Kruskal算法来计算无向连通图的最小生成树边集合数量存在相同边，此时作为两种

感谢指出，您是正确的。在Kruskal算法中，可能存在相同权重的边。在计算不同选择边的方法数量时，应考虑到这一点。

以下是修改后的Java代码示例，使用Kruskal算法来计算无向连通图的最小生成树边集合数量，同时考虑到相同权重的边：

import java.util.*;

class Edge {
    int u, v, weight;
    
    public Edge(int u, int v, int weight) {
        this.u = u;
        this.v = v;
        this.weight = weight;
    }
}

class UnionFind {
    int[] parent;
    int[] rank;
    
    public UnionFind(int n) {
        parent = new int[n];
        rank = new int[n];
        
        Arrays.fill(rank, 1);
        
        for (int i = 0; i < n; i++) {
            parent[i] = i;
        }
    }
    
    public int find(int x) {
        if (x != parent[x]) {
            parent[x] = find(parent[x]);
        }
        
        return parent[x];
    }
    
    public void union(int x, int y) {
        int rootX = find(x);
        int rootY = find(y);
        
        if (rootX != rootY) {
            if (rank[rootX] > rank[rootY]) {
                parent[rootY] = rootX;
            } else if (rank[rootX] < rank[rootY]) {
                parent[rootX] = rootY;
            } else {
                parent[rootY] = rootX;
                rank[rootX]++;
            }
        }
    }
}

public class MinimumSpanningTreeCount {
    
    public static long countMSTs(int n, Edge[] edges) {
        Arrays.sort(edges, Comparator.comparingInt(e -> e.weight));
        
        UnionFind uf = new UnionFind(n);
        
        long count = 1;
        int i = 0;
        while (i < edges.length) {
            int j = i;
            while (j < edges.length && edges[j].weight == edges[i].weight) {
                j++;
            }
            
            int sameWeightCount = j - i;
            
            Set<Integer> parents = new HashSet<>();
            for (int k = i; k < j; k++) {
                int u = edges[k].u;
                int v = edges[k].v;
                
                if (uf.find(u) != uf.find(v)) {
                    parents.add(uf.find(u));
                    parents.add(uf.find(v));
                }
            }
            
            for (int parent : parents) {
                count *= (long) Math.pow(2, sameWeightCount - uf.rank[parent]);
            }
            
            for (int k = i; k < j; k++) {
                int u = edges[k].u;
                int v = edges[k].v;
                
                if (uf.find(u) != uf.find(v)) {
                    uf.union(u, v);
                }
            }
            
            i = j;
        }
        
        return count;
    }
    
    public static void main(String[] args) {
        int n = 4;
        int m = 5;
        
        Edge[] edges = new Edge[m];
        edges[0] = new Edge(0, 1, 1);
        edges[1] = new Edge(0, 2, 2);
        edges[2] = new Edge(0, 3, 3);
        edges[3] = new Edge(1, 2, 4);
        edges[4] = new Edge(2, 3, 5);
        
        long count = countMSTs(n, edges);
        System.out.println("Different methods of choosing edges: " + count);
    }
}

这个修改后的代码在计算不同选择边的方法数量时，考虑了相同权重的边。它将具有相同权重的边分组，并在计算数量时，根据连通分量的大小进行了调整。通过这种方式，可以正确计算无向连通图的最小生成树边集合数量。