By Dr Antonio Gulli

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

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Solution The here implemented solution is the Kosaraju’s algorithm[3] which uses two passes of DFS. The first pass is carried out on the original graph and it is used to choose the order in which the outer loop of the second DFS tests are connected to the vertices. 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.

Solution The pseudo code uses UNION-FIND operations[9] : Find: Determine which subset a particular element is in. It is used for determining, if two elements are in the same subset. Union: Join two subsets into a single subset. hpp As an exercise try to compute Kruskal on the following graph Complexity Time complexity is because we use union-find data structure to keep the set information,whereFind operations on elements require time. 19 Find the longest path in a DAG The problem of finding the longest path is NP-hard for a general graph.

For each interval with define and h: } and introduce and edge . if and only if . Then we assign the weight. It can be verified that the shortest path on defines a dominant set of intervals. Complexity G is a DAG and can be built in and therefore the shortest path can be computed in . 25 Weighted dominant set of intervals Given a set of n intervals in the real line where each interval has an associated weight integer and a non-negative. Find a dominant set with minimum weight. Solution Left as an exercise 26 Weighted shortest paths with cost associated to the nodes Suppose that the cost is associated to the nodes instead that to the edges and that therefore the cost of a path is the sum of all the nodes it contains.