Count nodes within K-distance from all nodes in a set

Given an undirected tree with some marked nodes and a positive number K. We need to print the count of all such nodes which have distance from all marked nodes less than K that means every node whose distance from all marked nodes is less than K, should be counted in the result.
Examples:



In above tree we can see that node with index 
0, 2, 3, 5, 6, 7 have distances less than 3
from all the marked nodes.
so answer will be 6



We can solve this problem using breadth first search. Main thing to observe in this problem is that if we find two marked nodes which are at largest distance from each other considering all pairs of marked nodes then if a node is at a distance less than K from both of these two nodes then it will be at a distance less than K from all the marked nodes because these two nodes represents the extreme limit of all marked nodes, if a node lies in this limit then it will be at a distance less than K from all marked nodes otherwise not.
As in above example, node-1 and node-4 are most distant marked node so nodes which are at distance less than 3 from these two nodes will also be at distance less than 3 from node 2 also. Now first distant marked node we can get by doing a bfs from any random node, second distant marked node we can get by doing another bfs from marked node we just found from the first bfs and in this bfs we can also found distance of all nodes from first distant marked node and to find distance of all nodes from second distant marked node we will do one more bfs, so after doing these three bfs we can get distance of all nodes from two extreme marked nodes which can be compared with K to know which nodes fall in K-distance range from all marked nodes.

C++

filter_none

edit
close

play_arrow

link
brightness_4
code

//  C++ program to count nodes inside K distance 
// range from marked nodes
#include <bits/stdc++.h>
using namespace std;
  
// Utility bfs method to fill distance vector and returns
// most distant marked node from node u
int bfsWithDistance(vector<int> g[], bool mark[], int u, 
                                        vector<int>& dis)
{
    int lastMarked;
    queue<int> q;
  
    //  push node u in queue and initialize its distance as 0
    q.push(u);
    dis[u] = 0;
  
    //  loop untill all nodes are processed
    while (!q.empty())
    {
        u = q.front();      q.pop();
        //  if node is marked, update lastMarked variable
        if (mark[u])
            lastMarked = u;
  
        // loop over all neighbors of u and update their 
        // distance before pushing in queue
        for (int i = 0; i < g[u].size(); i++)
        {
            int v = g[u][i];
              
            //  if not given value already
            if (dis[v] == -1)
            {
                dis[v] = dis[u] + 1;
                q.push(v);
            }
        }
    }
    //  return last updated marked value
    return lastMarked;
}
  
// method returns count of nodes which are in K-distance 
// range from marked nodes
int nodesKDistanceFromMarked(int edges[][2], int V, 
                             int marked[], int N, int K)
{
    //  vertices in a tree are one more than number of edges
    V = V + 1;
    vector<int> g[V];
  
    //  fill vector for graph
    int u, v;    
    for (int i = 0; i < (V - 1); i++)
    {
        u = edges[i][0];
        v = edges[i][1];
  
        g[u].push_back(v);
        g[v].push_back(u);
    }
  
    //  fill boolean array mark from marked array
    bool mark[V] = {false};
    for (int i = 0; i < N; i++)
        mark[marked[i]] = true;
  
    //  vectors to store distances
    vector<int> tmp(V, -1), dl(V, -1), dr(V, -1);
  
    //  first bfs(from any random node) to get one 
    // distant marked node
    u = bfsWithDistance(g, mark, 0, tmp);
  
    /*  second bfs to get other distant marked node 
        and also dl is filled with distances from first
        chosen marked node */
    u = bfsWithDistance(g, mark, u, dl);
  
    //  third bfs to fill dr by distances from second 
    // chosen marked node
    bfsWithDistance(g, mark, u, dr);
  
    int res = 0;
    //  loop over all nodes
    for (int i = 0; i < V; i++)
    {
        // increase res by 1, if current node has distance 
        // less than K from both extreme nodes
        if (dl[i] <= K && dr[i] <= K)
            res++;
    }
    return res;
}
  
// Driver code to test above methods
int main()
{
    int edges[][2] = 
    {
        {1, 0}, {0, 3}, {0, 8}, {2, 3},
        {3, 5}, {3, 6}, {3, 7}, {4, 5},
        {5, 9}
    };
    int V = sizeof(edges) / sizeof(edges[0]);
      
    int marked[] = {1, 2, 4};
    int N = sizeof(marked) / sizeof(marked[0]);
  
    int K = 3;
    cout << nodesKDistanceFromMarked(edges, V, marked, N, K);
    return 0;
}

chevron_right


Python3

# Python3 program to count nodes inside
# K distance range from marked nodes
import queue

# Utility bfs method to fill distance
# vector and returns most distant
# marked node from node u
def bfsWithDistance(g, mark, u, dis):
lastMarked = 0
q = queue.Queue()

# push node u in queue and initialize
# its distance as 0
q.put(u)
dis[u] = 0

# loop untill all nodes are processed
while (not q.empty()):
u = q.get()

# if node is marked, update
# lastMarked variable
if (mark[u]):
lastMarked = u

# loop over all neighbors of u and
# update their distance before
# pushing in queue
for i in range(len(g[u])):
v = g[u][i]

# if not given value already
if (dis[v] == -1):
dis[v] = dis[u] + 1
q.put(v)

# return last updated marked value
return lastMarked

# method returns count of nodes which
# are in K-distance range from marked nodes
def nodesKDistanceFromMarked(edges, V, marked, N, K):

# vertices in a tree are one
# more than number of edges
V = V + 1
g = [[] for i in range(V)]

# fill vector for graph
u, v = 0, 0
for i in range(V – 1):
u = edges[i][0]
v = edges[i][1]

g[u].append(v)
g[v].append(u)

# fill boolean array mark from
# marked array
mark = [False] * V
for i in range(N):
mark[marked[i]] = True

# vectors to store distances
tmp = [-1] * V
dl = [-1] * V
dr = [-1] * V

# first bfs(from any random node)
# to get one distant marked node
u = bfsWithDistance(g, mark, 0, tmp)

# second bfs to get other distant
# marked node and also dl is filled
# with distances from first chosen
# marked node
u = bfsWithDistance(g, mark, u, dl)

# third bfs to fill dr by distances
# from second chosen marked node
bfsWithDistance(g, mark, u, dr)

res = 0

# loop over all nodes
for i in range(V):

# increase res by 1, if current node
# has distance less than K from both
# extreme nodes
if (dl[i] <= K and dr[i] <= K): res += 1 return res # Driver Code if __name__ == '__main__': edges = [[1, 0], [0, 3], [0, 8], [2, 3], [3, 5], [3, 6], [3, 7], [4, 5], [5, 9]] V = len(edges) marked = [1, 2, 4] N = len(marked) K = 3 print(nodesKDistanceFromMarked(edges, V, marked, N, K)) # This code is contributed by PranchalK [tabbyending] Output:

6

This article is contributed by Utkarsh Trivedi. If you like GeeksforGeeks and would like to contribute, you can also write an article using contribute.geeksforgeeks.org or mail your article to contribute@geeksforgeeks.org. See your article appearing on the GeeksforGeeks main page and help other Geeks.

Please write comments if you find anything incorrect, or you want to share more information about the topic discussed above.



My Personal Notes arrow_drop_up

Improved By : PranchalK



Article Tags :
Practice Tags :


1


Please write to us at contribute@geeksforgeeks.org to report any issue with the above content.