话不多说上代码

部分代码参考 https://www.cnblogs.com/tp25959/p/10815931.html
这套代码也能把 Kruskal算法求最小生成树 过掉。

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.util.ArrayList;

class Edge
{
    int a,b,c;  //c是边权

    Edge(int a, int b, int c) {
        this.a = a;
        this.b = b;
        this.c = c;
    }

}

public class Main {

    static int N = 510, M = (int) (1e5 + 10);

    static int n, m;
    static int[] p = new int[N];  //并查集数组
    static ArrayList<Edge> e = new ArrayList<>();  //用数组的话就是e[M]

    static int find(int x)  //并查集
    {
        if (p[x] != x) p[x] = find(p[x]);
        return p[x];
    }

    public static void main(String[] args) throws IOException {

        BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
        String[] input = br.readLine().split(" ");

        n = Integer.parseInt(input[0]);
        m = Integer.parseInt(input[1]);

        for (int i = 1; i <= n; i ++ ) p[i] = i;  //初始化并查集

        for (int i = 1; i <= m; i ++ )
        {
            int u, v, w;
            input = br.readLine().split(" ");
            u = Integer.parseInt(input[0]);
            v = Integer.parseInt(input[1]);
            w = Integer.parseInt(input[2]);

            e.add(new Edge(u, v, w));
        }

        e.sort((o1, o2)->{
            return o1.c - o2.c;  //按边权从小到大排序
        });

        int res = 0;
        for (int i = 0; i < m; i ++ )  //m其实就是e.size()
        {
            int a = e.get(i).a, b = e.get(i).b, c = e.get(i).c;
            if (find(a) != find(b))
            {
                res += c;
                p[find(a)] = find(b);
            }
        }

        //最小生成树存在 ⇆ 图连通。所以统计 root 结点个数，大于1就不连通。
        int cnt = 0;
        for (int i = 1; i <= n; i++) 
        if (p[i] == i)
        {
            cnt++;
            if (cnt > 1) break;
        }

        if(cnt == 1) System.out.println(res);
        else System.out.println("impossible");

    }
}