data in the order given and also provide me the binary tree like efficient searching setup, with little overhead?
Any other data structure suggestion will also be appreciated.
Thanks.
EDIT:
Some details: The tree is a very simple "hand made" tree that can be considered very very basic. The thing is, there are thousands of names and other text that will be entered as data that I want to search but I don't want to traverse the nodes in a traditional way and need a fast search like binary search.
Also, importantly, the user must be able to see the structure he has entered and NOT the sorted one. So I cant keep it sorted to support the search. That is why I said I don't want to have thousands of nodes twice.
If you don't want to change your trees hierarchy use a map to store pointers to vertexes: std::map<SearchKeyType,Vertex*> M.
Every time when you will add vertex to your tree you need to add it to your map too. It's very easy: M[key]=&vertex. To find an element use M.find(key);, or M[key]; if you are sure that key exists.
If your tree has duplicate keys, then you should use a multimap.
Edit: If your key's size is too big, than you can use pointer to key instead of key:
inline bool comparisonFunction(SearchKeyType * arg1,SearchKeyType * arg2);
std::map<SearchKeyType *, Vertex *, comparisonFunction> M;
inline bool comparisonFunction(SearchKeyType * arg1,SearchKeyType * arg2)
{
return (*arg1)<(*arg2);
}
to search Element with value V you must write following:
Vertex * v = M[&V]; // assuming that element V exists in M