Create a spanning tree using the breadth-first search algorithm. Start at A (ie 0). What is the maximum number of

edges needed to connect any other vertex to vertex 'A'?

vector <int> v[10] ;   //Vector for maintaining adjacency list explained above

   int level[10]; //To determine the level of each node

   bool vis[10]; //Mark the node if visited  

   void bfs(int s) {

       queue <int> q;

       q.push(s);

       level[ s ] = 0 ;  //Setting the level of the source node as 0

       vis[ s ] = true;

       while(!q.empty())

       {

           int p = q.front();

           q.pop();

           for(int i = 0;i < v[ p ].size() ; i++)

           {

               if(vis[ v[ p ][ i ] ] == false)

               {

           //Setting the level of each node with an increment in the level of parent node

                   level[ v[ p ][ i ] ] = level[ p ]+1;                  

                    q.push(v[ p ][ i ]);

                    vis[ v[ p ][ i ] ] = true;

     }

           }

       }

   }

{x = -1/2 ,   y = -1

step-by-step explanation:

solve the following system:

{2 x + y = -2

{10 x - 3 y = -2

in the first equation, look to solve for y:

{2 x + y = -2

{10 x - 3 y = -2

subtract 2 x from both sides:

{y = -2 x - 2

{10 x - 3 y = -2

substitute y = -2 x - 2 into the second equation:

{y = -2 x - 2

{10 x - 3 (-2 x - 2) = -2

10 x - 3 (-2 x - 2) = (6 x + 6) + 10 x = 16 x + 6:

{y = -2 x - 2

{16 x + 6 = -2

in the second equation, look to solve for x:

{y = -2 x - 2

{16 x + 6 = -2

subtract 6 from both sides:

{y = -2 x - 2

{16 x = -8

divide both sides by 16:

{y = -2 x - 2

{x = -1/2

substitute x = -1/2 into the first equation:

{y = -1

{x = -1/2

collect results in alphabetical order:

answer:   {x = -1/2 ,   y = -1