# Read e-book online A Collection of Graph Programming Interview Questions Solved PDF

By Dr Antonio Gulli

ISBN-10: 1497484464

ISBN-13: 9781497484467

A set of Graph Programming Interview Questions Solved in C++

**Read Online or Download A Collection of Graph Programming Interview Questions Solved in C++ PDF**

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The second DFS is on the transpose graph of the original graph. Each recursive exploration finds a single new strongly connected component. visited[id]) { dfsUtil(g, id, visited); std::cout << "end component id=" << componentID++ << std::endl; } } } Complexity Time complexity is here 14 Covering DFS Trees Given a graph it is possible to define a path tree generated by a DFS visit. If G is not (strongly) connected then we need to restart the DFS from each node and the visit will generate a forest of trees, If the graph is direc,t the visit will respect the direction of the edges, while if the graph is not direct, the visit will produce a random orientation of each edge as a consequence of the direction of the visit.

Rend(); ++pathIterator) { std::cout << nameMap[boost::source(*pathIterator, g)] << " -> " << nameMap[boost::target(*pathIterator, g)] << " = " << boost::get(boost::edge_weight, g, *pathIterator) << std::endl; } std::cout << "Distance: " << distanceMap[v2] << std::endl; } A useful exercise is to use Dijkstra for computing the shortest path on the graph represented in picture. Complexity Time complexity is , if the priority queue is implemented using Fibonacci heaps. Fibonacci heap is a priority queue which implements extracting minimum efficiently.

Code This implementation is left as an exercise. Complexity The complexity here is the same as DFS. 7 Given two nodes in an undirected graph, find the path connecting them Solution A simple solution is to visit the graph with DFS order starting from one of the two nodes and stopping the visit as soon as the other node is found. During this process the visited nodes are stored into a stack. At the end the stack will contain the desired path. Code This implementation is left as an exercise. Complexity The complexity is the same as DFS.

### A Collection of Graph Programming Interview Questions Solved in C++ by Dr Antonio Gulli

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