prim's algorithm for minimum spanning tree code example

Example 1: prim's algorithm python

def empty_graph(n):
    res = []
    for i in range(n):
        res.append([0]*n)
    return res
def convert(graph):
    matrix = []
    for i in range(len(graph)): 
        matrix.append([0]*len(graph))
        for j in graph[i]:
            matrix[i][j] = 1
    return matrix
def prims_algo(graph):
    graph1 = convert(graph)
    n = len(graph1)
    tree = empty_graph(n)
    con =[0]
    while len(con) < n :
        found = False
        for i in con:
            for j in range(n):
                if j not in con and graph1[i][j] == 1:
                    tree[i][j] =1
                    tree[j][i] =1
                    con += [j]
                    found  = True
                    break
            if found :
                break
    return tree
matrix = [[0, 1, 1, 1, 0, 1, 1, 0, 0],
          [1, 0, 0, 1, 0, 0, 1, 1, 0],
          [1, 0, 0, 1, 0, 0, 0, 0, 0],
          [1, 1, 1, 0, 1, 0, 0, 0, 0],
          [0, 0, 0, 1, 0, 1, 0, 0, 1],
          [1, 0, 0, 0, 1, 0, 0, 0, 1],
          [1, 1, 0, 0, 0, 0, 0, 0, 0],
          [0, 1, 0, 0, 0, 0, 0, 0, 0],
          [0, 0, 0, 0, 1, 1, 0, 0, 0]]

lst = [[1,2,3,5,6],[0,3,6,7],[0,3],[0,1,2,4],[3,5,8],[0,4,8],[0,1],[1],[4,5]]
print("From graph to spanning tree:\n")
print(prims_algo(lst))

Example 2: prims c++

#include <iostream>
#include <vector>
#include <queue>
#include <functional>
#include <utility>

using namespace std;
const int MAX = 1e4 + 5;
typedef pair<long long, int> PII;
bool marked[MAX];
vector <PII> adj[MAX];

long long prim(int x)
{
    priority_queue<PII, vector<PII>, greater<PII> > Q;
    int y;
    long long minimumCost = 0;
    PII p;
    Q.push(make_pair(0, x));
    while(!Q.empty())
    {
        // Select the edge with minimum weight
        p = Q.top();
        Q.pop();
        x = p.second;
        // Checking for cycle
        if(marked[x] == true)
            continue;
        minimumCost += p.first;
        marked[x] = true;
        for(int i = 0;i < adj[x].size();++i)
        {
            y = adj[x][i].second;
            if(marked[y] == false)
                Q.push(adj[x][i]);
        }
    }
    return minimumCost;
}

int main()
{
    int nodes, edges, x, y;
    long long weight, minimumCost;
    cin >> nodes >> edges;
    for(int i = 0;i < edges;++i)
    {
        cin >> x >> y >> weight;
        adj[x].push_back(make_pair(weight, y));
        adj[y].push_back(make_pair(weight, x));
    }
    // Selecting 1 as the starting node
    minimumCost = prim(1);
    cout << minimumCost << endl;
    return 0;
}

Example 3: prims minimum spanning tree

import math
def empty_tree (n):
    lst = []
    for i in range(n):
        lst.append([0]*n)
    return lst
def min_extension (con,graph,n):
    min_weight = math.inf
    for i in con:
        for j in range(n):
            if j not in con and 0 < graph[i][j] < min_weight:
                min_weight = graph[i][j]
                v,w = i,j
    return v,w
            
def min_span(graph):
    con = [0]
    n = len(graph)
    tree = empty_tree(n)
    while len(con) < n :
        i ,j  = min_extension(con,graph,n)
        tree[i][j],tree[j][i] = graph[i][j], graph[j][i]
        con += [j]
    return tree

def find_weight_of_edges(graph):
    tree = min_span(graph)
    lst = []
    lst1 = []
    x = 0
    for i in tree:
        lst += i 
    for i in lst:
        if i not in lst1:
            lst1.append(i)
            x += i
    return x

graph = [[0,1,0,0,0,0,0,0,0],
         [1,0,3,4,0,3,0,0,0],
         [0,3,0,0,0,4,0,0,0],
         [0,4,0,0,2,9,1,0,0],
         [0,0,0,2,0,6,0,0,0],
         [0,3,4,9,6,0,0,0,6],
         [0,0,0,1,0,0,0,2,8],
         [0,0,0,0,0,0,2,0,3],
         [0,0,0,0,0,6,8,3,0]]
graph1 = [[0,3,5,0,0,6],
          [3,0,4,1,0,0],
          [5,4,0,4,5,2],
          [0,1,4,0,6,0],
          [0,0,5,6,0,8],
          [6,0,2,0,8,0]]
print(min_span(graph1))
print("Total weight of the tree is: " + str(find_weight_of_edges(graph1)))