BFS traversal of graph

PROBLEM :

Write a function to print the bredth first traversal for a graph from a given source s.

Input:
The task is to complete the function BFS which takes 3 arguments an integer denoting the starting node (s) of the bfs travel , a  graph (g)  and an array of visited nodes (vis)  which are initially all set to false .
There are multiple test cases. For each test case, this method will be called individually.

Output:
The function should print the breath first traversal for the graph from the given source.

Note : The expected output button always produces BFS starting from node 1.

Constraints:
1 <=T<= 100
1 <=Number of  edges<= 100

Example:
Input:
1
4
1 2 1 3 1 4 3 5

Output:
1 2 3 4 5    //bfs from node 1

There is  one test cases.  First line of each test case represent an integer N denoting no of edges and then in the next line N pairs of values a and b are fed which denotes there is a unidirectional edge from a to b .

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SIMPLE c++ IMPLEMENTATION :
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void bfs(int s,vector<int> adj[],bool vis[])
{
    queue<int> que ;
    que.push(s) ;
    vis[s]=true ;
   
    while(!que.empty())
    {
        int curr=que.front() ;
        que.pop() ;
       
        cout<<curr<<" " ;
        vis[curr]=true ;
       
        vector<int>::iterator it ;
        for(it=adj[curr].begin();it!=adj[curr].end();it++)
        {
            if(!vis[*it])
            {
                vis[*it]=true ;
                que.push(*it) ;
            }
        }
    }
}

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