kruskals algo

#include<iostream.h>
#include<conio.h>
#include<stdlib.h>
int cost[10][10],i,j,k,n,m,c,visit,visited[10],l,v,count,count1,vst,p;

int main()
{
    int dup1,dup2;
    cout<<"enter no of vertices:\n";
    cin>>n;
    cout<<"enter no of edges:\n";
    cin>>m;
    cout<<"EDGE Cost:\n";
    for(k=1;k<=m;k++)
    {
        cin>>i>>j>>c;
        cost[i][j]=c;
        cost[j][i]=c;
    }
    for(i=1;i<=n;i++)
        for(j=1;j<=n;j++)
            if(cost[i][j]==0)
                cost[i][j]=31999;
            visit=0;
            while(visit<n)
            {
                v=31999;
                for(i=1;i<=n;i++)
                    for(j=1;j<=n;j++)
                        if(cost[i][j]!=31999 && cost[i][j]<v  && cost[i][j]!=-1 )
                        {
                            int count =0;
                            for(p=1;p<=n;p++)
                            {
                                if(visited[p]==i || visited[p]==j)
                                    count++;
                            }
                            if(count >= 2)
                            {
                                for(p=1;p<=n;p++)
                                    if(cost[i][p]!=31999 && p!=j)
                                        dup1=p;
                                    for(p=1;p<=n;p++)
                                        if(cost[j][p]!=31999 && p!=i)
                                            dup2=p;

                                    if(cost[dup1][dup2]==-1)
                                        continue;
                            }
                            l=i;
                            k=j;
                            v=cost[i][j];
                        }
                        cout<<"edge from "<<l<<"-->"<<k<<endl;
                        cost[l][k]=-1;
                        cost[k][l]=-1;
                        visit++;
                        int count=0;
                        count1=0;
                        for(i=1;i<=n;i++)
                        {
                            if(visited[i]==l)
                                count++;
                            if(visited[i]==k)
                                count1++;
                        }
                        if(count==0)
                            visited[++vst]=l;
                        if(count1==0)
                            visited[++vst]=k;
            }
            return 0;
}
OUTPUT:
enter no of vertices:
6
enter no of edges:
10
EDGE Cost:
1 2 12
2 4 2
2 3 1
3 4 6
1 5 15
1 6 17
2 6 3
4 5 14
4 6 10
5 6 19
edge from 2-->3
edge from 2-->4
edge from 2-->6
edge from 4-->6
edge from 1-->2
edge from 4-->5
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