📜  Kruskal算法(邻接矩阵的简单实现)

📅  最后修改于: 2021-05-31 17:04:19             🧑  作者: Mango

以下是使用Kruskal算法查找MST的步骤

在之前的文章中,我们讨论了Kruskal算法的一种实现。在这篇文章中,讨论了邻接矩阵的一种更简单的实现。

C++
// Simple C++ implementation for Kruskal's
// algorithm
#include 
using namespace std;
  
#define V 5
int parent[V];
  
// Find set of vertex i
int find(int i)
{
    while (parent[i] != i)
        i = parent[i];
    return i;
}
  
// Does union of i and j. It returns
// false if i and j are already in same
// set.
void union1(int i, int j)
{
    int a = find(i);
    int b = find(j);
    parent[a] = b;
}
  
// Finds MST using Kruskal's algorithm
void kruskalMST(int cost[][V])
{
    int mincost = 0; // Cost of min MST.
  
    // Initialize sets of disjoint sets.
    for (int i = 0; i < V; i++)
        parent[i] = i;
  
    // Include minimum weight edges one by one
    int edge_count = 0;
    while (edge_count < V - 1) {
        int min = INT_MAX, a = -1, b = -1;
        for (int i = 0; i < V; i++) {
            for (int j = 0; j < V; j++) {
                if (find(i) != find(j) && cost[i][j] < min) {
                    min = cost[i][j];
                    a = i;
                    b = j;
                }
            }
        }
  
        union1(a, b);
        printf("Edge %d:(%d, %d) cost:%d \n",
               edge_count++, a, b, min);
        mincost += min;
    }
    printf("\n Minimum cost= %d \n", mincost);
}
  
// driver program to test above function
int main()
{
    /* Let us create the following graph
          2    3
      (0)--(1)--(2)
       |   / \   |
      6| 8/   \5 |7
       | /     \ |
      (3)-------(4)
            9          */
    int cost[][V] = {
        { INT_MAX, 2, INT_MAX, 6, INT_MAX },
        { 2, INT_MAX, 3, 8, 5 },
        { INT_MAX, 3, INT_MAX, INT_MAX, 7 },
        { 6, 8, INT_MAX, INT_MAX, 9 },
        { INT_MAX, 5, 7, 9, INT_MAX },
    };
  
    // Print the solution
    kruskalMST(cost);
  
    return 0;
}


Java
// Simple Java implementation for Kruskal's
// algorithm
import java.util.*;
  
class GFG 
{
  
static int V = 5;
static int[] parent = new int[V];
static int INF = Integer.MAX_VALUE;
  
// Find set of vertex i
static int find(int i)
{
    while (parent[i] != i)
        i = parent[i];
    return i;
}
  
// Does union of i and j. It returns
// false if i and j are already in same
// set.
static void union1(int i, int j)
{
    int a = find(i);
    int b = find(j);
    parent[a] = b;
}
  
// Finds MST using Kruskal's algorithm
static void kruskalMST(int cost[][])
{
    int mincost = 0; // Cost of min MST.
  
    // Initialize sets of disjoint sets.
    for (int i = 0; i < V; i++)
        parent[i] = i;
  
    // Include minimum weight edges one by one
    int edge_count = 0;
    while (edge_count < V - 1)
    {
        int min = INF, a = -1, b = -1;
        for (int i = 0; i < V; i++)
        {
            for (int j = 0; j < V; j++) 
            {
                if (find(i) != find(j) && cost[i][j] < min) 
                {
                    min = cost[i][j];
                    a = i;
                    b = j;
                }
            }
        }
  
        union1(a, b);
        System.out.printf("Edge %d:(%d, %d) cost:%d \n",
            edge_count++, a, b, min);
        mincost += min;
    }
    System.out.printf("\n Minimum cost= %d \n", mincost);
}
  
// Driver code
public static void main(String[] args) 
{
/* Let us create the following graph
        2 3
    (0)--(1)--(2)
    | / \ |
    6| 8/ \5 |7
    | /     \ |
    (3)-------(4)
            9         */
    int cost[][] = {
        { INF, 2, INF, 6, INF },
        { 2, INF, 3, 8, 5 },
        { INF, 3, INF, INF, 7 },
        { 6, 8, INF, INF, 9 },
        { INF, 5, 7, 9, INF },
    };
  
    // Print the solution
    kruskalMST(cost);
    }
}
  
// This code contributed by Rajput-Ji


Python3
# Python implementation for Kruskal's
# algorithm
  
# Find set of vertex i
def find(i):
    while parent[i] != i:
        i = parent[i]
    return i
  
# Does union of i and j. It returns
# false if i and j are already in same 
# set. 
def union(i, j):
    a = find(i)
    b = find(j)
    parent[a] = b
  
# Finds MST using Kruskal's algorithm 
def kruskalMST(cost):
    mincost = 0 # Cost of min MST
  
    # Initialize sets of disjoint sets
    for i in range(V):
        parent[i] = i
  
    # Include minimum weight edges one by one 
    edge_count = 0
    while edge_count < V - 1:
        min = INF
        a = -1
        b = -1
        for i in range(V):
            for j in range(V):
                if find(i) != find(j) and cost[i][j] < min:
                    min = cost[i][j]
                    a = i
                    b = j
        union(a, b)
        print('Edge {}:({}, {}) cost:{}'.format(edge_count, a, b, min))
        edge_count += 1
        mincost += min
  
    print("Minimum cost= {}".format(mincost))
  
  
# Driver code 
# Let us create the following graph 
#         2 3 
#     (0)--(1)--(2) 
#     | / \ | 
#     6| 8/ \5 |7 
#     | /     \ | 
#     (3)-------(4) 
#         9
  
V = 5
parent = [i for i in range(V)]
INF = float('inf')
cost = [[INF, 2, INF, 6, INF],
        [2, INF, 3, 8, 5],
        [INF, 3, INF, INF, 7],
        [6, 8, INF, INF, 9],
        [INF, 5, 7, 9, INF]]
  
# Print the solution 
kruskalMST(cost)
  
# This code is contributed by ng24_7


C#
// Simple C# implementation for Kruskal's
// algorithm
using System;     
  
class GFG 
{
  
static int V = 5;
static int[] parent = new int[V];
static int INF = int.MaxValue;
  
// Find set of vertex i
static int find(int i)
{
    while (parent[i] != i)
        i = parent[i];
    return i;
}
  
// Does union of i and j. It returns
// false if i and j are already in same
// set.
static void union1(int i, int j)
{
    int a = find(i);
    int b = find(j);
    parent[a] = b;
}
  
// Finds MST using Kruskal's algorithm
static void kruskalMST(int [,]cost)
{
    int mincost = 0; // Cost of min MST.
  
    // Initialize sets of disjoint sets.
    for (int i = 0; i < V; i++)
        parent[i] = i;
  
    // Include minimum weight edges one by one
    int edge_count = 0;
    while (edge_count < V - 1)
    {
        int min = INF, a = -1, b = -1;
        for (int i = 0; i < V; i++)
        {
            for (int j = 0; j < V; j++) 
            {
                if (find(i) != find(j) && cost[i, j] < min) 
                {
                    min = cost[i, j];
                    a = i;
                    b = j;
                }
            }
        }
  
        union1(a, b);
        Console.Write("Edge {0}:({1}, {2}) cost:{3} \n",
            edge_count++, a, b, min);
        mincost += min;
    }
    Console.Write("\n Minimum cost= {0} \n", mincost);
}
  
// Driver code
public static void Main(String[] args) 
{
/* Let us create the following graph
        2 3
    (0)--(1)--(2)
    | / \ |
    6| 8/ \5 |7
    | /     \ |
    (3)-------(4)
            9         */
    int [,]cost = {
        { INF, 2, INF, 6, INF },
        { 2, INF, 3, 8, 5 },
        { INF, 3, INF, INF, 7 },
        { 6, 8, INF, INF, 9 },
        { INF, 5, 7, 9, INF },
    };
  
    // Print the solution
    kruskalMST(cost);
}
}
  
/* This code contributed by PrinciRaj1992 */


输出:
Edge 0:(0, 1) cost:2 
Edge 1:(1, 2) cost:3 
Edge 2:(1, 4) cost:5 
Edge 3:(0, 3) cost:6 

 Minimum cost= 16

请注意,上述解决方案效率不高。这个想法是为邻接矩阵表示提供一个简单的实现。请参阅下面的有效实施。
Kruskal的最小生成树算法贪婪的算法2
C++中使用STL的Kruskal最小生成树

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