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Minimum Spanning Tree

What is a Spanning Tree?

Given an undirected and connected graph G=(V, E), a spanning tree of the graph G is a tree that spans G (that is, it includes every vertex of G) and is a subgraph of G (every edge in the tree belongs to G)

What is a Minimum Spanning Tree?

The cost of the spanning tree is the sum of the weights of all the edges in the tree. There can be many spanning trees. The minimum spanning tree is the spanning tree where the cost is minimum among all the spanning trees. There also can be many minimum spanning trees.

Kruskal’s Algorithm

Kruskal’s Algorithm builds the spanning tree by adding edges one by one into a growing spanning tree. Kruskal's algorithm follows a greedy approach as in each iteration it finds an edge that has the least weight and adds it to the growing spanning tree.

Algorithm Steps:

  • Sort the graph edges with respect to their weights.

  • Start adding edges to the MST from the edge with the smallest weight until the edge of the largest weight.

  • Only add edges that don't form a cycle, edges that connect only disconnected components.

So now the question is how to check if 2 vertices are connected or not?

This could be done using DFS which starts from the first vertex, then check if the second vertex is visited or not. But DFS will make time complexity large as it has an order of O(V+E) where V is the number of vertices, E is the number of edges. So the best solution is "Disjoint Sets": Disjoint sets are sets whose intersection is the empty set so it means that they don't have any element in common.

minimum spanning tree with total cost 11 ( = 1 + 2 + 3 + 5).

C++ code:

using namespace std;
const int MAX = 1e4 + 5;
int id[MAX], nodes, edges;
pair <long long, pair<int, int> > p[MAX];

int root(int x)
    while(id[x] != x)
        id[x] = id[id[x]];
        x = id[x];
    return x;

void union1(int x, int y)
    int p = root(x);
    int q = root(y);
    id[p] = id[q];
long long kruskal(pair<long long, pair<int, int> > p[])
    int x, y;
    long long cost, minimumCost = 0;
    for(int i = 0;i < edges;++i)
        x = p[i].second.first;
        y = p[i].second.second;
        cost = p[i].first;
        if(root(x) != root(y))
            minimumCost += cost;  //can add any behaviour
            union1(x, y);
    return minimumCost;

int main()
    int x, y;
    long long weight, cost, minimumCost;
    for(int i = 0;i < MAX;++i)
        id[i] = i;
    cin >> nodes >> edges;
    for(int i = 0;i < edges;++i)
        cin >> x >> y >> weight;
        p[i] = make_pair(weight, make_pair(x, y));
    sort(p, p + edges);
    minimumCost = kruskal(p);
    cout << minimumCost << endl;
    return 0;

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