以下是使用 Kruskal 算法查找 MST 的步骤
1. Sort all the edges in non-decreasing order of their weight.
2. Pick the smallest edge. Check if it forms a cycle with the spanning tree formed so far. If cycle is not formed, include this edge. Else, discard it.
3. Repeat step#2 until there are (V-1) edges in the spanning tree.
我们在上一篇文章中讨论了 Kruskal 算法的一种实现。在这篇文章中，讨论了一种更简单的邻接矩阵实现。
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-JiPython3
# 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_7C#
// 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 */Javascript
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现。
Kruskal 的最小生成树算法 |贪婪算法2
在 C++ 中使用 STL 的 Kruskal 最小生成树
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