find min spanning tree code example

Example 1: find the graph is minimal spanig tree or not

#include 
#include 
#include 
#include 

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

void initialize()
{
    for(int i = 0;i < MAX;++i)
        id[i] = i;
}

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 > p[])
{
    int x, y;
    long long cost, minimumCost = 0;
    for(int i = 0;i < edges;++i)
    {
        // Selecting edges one by one in increasing order from the beginning
        x = p[i].second.first;
        y = p[i].second.second;
        cost = p[i].first;
        // Check if the selected edge is creating a cycle or not
        if(root(x) != root(y))
        {
            minimumCost += cost;
            union1(x, y);
        }    
    }
    return minimumCost;
}

int main()
{
    int x, y;
    long long weight, cost, minimumCost;
    initialize();
    cin >> nodes >> edges;
    for(int i = 0;i < edges;++i)
    {
        cin >> x >> y >> weight;
        p[i] = make_pair(weight, make_pair(x, y));
    }
    // Sort the edges in the ascending order
    sort(p, p + edges);
    minimumCost = kruskal(p);
    cout << minimumCost << endl;
    return 0;
}

Example 2: kruskal's algorithm

#include

using namespace std;

int  main()
{
	int n = 9;
	
	int mat[9][9] = {
	{100,4,100,100,100,100,100,8,100},
	{4,100,8,100,100,100,100,100,100},
	{100,8,100,7,100,4,100,100,2},
	{100,100,7,100,9,14,100,100,100},
	{100,100,100,9,100,10,100,100,100},
	{100,100,4,14,10,100,2,100,100},
	{100,100,100,100,100,2,100,1,6},
	{8,100,100,100,100,100,1,100,7},
	{100,100,2,100,100,100,6,7,100}};
	
	int parent[n];
	
	int edges[100][3];
	int count = 0;
	
	for(int i=0;i edges[j+1][2])
				{
					int t1=edges[j][0], t2=edges[j][1], t3=edges[j][2];
					
					edges[j][0] = edges[j+1][0];
					edges[j][1] = edges[j+1][1];
					edges[j][2] = edges[j+1][2];
					
					edges[j+1][0] = t1;
					edges[j+1][1] = t2;
					edges[j+1][2] = t3;
				}
				
	int mst[n-1][2];
	int mstVal = 0;
	int l = 0;
	
	cout< "<