diff --git a/体系学习班/class16/Code04_KruskalNowCoder.java b/体系学习班/class16/Code04_KruskalNowCoder.java
new file mode 100644
index 0000000..30fff69
--- /dev/null
+++ b/体系学习班/class16/Code04_KruskalNowCoder.java
@@ -0,0 +1,99 @@
+package class16;
+
+// 课上没讲这个实现
+// 因为是一样的，都是用Kruskal算法实现最小生成树，只不过是牛客网的测试数据
+// 测试链接 : https://www.nowcoder.com/questionTerminal/c23eab7bb39748b6b224a8a3afbe396b
+// 请同学们务必参考如下代码中关于输入、输出的处理
+// 这是输入输出处理效率很高的写法
+// 提交以下所有代码，把主类名改成Main，可以直接通过
+import java.io.BufferedReader;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.io.OutputStreamWriter;
+import java.io.PrintWriter;
+import java.io.StreamTokenizer;
+import java.util.Arrays;
+
+public class Code04_KruskalNowCoder {
+
+	public static int MAXM = 100001;
+
+	public static int[][] edges = new int[MAXM][3];
+
+	public static void main(String[] args) throws IOException {
+		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+		StreamTokenizer in = new StreamTokenizer(br);
+		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
+		while (in.nextToken() != StreamTokenizer.TT_EOF) {
+			int n = (int) in.nval;
+			in.nextToken();
+			int m = (int) in.nval;
+			for (int i = 0; i < m; i++) {
+				in.nextToken();
+				edges[i][0] = (int) in.nval;
+				in.nextToken();
+				edges[i][1] = (int) in.nval;
+				in.nextToken();
+				edges[i][2] = (int) in.nval;
+			}
+			Arrays.sort(edges, 0, m, (a, b) -> a[2] - b[2]);
+			build(n);
+			int ans = 0;
+			for (int[] edge : edges) {
+				if (union(edge[0], edge[1])) {
+					ans += edge[2];
+				}
+			}
+			out.println(ans);
+			out.flush();
+		}
+	}
+
+	// 下面是并查集结构
+	public static int MAXN = 10001;
+
+	public static int[] father = new int[MAXN];
+
+	public static int[] size = new int[MAXN];
+
+	public static int[] help = new int[MAXN];
+
+	public static void build(int n) {
+		for (int i = 1; i <= n; i++) {
+			father[i] = i;
+			size[i] = 1;
+		}
+	}
+
+	private static int find(int i) {
+		int size = 0;
+		while (i != father[i]) {
+			help[size++] = i;
+			i = father[i];
+		}
+		while (size > 0) {
+			father[help[--size]] = i;
+		}
+		return i;
+	}
+
+	// 如果i和j，原本是一个集合，返回false
+	// 如果i和j，不是一个集合，合并，然后返回true
+	public static boolean union(int i, int j) {
+		int fi = find(i);
+		int fj = find(j);
+		if (fi != fj) {
+			if (size[fi] >= size[fj]) {
+				father[fj] = fi;
+				size[fi] += size[fj];
+			} else {
+				father[fi] = fj;
+				size[fj] += size[fi];
+			}
+			return true;
+		} else {
+			return false;
+		}
+	}
+
+}
diff --git a/体系学习班/class16/Code05_PrimNowCoder.java b/体系学习班/class16/Code05_PrimNowCoder.java
new file mode 100644
index 0000000..cb604e0
--- /dev/null
+++ b/体系学习班/class16/Code05_PrimNowCoder.java
@@ -0,0 +1,66 @@
+package class16;
+
+// 课上没讲这个实现
+// 因为是一样的，都是用Prim算法实现最小生成树，只不过是牛客网的测试数据
+// 测试链接 : https://www.nowcoder.com/questionTerminal/c23eab7bb39748b6b224a8a3afbe396b
+// 请同学们务必参考如下代码中关于输入、输出的处理
+// 这是输入输出处理效率很高的写法
+// 提交以下所有代码，把主类名改成Main，可以直接通过
+import java.io.BufferedReader;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.io.OutputStreamWriter;
+import java.io.PrintWriter;
+import java.io.StreamTokenizer;
+import java.util.ArrayList;
+import java.util.PriorityQueue;
+
+public class Code05_PrimNowCoder {
+
+	public static void main(String[] args) throws IOException {
+		BufferedReader br = new BufferedReader(new InputStreamReader(System.in));
+		StreamTokenizer in = new StreamTokenizer(br);
+		PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));
+		while (in.nextToken() != StreamTokenizer.TT_EOF) {
+			ArrayList<ArrayList<int[]>> graph = new ArrayList<>();
+			int n = (int) in.nval;
+			for (int i = 0; i <= n; i++) {
+				graph.add(new ArrayList<>());
+			}
+			in.nextToken();
+			int m = (int) in.nval;
+			for (int i = 0; i < m; i++) {
+				in.nextToken();
+				int A = (int) in.nval;
+				in.nextToken();
+				int B = (int) in.nval;
+				in.nextToken();
+				int cost = (int) in.nval;
+				graph.get(A).add(new int[] { B, cost });
+				graph.get(B).add(new int[] { A, cost });
+			}
+			PriorityQueue<int[]> heap = new PriorityQueue<>((a, b) -> a[1] - b[1]);
+			boolean[] visited = new boolean[n + 1];
+			for (int[] edge : graph.get(1)) {
+				heap.add(edge);
+			}
+			visited[1] = true;
+			int ans = 0;
+			while (!heap.isEmpty()) {
+				int[] edge = heap.poll();
+				int next = edge[0];
+				int cost = edge[1];
+				if (!visited[next]) {
+					visited[next] = true;
+					ans += cost;
+					for (int[] e : graph.get(next)) {
+						heap.add(e);
+					}
+				}
+			}
+			out.println(ans);
+			out.flush();
+		}
+	}
+
+}