Minimum spanning tree template (java)

Prim algorithm

Time complexity O(n^2)

class Solution {
    
    
    public int minCostConnectPoints(int[][] points) {
    
    
        //
        int res=0;
        int n=points.length;
        int[][] arr = new int[n][n];
        //构建邻接矩阵
        for(int i=0;i<n;i++){
    
    
            for(int j=i+1;j<n;j++){
    
    
                int dis = Math.abs(points[i][0]-points[j][0])+Math.abs(points[i][1]-points[j][1]);
                arr[i][j]=arr[j][i]=dis;
            }
        }
        //cost[i]表示‘当前节点集合’抵达节点i的最短边
        int[] cost = new int[n];
        for(int i=0;i<n;i++){
    
    
            cost[i]=arr[0][i];
        }
        for(int i=0;i<n;i++)
            if(i==n-1)
                System.out.println(cost[i]);
            else 
                System.out.print(cost[i]+" ");
            
        Set<Integer>set = new HashSet();
        set.add(0);
        while(set.size()<n){
    
    
            int next=-1;
            int minEdge=Integer.MAX_VALUE;
            //找出离当前节点的最短边
            for(int i=1;i<n;i++){
    
    
                if(!set.contains(i)&&cost[i]<minEdge){
    
    
                    next=i;
                    minEdge=cost[i];
                }
            }
            //加入到节点集
            set.add(next);
            res+=cost[next];
            //加入next更新
            for(int i=0;i<n;i++){
    
    
                if(!set.contains(i)&&cost[i]>arr[next][i])
                    cost[i]=arr[next][i];
            }
            for(int i=0;i<n;i++)
            if(i==n-1)
                System.out.println(cost[i]);
            else 
                System.out.print(cost[i]+" ");
        }
        return res;

    }
}