88 lines
2.3 KiB
C++
88 lines
2.3 KiB
C++
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// Program to detect whether a graph contains a cycle. The vertices of the graph are
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// placed in disjoint sets. As each edge is iterated through, the subset of the two
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// vertices is determined. If the subsets are the same, then a cycle is found.
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// Otherwise, the algorithm will join the two subsets together and repeat the process
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// until each edge is iterated through or a cycle is found.
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#include <algorithm>
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#include <iostream>
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#include <unordered_set>
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#include <utility>
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#include <vector>
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// Edge list implementation of unweighted, undirected graph
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class Graph
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{
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private:
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std::vector<std::pair<int, int>> edgeList;
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std::unordered_set<int> uniqueVertices; // Keep track of the number of unique vertices
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public:
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void insertEdge(int from, int to);
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bool isCycle();
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int findSet(std::vector<int>& parent, int i);
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void unionSet(std::vector<int>& parent, int x, int y);
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};
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void Graph::insertEdge(int from, int to)
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{
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edgeList.push_back(std::make_pair(from, to));
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uniqueVertices.insert(from);
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uniqueVertices.insert(to);
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}
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bool Graph::isCycle()
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{
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// Initialize parent vector as disjoint sets
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std::vector<int> parent(uniqueVertices.size(), -1);
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// Iterate through all graph edges
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for (auto i : edgeList)
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{
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// Find the subset of both vertices of an edge
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int x = findSet(parent, i.first);
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int y = findSet(parent, i.second);
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// If the subsets are the same, then there is a cycle in the graph
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if (x == y)
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{
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return true;
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}
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unionSet(parent, x, y);
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}
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return false;
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}
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int Graph::findSet(std::vector<int>& parent, int i)
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{
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if (parent[i] == -1)
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{
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return i;
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}
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return findSet(parent, parent[i]);
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}
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void Graph::unionSet(std::vector<int>& parent, int x, int y)
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{
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parent[x] = y;
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}
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// Sample test case
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// Time complexity is O(E) where E is the number of edges
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int main()
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{
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Graph graph;
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graph.insertEdge(0, 1);
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graph.insertEdge(0, 2);
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graph.insertEdge(1, 2);
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if (graph.isCycle())
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{
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std::cout << "Graph contains a cycle" << std::endl;
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}
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else{
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std::cout << "Graph does not contain a cycle" << std::endl;
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}
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return 0;
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}
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