Prim算法
时间复杂度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;
}
}