Print the lexicographically smallest BFS of the graph starting from 1

Given a connected graph with N vertices and M edges. The task is to print the lexicographically smallest BFS traversal of the graph starting from 1.

Note: The vertices are numbered from 1 to N.

Examples:



Input: N = 5, M = 5 
       Edges: 
       1 4
       3 4
       5 4
       3 2
       1 5 
Output: 1 4 3 2 5 
Start from 1, go to 4, then to 3 and then to 2 and to 5. 

Input: N = 3, M = 2 
       Edges: 
       1 2 
       1 3 
Output: 1 2 3 

Approach: Instead of doing a normal BFS traversal on the graph, we can use a priority queue(min heap) instead of a simple queue. When a node is visited add its adjacent nodes into the priority queue. Every time, we visit a new node, it will be the one with the smallest index in the priority queue. Print the nodes when every time we visit them starting from 1.

Below is the implementation of the above approach:

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// C++ program to print the lexcicographically
// smallest path starting from 1
  
#include <bits/stdc++.h>
using namespace std;
  
// Function to print the smallest lexicographically
// BFS path starting from 1
void printLexoSmall(vector<int> adj[], int n)
{
    // Visited array
    bool vis[n + 1];
    memset(vis, 0, sizeof vis);
  
    // Minimum Heap
    priority_queue<int, vector<int>, greater<int> > Q;
  
    // First one visited
    vis[1] = true;
    Q.push(1);
  
    // Iterate till all nodes are visited
    while (!Q.empty()) {
  
        // Get the top element
        int now = Q.top();
  
        // Pop the element
        Q.pop();
  
        // Print the current node
        cout << now << " ";
  
        // Find adjacent nodes
        for (auto p : adj[now]) {
  
            // If not visited
            if (!vis[p]) {
  
                // Push
                Q.push(p);
  
                // Mark as visited
                vis[p] = true;
            }
        }
    }
}
  
// Function to insert edges in the graph
void insertEdges(int u, int v, vector<int> adj[])
{
    adj[u].push_back(v);
    adj[v].push_back(u);
}
  
// Driver Code
int main()
{
    int n = 5, m = 5;
    vector<int> adj[n + 1];
  
    // Insert edges
    insertEdges(1, 4, adj);
    insertEdges(3, 4, adj);
    insertEdges(5, 4, adj);
    insertEdges(3, 2, adj);
    insertEdges(1, 5, adj);
  
    // Function call
    printLexoSmall(adj, n);
  
    return 0;
}

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Output:

1 4 3 2 5


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Striver(underscore)79 at Codechef and codeforces D

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