Problem of the day
class graphNode
{
public:
int data;
vector<graphNode*> neighbours;
}
1. Nodes are numbered from 1 to N.
2. Your solution will run on multiple test cases. If you are using global variables make sure to clear them.
The first line of input contains an integer 'T' representing the number of the test case. Then the test cases are as follows.
The first line of each test case contains a single integer ‘N’ representing the number of nodes in the graph.
The second line of each test case contains a single integer ‘M’ representing the number of edges.
The next ‘M’ lines in each test case contain two integers ‘U’ and ‘V’ separated by a single space denoting an undirected edge between nodes U and V.
For each test case, print a single line containing "true" if the graph is cloned correctly otherwise it will print "false".
The output of each test case will be printed in a separate line.
You do not need to print anything; It has already been taken care of. Just implement the given function.
1 <= T <= 5
2 <= N <= 100000
1 <= M <= min(N(N-1)/2,100000)
1 <= E[i][0], E[i][1] <= N
Where ‘N’ is the number of nodes in the given graph, ‘M’ denotes the number of edges and ‘E’ denotes the edge matrix.
Time Limit: 1 sec.
2
5
6
1 2
4 1
2 4
3 4
5 2
1 3
3
2
1 2
1 3
true
true
In the first test case, the returned graph contains 5 nodes and 6 edges which are:
1 2
4 1
2 4
3 4
5 2
1 3
Since it is similar to the given graph with different address nodes then the solution is correct.
In the second test case, the returned graph contains 3 nodes and 2 edges which are:
1 2
1 3
Since it is similar to the given graph with different address nodes then the solution is correct.
2
5
4
1 2
2 3
3 4
4 5
2
1
1 2
true
true
In the first test case, the returned graph contains 5 nodes and 4 edges which are:
1 2
2 3
3 4
4 5
Since it is similar to the given graph with different address nodes then the solution is correct.
In the second test case, the returned graph contains 2 nodes and 1 edge which is:
1 2
Since it is similar to the given graph with different address nodes then the solution is correct.