BFS using STL for competitive coding

Last Updated : 19 May, 2024

A STL based simple implementation of BFS using queue and vector in STL. The adjacency list is represented using vectors of vector.

In BFS, we start with a node.

Create a queue and enqueue source into it.
Mark source as visited.
While queue is not empty, do following
Dequeue a vertex from queue. Let this be f.
Print f
Enqueue all not yet visited adjacent of f and mark them visited.

Below is an example BFS starting from source vertex 1. Note that there can be multiple BFSs possible for a graph (even from a particular vertex).

For more details of BFS, refer this post .
The code here is simplified such that it could be used in competitive coding.

Implementation:

CPP

// A Quick implementation of BFS using
// vectors and queue
#include <bits/stdc++.h>
#define pb push_back

using namespace std;

vector<bool> v;
vector<vector<int> > g;

void edge(int a, int b)
{
    g[a].pb(b);

    // for undirected graph add this line
    // g[b].pb(a);
}

void bfs(int u)
{
    queue<int> q;

    q.push(u);
    v[u] = true;

    while (!q.empty()) {

        int f = q.front();
        q.pop();

        cout << f << " ";

        // Enqueue all adjacent of f and mark them visited
        for (auto i = g[f].begin(); i != g[f].end(); i++) {
            if (!v[*i]) {
                q.push(*i);
                v[*i] = true;
            }
        }
    }
}

// Driver code
int main()
{
    int n, e;
    cin >> n >> e;

    v.assign(n, false);
    g.assign(n, vector<int>());

    int a, b;
    for (int i = 0; i < e; i++) {
        cin >> a >> b;
        edge(a, b);
    }

    for (int i = 0; i < n; i++) {
        if (!v[i])
            bfs(i);
    }

    return 0;
}

Java

import java.util.ArrayDeque;
import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;
import java.util.Queue;

public class Main {
    static void bfs(Map<Integer, List<Integer> > graph,
                    int root)
    {
        // Set to keep track of visited vertices
        boolean[] visited = new boolean[graph.size()];
        // Queue for BFS traversal
        Queue<Integer> queue = new ArrayDeque<>();

        // Enqueue the root vertex
        queue.add(root);
        // Mark root as visited
        visited[root] = true;

        // BFS traversal
        while (!queue.isEmpty()) {
            // Dequeue a vertex from the queue
            int vertex = queue.poll();
            System.out.print(vertex + " ");

            // Visit all adjacent vertices of the dequeued
            // vertex
            for (int neighbour : graph.getOrDefault(
                     vertex, new ArrayList<>())) {
                // If neighbour has not been visited, mark
                // it as visited and enqueue it
                if (!visited[neighbour]) {
                    visited[neighbour] = true;
                    queue.add(neighbour);
                }
            }
        }
    }

    public static void main(String[] args)
    {
        // Create a map to use as an adjacency list
        Map<Integer, List<Integer> > graph
            = new HashMap<>();

        // Define the edges
        int[][] edges
            = { { 0, 1 }, { 0, 2 }, { 0, 3 }, { 0, 4 },
                { 1, 5 }, { 2, 5 }, { 3, 6 }, { 4, 6 },
                { 5, 7 }, { 6, 7 } };

        // Create the graph
        for (int[] edge : edges) {
            int a = edge[0];
            int b = edge[1];
            graph.computeIfAbsent(a, k -> new ArrayList<>())
                .add(b);
            graph.computeIfAbsent(b, k -> new ArrayList<>())
                .add(a);
        }

        // Perform BFS starting from vertex 0
        System.out.println("BFS starting from vertex 0:");
        bfs(graph, 0);
    }
}

Python

from collections import defaultdict, deque


def bfs(graph, root):
    visited = set()
    queue = deque([root])

    while queue:
        # Dequeue a vertex from queue
        vertex = queue.popleft()
        print(vertex, end=" ")

        # If not visited, mark it as visited, and enqueue it
        for neighbour in graph[vertex]:
            if neighbour not in visited:
                queue.append(neighbour)
                visited.add(neighbour)


# Create a dictionary to use as an adjacency list
graph = defaultdict(list)

edges = [(0, 1), (0, 2), (0, 3), (0, 4), (1, 5),
         (2, 5), (3, 6), (4, 6), (5, 7), (6, 7)]

# Create the graph
for edge in edges:
    a, b = edge
    graph[a].append(b)
    graph[b].append(a)

# Perform BFS
print("BFS starting from vertex 0:")
bfs(graph, 0)

JavaScript

function bfs(graph, root) {
    // Set to keep track of visited vertices
    let visited = new Array(Object.keys(graph).length).fill(false);
    // Queue for BFS traversal
    let queue = [];

    // Enqueue the root vertex
    queue.push(root);
    // Mark root as visited
    visited[root] = true;

    // BFS traversal
    while (queue.length !== 0) {
        // Dequeue a vertex from the queue
        let vertex = queue.shift();
        process.stdout.write(vertex + " ");

        // Visit all adjacent vertices of the dequeued vertex
        for (let neighbour of graph[vertex] || []) {
            // If neighbour has not been visited, mark it as visited and enqueue it
            if (!visited[neighbour]) {
                visited[neighbour] = true;
                queue.push(neighbour);
            }
        }
    }
}

// Define the edges
let edges = [
    [0, 1], [0, 2], [0, 3], [0, 4],
    [1, 5], [2, 5], [3, 6], [4, 6],
    [5, 7], [6, 7]
];

// Create the graph
let graph = {};
for (let edge of edges) {
    let [a, b] = edge;
    graph[a] = graph[a] || [];
    graph[b] = graph[b] || [];
    graph[a].push(b);
    graph[b].push(a);
}

// Perform BFS starting from vertex 0
console.log("BFS starting from vertex 0:");
bfs(graph, 0);

Input:
8 10
0 1
0 2
0 3
0 4
1 5
2 5
3 6
4 6
5 7
6 7

Output:
0 1 2 3 4 5 6 7

Time Complexity: O(V+E) – we traverse all vertices at least once and check every edge.
Auxiliary Space: O(V) – for using a queue to store vertices.

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