prim's algorithm minimum spanning tree java code 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;
}

Tags:

C Example