It is just that i am unable to understand how complexity(time) of bfs/dfs are O(v+e).
why not O(v*e)???since most of the times i have seen we just multiply inner loop and outer loop iterations and compute complexity so what is the difference we are having here.
P.S: I have already seen link: http://stackoverflow.com/questions/6850357/explanation-of-runtimes-of-bfs-and-dfs
and CLRS but didn’t get it.
I think u didn’t go through the link contain correct explaination why the time complexity of dfs and bfs is O(v+e) hope this help 
DFS(analysis):
*Setting/getting a vertex/edge label takes O(1) time
*Each vertex is labeled twice
->once as UNEXPLORED
->once as VISITED
*Each edge is labeled twice
->once as UNEXPLORED
->once as DISCOVERY or BACK
*Method incidentEdges is called once for each vertex
*DFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
*Recall that Σv deg(v) = 2m
BFS(analysis):
*Setting/getting a vertex/edge label takes O(1) time
*Each vertex is labeled twice
–>once as UNEXPLORED
–>once as VISITED
*Each edge is labeled twice
–>once as UNEXPLORED
–>once as DISCOVERY or CROSS
*Each vertex is inserted once into a sequence Li
*Method incidentEdges is called once for each vertex
*BFS runs in O(n + m) time provided the graph is represented by the adjacency list structure
*Recall that Σv deg(v) = 2m