diff --git a/MCA算法突击课/第02期/mca_02/Code01_MaxAnimalNumber.java b/MCA算法突击课/第02期/mca_02/Code01_MaxAnimalNumber.java
new file mode 100644
index 0000000..37f7705
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code01_MaxAnimalNumber.java
@@ -0,0 +1,119 @@
+package 第02期.mca_02;
+
+// 每周有营养的大厂算法面试题
+// 课时205
+// 有n个动物重量分别是a1、a2、a3.....an，
+// 这群动物一起玩叠罗汉游戏，
+// 规定从左往右选择动物，每只动物左边动物的总重量不能超过自己的重量
+// 返回最多能选多少个动物，求一个高效的算法。
+// 比如有7个动物，从左往右重量依次为：1，3，5，7，9，11，21
+// 则最多能选5个动物：1，3，5，9，21
+// 注意本题给的例子是有序的，但是实际给定的动物数组，可能是无序的，
+// 要求从左往右选动物，且不能打乱原始数组
+public class Code01_MaxAnimalNumber {
+
+	// 普通动态规划
+	// 非常一般的方法
+	// 来自背包的思路
+	public static int maxAnimals1(int[] arr) {
+		int sum = 0;
+		for (int num : arr) {
+			sum += num;
+		}
+		int[][] dp = new int[arr.length][sum + 1];
+		for (int i = 0; i < arr.length; i++) {
+			for (int j = 0; j <= sum; j++) {
+				dp[i][j] = -1;
+			}
+		}
+		return process1(arr, 0, 0, dp);
+	}
+
+	public static int process1(int[] arr, int index, int pre, int[][] dp) {
+		if (index == arr.length) {
+			return 0;
+		}
+		if (dp[index][pre] != -1) {
+			return dp[index][pre];
+		}
+		int p1 = process1(arr, index + 1, pre, dp);
+		int p2 = 0;
+		if (arr[index] >= pre) {
+			p2 = 1 + process1(arr, index + 1, pre + arr[index], dp);
+		}
+		int ans = Math.max(p1, p2);
+		dp[index][pre] = ans;
+		return ans;
+	}
+
+	// 最优解
+	// 如果arr长度为N，时间复杂度O(N*logN)
+	public static int maxAnimals2(int[] arr) {
+		if (arr == null || arr.length == 0) {
+			return 0;
+		}
+		// ends数组
+		int[] ends = new int[arr.length + 1];
+		ends[0] = 0;
+		int endsSize = 1;
+		int max = 1;
+		for (int i = 0; i < arr.length; i++) {
+			int l = 0;
+			int r = endsSize - 1;
+			int m = 0;
+			int find = 0;
+			while (l <= r) {
+				m = (l + r) / 2;
+				if (ends[m] <= arr[i]) {
+					find = m;
+					l = m + 1;
+				} else {
+					r = m - 1;
+				}
+			}
+			if (find == endsSize - 1) {
+				ends[endsSize] = ends[endsSize - 1] + arr[i];
+				endsSize++;
+			} else {
+				if (ends[find + 1] > ends[find] + arr[i]) {
+					ends[find + 1] = ends[find] + arr[i];
+				}
+			}
+			max = Math.max(max, find + 1);
+		}
+		return max;
+	}
+
+	public static int[] randomArray(int n, int v) {
+		int[] arr = new int[n];
+		for (int i = 0; i < n; i++) {
+			arr[i] = (int) (Math.random() * v) + 1;
+		}
+		return arr;
+	}
+
+	public static void main(String[] args) {
+		int N = 100;
+		int V = 1000;
+		int testTime = 2000;
+		System.out.println("测试开始");
+		for (int i = 0; i < testTime; i++) {
+			int n = (int) (Math.random() * N) + 1;
+			int[] arr = randomArray(n, V);
+			int ans1 = maxAnimals1(arr);
+			int ans2 = maxAnimals2(arr);
+			if (ans1 != ans2) {
+				System.out.println("出错了");
+				for (int num : arr) {
+					System.out.print(num + " ");
+				}
+				System.out.println();
+				System.out.println(ans1);
+				System.out.println(ans2);
+				break;
+			}
+		}
+		System.out.println("测试结束");
+	}
+
+}
diff --git a/MCA算法突击课/第02期/mca_02/Code02_WindPrevent.java b/MCA算法突击课/第02期/mca_02/Code02_WindPrevent.java
new file mode 100644
index 0000000..894294d
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code02_WindPrevent.java
@@ -0,0 +1,151 @@
+package 第02期.mca_02;
+
+// 每周有营养的大厂算法面试题
+// 课时207
+// 来自真实笔试
+// 给定一个二维数组matrix，数组中的每个元素代表一棵树的高度。
+// 你可以选定连续的若干行组成防风带，防风带每一列的防风高度为这一列的最大值
+// 防风带整体的防风高度为，所有列防风高度的最小值。
+// 比如，假设选定如下三行
+// 1 5 4
+// 7 2 6
+// 2 3 4
+// 1、7、2的列，防风高度为7
+// 5、2、3的列，防风高度为5
+// 4、6、4的列，防风高度为6
+// 防风带整体的防风高度为5，是7、5、6中的最小值
+// 给定一个正数k，k <= matrix的行数，表示可以取连续的k行，这k行一起防风。
+// 求防风带整体的防风高度最大值
+public class Code02_WindPrevent {
+
+	public static int bestHeight1(int[][] matrix, int k) {
+		int n = matrix.length;
+		int m = matrix[0].length;
+		int ans = 0;
+		for (int startRow = 0; startRow < n; startRow++) {
+			int bottleNeck = Integer.MAX_VALUE;
+			for (int col = 0; col < m; col++) {
+				int height = 0;
+				for (int endRow = startRow; endRow < n && (endRow - startRow + 1 <= k); endRow++) {
+					height = Math.max(height, matrix[endRow][col]);
+				}
+				bottleNeck = Math.min(bottleNeck, height);
+			}
+			ans = Math.max(ans, bottleNeck);
+		}
+		return ans;
+	}
+
+//	public static class WindowManager {
+//
+//		// 建立出m个窗口！
+//		public WindowManager(int m) {
+//
+//		}
+//
+//		public void addRow(int[][] matrix, int row) {
+//
+//		}
+//
+//		public void deleteRow(int[][] matrix, int row) {
+//
+//		}
+//
+//		public int getAllWindowMaxMin() {
+//			return 100;
+//		}
+//
+//	}
+//
+//	public static int bestWindHeight(int[][] matrix, int k) {
+//		int n = matrix.length;
+//		int m = matrix[0].length;
+//		k = Math.min(k, n);
+//		WindowManager windowManager = new WindowManager(m);
+//		for (int i = 0; i < k - 1; i++) {
+//			windowManager.addRow(matrix, i);
+//		}
+//		int ans = 0;
+//		for (int i = k - 1; i < n; i++) {
+//			windowManager.addRow(matrix, i);
+//			int cur = windowManager.getAllWindowMaxMin();
+//			ans = Math.max(ans, cur);
+//			windowManager.deleteRow(matrix, i - k + 1);
+//		}
+//		return ans;
+//	}
+
+	public static int bestHeight2(int[][] matrix, int k) {
+		int n = matrix.length;
+		int m = matrix[0].length;
+		int[][] windowMaxs = new int[m][n];
+		int[][] windowLR = new int[m][2];
+		for (int i = 0; i < k; i++) {
+			addRow(matrix, m, i, windowMaxs, windowLR);
+		}
+		int ans = bottleNeck(matrix, m, windowMaxs, windowLR);
+		for (int i = k; i < n; i++) {
+			addRow(matrix, m, i, windowMaxs, windowLR);
+			deleteRow(m, i - k, windowMaxs, windowLR);
+			ans = Math.max(ans, bottleNeck(matrix, m, windowMaxs, windowLR));
+		}
+		return ans;
+	}
+
+	public static void addRow(int[][] matrix, int m, int row, int[][] windowMaxs, int[][] windowLR) {
+		for (int col = 0; col < m; col++) {
+			while (windowLR[col][0] != windowLR[col][1]
+					&& matrix[windowMaxs[col][windowLR[col][1] - 1]][col] <= matrix[row][col]) {
+				windowLR[col][1]--;
+			}
+			windowMaxs[col][windowLR[col][1]++] = row;
+		}
+	}
+
+	public static void deleteRow(int m, int row, int[][] windowMaxs, int[][] windowLR) {
+		for (int col = 0; col < m; col++) {
+			if (windowMaxs[col][windowLR[col][0]] == row) {
+				windowLR[col][0]++;
+			}
+		}
+	}
+
+	public static int bottleNeck(int[][] matrix, int m, int[][] windowMaxs, int[][] windowLR) {
+		int ans = Integer.MAX_VALUE;
+		for (int col = 0; col < m; col++) {
+			ans = Math.min(ans, matrix[windowMaxs[col][windowLR[col][0]]][col]);
+		}
+		return ans;
+	}
+
+	public static int[][] generateMatrix(int n, int m, int v) {
+		int[][] matrix = new int[n][m];
+		for (int i = 0; i < n; i++) {
+			for (int j = 0; j < m; j++) {
+				matrix[i][j] = (int) (Math.random() * v) + 1;
+			}
+		}
+		return matrix;
+	}
+
+	public static void main(String[] args) {
+		int nMax = 10;
+		int mMax = 10;
+		int vMax = 50;
+		int testTimes = 1000;
+		System.out.println("测试开始");
+		for (int i = 0; i < testTimes; i++) {
+			int n = (int) (Math.random() * nMax) + 1;
+			int m = (int) (Math.random() * mMax) + 1;
+			int[][] matrix = generateMatrix(n, m, vMax);
+			int k = (int) (Math.random() * n) + 1;
+			int ans1 = bestHeight1(matrix, k);
+			int ans2 = bestHeight2(matrix, k);
+			if (ans1 != ans2) {
+				System.out.println("出错了!");
+			}
+		}
+		System.out.println("测试结束");
+	}
+
+}
diff --git a/MCA算法突击课/第02期/mca_02/Code03_SwimInRisingWater.java b/MCA算法突击课/第02期/mca_02/Code03_SwimInRisingWater.java
new file mode 100644
index 0000000..7d397f0
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code03_SwimInRisingWater.java
@@ -0,0 +1,157 @@
+package 第02期.mca_02;
+
+import java.util.Arrays;
+import java.util.PriorityQueue;
+
+// 每周有营养的大厂算法面试题
+// 课时213
+// 在一个 n x n 的整数矩阵 grid 中，
+// 每一个方格的值 grid[i][j] 表示位置 (i, j) 的平台高度。
+// 当开始下雨时，在时间为 t 时，水池中的水位为 t 。
+// 你可以从一个平台游向四周相邻的任意一个平台，但是前提是此时水位必须同时淹没这两个平台。
+// 假定你可以瞬间移动无限距离，也就是默认在方格内部游动是不耗时的。
+// 当然，在你游泳的时候你必须待在坐标方格里面。
+// 你从坐标方格的左上平台 (0，0) 出发。
+// 返回 你到达坐标方格的右下平台 (n-1, n-1) 所需的最少时间 。
+// 测试链接 ：https://leetcode.cn/problems/swim-in-rising-water
+public class Code03_SwimInRisingWater {
+
+	// 并查集的解法
+	public static int swimInWater1(int[][] grid) {
+		// 行号
+		int n = grid.length;
+		// 列号
+		int m = grid[0].length;
+		// [0,0,5]
+		// [0,1,3]....
+		int[][] points = new int[n * m][3];
+		int pi = 0;
+		for (int i = 0; i < n; i++) {
+			for (int j = 0; j < m; j++) {
+				points[pi][0] = i;
+				points[pi][1] = j;
+				points[pi++][2] = grid[i][j];
+			}
+		}
+		// 所有格子小对象，生成好了!
+		// 排序！[a,b,c]  [d,e,f]
+		Arrays.sort(points, (a, b) -> a[2] - b[2]);
+		// 生成并查集！n * m
+		// 初始化的时候，把所有格子独自成一个集合！
+		UnionFind uf = new UnionFind(n, m);
+		int ans = 0;
+		for (int i = 0; i < points.length; i++) {
+			int r = points[i][0];
+			int c = points[i][1];
+			int v = points[i][2];
+			if (r > 0 && grid[r - 1][c] <= v) {
+				uf.union(r, c, r - 1, c);
+			}
+			if (r < n - 1 && grid[r + 1][c] <= v) {
+				uf.union(r, c, r + 1, c);
+			}
+			if (c > 0 && grid[r][c - 1] <= v) {
+				uf.union(r, c, r, c - 1);
+			}
+			if (c < m - 1 && grid[r][c + 1] <= v) {
+				uf.union(r, c, r, c + 1);
+			}
+			if (uf.isSameSet(0, 0, n - 1, m - 1)) {
+				ans = v;
+				break;
+			}
+		}
+		return ans;
+	}
+
+	public static class UnionFind {
+		public int col;
+		public int pointsSize;
+		public int[] father;
+		public int[] size;
+		public int[] help;
+
+		public UnionFind(int n, int m) {
+			col = m;
+			pointsSize = n * m;
+			father = new int[pointsSize];
+			size = new int[pointsSize];
+			help = new int[pointsSize];
+			for (int i = 0; i < pointsSize; i++) {
+				father[i] = i;
+				size[i] = 1;
+			}
+		}
+
+		private int find(int i) {
+			int hi = 0;
+			while (i != father[i]) {
+				help[hi++] = i;
+				i = father[i];
+			}
+			while (hi > 0) {
+				father[help[--hi]] = i;
+			}
+			return i;
+		}
+
+		private int index(int i, int j) {
+			return i * col + j;
+		}
+
+		public void union(int row1, int col1, int row2, int col2) {
+			int f1 = find(index(row1, col1));
+			int f2 = find(index(row2, col2));
+			if (f1 != f2) {
+				if (size[f1] >= size[f2]) {
+					father[f2] = f1;
+					size[f1] += size[f2];
+				} else {
+					father[f1] = f2;
+					size[f2] += size[f1];
+				}
+			}
+		}
+
+		public boolean isSameSet(int row1, int col1, int row2, int col2) {
+			return find(index(row1, col1)) == find(index(row2, col2));
+		}
+
+	}
+
+	// Dijkstra算法
+	public static int swimInWater2(int[][] grid) {
+		int n = grid.length;
+		int m = grid[0].length;
+		PriorityQueue<int[]> heap = new PriorityQueue<>((a, b) -> a[2] - b[2]);
+		boolean[][] visited = new boolean[n][m];
+		heap.add(new int[] { 0, 0, grid[0][0] });
+		int ans = 0;
+		while (!heap.isEmpty()) {
+			int r = heap.peek()[0];
+			int c = heap.peek()[1];
+			int v = heap.peek()[2];
+			heap.poll();
+			if (visited[r][c]) {
+				continue;
+			}
+			visited[r][c] = true;
+			if (r == n - 1 && c == m - 1) {
+				ans = v;
+				break;
+			}
+			add(grid, heap, visited, r - 1, c, v);
+			add(grid, heap, visited, r + 1, c, v);
+			add(grid, heap, visited, r, c - 1, v);
+			add(grid, heap, visited, r, c + 1, v);
+		}
+		return ans;
+	}
+
+	public static void add(int[][] grid, PriorityQueue<int[]> heap, boolean[][] visited, int r, int c, int preV) {
+		if (r >= 0 && r < grid.length && c >= 0 && c < grid[0].length && !visited[r][c]) {
+			heap.add(new int[] { r, c, preV + Math.max(0, grid[r][c] - preV) });
+		}
+	}
+
+}
diff --git a/MCA算法突击课/第02期/mca_02/Code04_TreeDye.java b/MCA算法突击课/第02期/mca_02/Code04_TreeDye.java
new file mode 100644
index 0000000..77c3afc
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code04_TreeDye.java
@@ -0,0 +1,163 @@
+package 第02期.mca_02;
+
+import java.util.ArrayList;
+import java.util.Arrays;
+
+// 每周有营养的大厂算法面试题
+// 课时230
+// 来自米哈游
+// 给定一个正数n，表示有多少个节点
+// 给定一个二维数组edges，表示所有无向边
+// edges[i] = {a, b} 表示a到b有一条无向边
+// edges一定表示的是一个无环无向图，也就是树结构
+// 每个节点可以染1、2、3三种颜色
+// 要求 : 非叶节点的相邻点一定要至少有两种和自己不同颜色的点
+// 返回一种达标的染色方案，也就是一个数组，表示每个节点的染色状况
+// 1 <= 节点数量 <= 10的5次方
+public class Code04_TreeDye {
+
+	// 1 2 3 1 2 3 1 2 3
+	public static int[] rule1 = { 1, 2, 3 };
+
+	// 1 3 2 1 3 2 1 3 2
+	public static int[] rule2 = { 1, 3, 2 };
+
+	public static int[] dye(int n, int[][] edges) {
+		ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
+		// 0 : { 2, 1 }
+		// 1 : { 0 }
+		// 2 : { 0 }
+		for (int i = 0; i < n; i++) {
+			graph.add(new ArrayList<>());
+		}
+		for (int[] edge : edges) {
+			// 0 -> 2
+			// 1 -> 0
+			graph.get(edge[0]).add(edge[1]);
+			graph.get(edge[1]).add(edge[0]);
+		}
+		// 选一个头节点！
+		int head = -1;
+		for (int i = 0; i < n; i++) {
+			if (graph.get(i).size() >= 2) {
+				head = i;
+				break;
+			}
+		}
+		// graph
+		// head
+		int[] colors = new int[n];
+		if (head == -1) { // 两个点，互相连一下
+			// 把colors，所有位置，都设置成1
+			Arrays.fill(colors, 1);
+		} else {
+			// dfs 染色了！
+			colors[head] = 1;
+			dfs(graph, graph.get(head).get(0), 1, rule1, colors);
+			for (int i = 1; i < graph.get(head).size(); i++) {
+				dfs(graph, graph.get(head).get(i), 1, rule2, colors);
+			}
+		}
+		return colors;
+	}
+
+	// 整个图结构，都在graph
+	// 当前来到的节点，是head号节点
+	// head号节点，在level层
+	// 染色的规则，rule {1,2,3...} {1,3,2...}
+	// 做的事情：以head为头的整颗树，每个节点，都染上颜色
+	// 填入到colors数组里去
+	public static void dfs(
+			ArrayList<ArrayList<Integer>> graph,
+			int head,
+			int level,
+			int[] rule, int[] colors) {
+
+		colors[head] = rule[level % 3];
+		for (int next : graph.get(head)) {
+			if (colors[next] == 0) {
+				dfs(graph, next, level + 1, rule, colors);
+			}
+		}
+	}
+
+	// 为了测试
+	// 生成无环无向图
+	public static int[][] randomEdges(int n) {
+		int[] order = new int[n];
+		for (int i = 0; i < n; i++) {
+			order[i] = i;
+		}
+		for (int i = n - 1; i >= 0; i--) {
+			swap(order, i, (int) (Math.random() * (i + 1)));
+		}
+		int[][] edges = new int[n - 1][2];
+		for (int i = 1; i < n; i++) {
+			edges[i - 1][0] = order[i];
+			edges[i - 1][1] = order[(int) (Math.random() * i)];
+		}
+		return edges;
+	}
+
+	// 为了测试
+	public static void swap(int[] arr, int i, int j) {
+		int tmp = arr[i];
+		arr[i] = arr[j];
+		arr[j] = tmp;
+	}
+
+	// 为了测试
+	public static boolean rightAnswer(int n, int[][] edges, int[] colors) {
+		ArrayList<ArrayList<Integer>> graph = new ArrayList<>();
+		for (int i = 0; i < n; i++) {
+			graph.add(new ArrayList<>());
+		}
+		for (int[] edge : edges) {
+			graph.get(edge[0]).add(edge[1]);
+			graph.get(edge[1]).add(edge[0]);
+		}
+		boolean[] hasColors = new boolean[4];
+		for (int i = 0, colorCnt = 1; i < n; i++, colorCnt = 1) {
+			if (colors[i] == 0) {
+				return false;
+			}
+			if (graph.get(i).size() <= 1) { // i号点是叶节点
+				continue;
+			}
+			hasColors[colors[i]] = true;
+			for (int near : graph.get(i)) {
+				if (!hasColors[colors[near]]) {
+					hasColors[colors[near]] = true;
+					colorCnt++;
+				}
+			}
+			if (colorCnt != 3) {
+				return false;
+			}
+			Arrays.fill(hasColors, false);
+		}
+		return true;
+	}
+
+	public static void main(String[] args) {
+//		int n = 7;
+//		int[][] edges = randomEdges(n);
+//		for (int[] edge : edges) {
+//			System.out.println(edge[0] + " , " + edge[1]);
+//		}
+
+		int N = 100;
+		int testTimes = 1000;
+		System.out.println("测试开始");
+		for (int i = 0; i < testTimes; i++) {
+			int n = (int) (Math.random() * N) + 1;
+			int[][] edges = randomEdges(n);
+			int[] ans = dye(n, edges);
+			if (!rightAnswer(n, edges, ans)) {
+				System.out.println("出错了");
+			}
+		}
+		System.out.println("测试结束");
+	}
+
+}
diff --git a/MCA算法突击课/第02期/mca_02/Code05_CutOrPoison.java b/MCA算法突击课/第02期/mca_02/Code05_CutOrPoison.java
new file mode 100644
index 0000000..e696928
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code05_CutOrPoison.java
@@ -0,0 +1,121 @@
+package 第02期.mca_02;
+
+// 每周有营养的大厂算法面试题
+// 课时235
+// 来自学员问题
+// 给定怪兽的血量为hp
+// 第i回合如果用刀砍，怪兽在这回合会直接掉血，没有后续效果
+// 第i回合如果用毒，怪兽在这回合不会掉血，
+// 但是之后每回合都会掉血，并且所有中毒的后续效果会叠加
+// 给定的两个数组cuts、poisons，两个数组等长，长度都是n
+// 表示你在n回合内的行动，
+// 每一回合的刀砍的效果由cuts[i]表示
+// 每一回合的中毒的效果由poisons[i]表示
+// 如果你在n个回合内没有直接杀死怪兽，意味着你已经无法有新的行动了
+// 但是怪兽如果有中毒效果的话，那么怪兽依然会在hp耗尽的那回合死掉
+// 返回你最快能在多少回合内将怪兽杀死
+// 数据范围 : 
+// 1 <= n <= 10的5次方
+// 1 <= hp <= 10的9次方
+// 1 <= cuts[i]、poisons[i] <= 10的9次方
+public class Code05_CutOrPoison {
+
+	// 不算好的方法
+	// 为了验证
+	public static int fast1(int[] cuts, int[] poisons, int hp) {
+		int sum = 0;
+		for (int num : poisons) {
+			sum += num;
+		}
+		int[][][] dp = new int[cuts.length][hp + 1][sum + 1];
+		return process1(cuts, poisons, 0, hp, 0, dp);
+	}
+
+	public static int process1(int[] cuts, int[] poisons, int index, int restHp, int poisonEffect, int[][][] dp) {
+		restHp -= poisonEffect;
+		if (restHp <= 0) {
+			return index + 1;
+		}
+		// restHp > 0
+		if (index == cuts.length) {
+			if (poisonEffect == 0) {
+				return Integer.MAX_VALUE;
+			} else {
+				return cuts.length + 1 + (restHp + poisonEffect - 1) / poisonEffect;
+			}
+		}
+		if (dp[index][restHp][poisonEffect] != 0) {
+			return dp[index][restHp][poisonEffect];
+		}
+		int p1 = restHp <= cuts[index] ? (index + 1)
+				: process1(cuts, poisons, index + 1, restHp - cuts[index], poisonEffect, dp);
+		int p2 = process1(cuts, poisons, index + 1, restHp, poisonEffect + poisons[index], dp);
+		int ans = Math.min(p1, p2);
+		dp[index][restHp][poisonEffect] = ans;
+		return ans;
+	}
+
+	// 真正想实现的方法
+	// O(N * log(hp))
+	public static int fast2(int[] cuts, int[] poisons, int hp) {
+		// 怪兽可能的最快死亡回合
+		int l = 1;
+		// 怪兽可能的最晚死亡回合
+		int r = hp + 1;
+		int m = 0;
+		int ans = Integer.MAX_VALUE;
+		while (l <= r) {
+			m = l + ((r - l) >> 1);
+			if (ok(cuts, poisons, hp, m)) {
+				ans = m;
+				r = m - 1;
+			} else {
+				l = m + 1;
+			}
+		}
+		return ans;
+	}
+
+	public static boolean ok(int[] cuts, int[] posions, long hp, int limit) {
+		int n = Math.min(cuts.length, limit);
+		for (int i = 0, j = 1; i < n; i++, j++) {
+			hp -= Math.max((long) cuts[i], (long) (limit - j) * (long) posions[i]);
+			if (hp <= 0) {
+				return true;
+			}
+		}
+		return false;
+	}
+
+	// 为了测试
+	public static int[] randomArray(int n, int v) {
+		int[] ans = new int[n];
+		for (int i = 0; i < n; i++) {
+			ans[i] = (int) (Math.random() * v) + 1;
+		}
+		return ans;
+	}
+
+	// 为了测试
+	public static void main(String[] args) {
+		int N = 30;
+		int cutV = 20;
+		int posionV = 10;
+		int hpV = 200;
+		int testTimes = 10000;
+		System.out.println("测试开始");
+		for (int i = 0; i < testTimes; i++) {
+			int n = (int) (Math.random() * N) + 1;
+			int[] cuts = randomArray(n, cutV);
+			int[] posions = randomArray(n, posionV);
+			int hp = (int) (Math.random() * hpV) + 1;
+			int ans1 = fast1(cuts, posions, hp);
+			int ans2 = fast2(cuts, posions, hp);
+			if (ans1 != ans2) {
+				System.out.println("出错了!");
+			}
+		}
+		System.out.println("测试结束");
+	}
+
+}
diff --git a/MCA算法突击课/第02期/mca_02/Code06_WhoWin21Balls.java b/MCA算法突击课/第02期/mca_02/Code06_WhoWin21Balls.java
new file mode 100644
index 0000000..361a96d
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code06_WhoWin21Balls.java
@@ -0,0 +1,162 @@
+package 第02期.mca_02;
+
+// 每周有营养的大厂算法面试题
+// 课时171
+// 来自微众
+// 人工智能岗
+// 一开始有21个球，甲和乙轮流拿球，甲先、乙后
+// 每个人在自己的回合，一定要拿不超过3个球，不能不拿
+// 最终谁的总球数为偶数，谁赢
+// 请问谁有必胜策略
+public class Code06_WhoWin21Balls {
+
+	// balls = 21
+	// ball是奇数
+	public static String win(int balls) {
+		return process(0, balls, 0, 0);
+	}
+
+	// 憋递归！
+	// turn 谁的回合！
+	// turn == 0 甲回合
+	// turn == 1 乙回合
+	// rest剩余球的数量
+	// 之前，jiaBalls、yiBalls告诉你！
+	// 当前，根据turn，知道是谁的回合！
+	// 当前，还剩多少球，rest
+	// 返回：谁会赢！
+	public static String process(int turn, int rest, int jia, int yi) {
+		if (rest == 0) {
+			return (jia & 1) == 0 ? "甲" : "乙";
+		}
+		// rest > 0, 还剩下球！
+		if (turn == 0) { // 甲的回合！
+			// 甲，自己赢！甲赢！
+			for (int pick = 1; pick <= Math.min(rest, 3); pick++) {
+				// pick 甲当前做的选择
+				if (process(1, rest - pick, jia + pick, yi).equals("甲")) {
+					return "甲";
+				}
+			}
+			return "乙";
+		} else {
+			for (int pick = 1; pick <= Math.min(rest, 3); pick++) {
+				// pick 甲当前做的选择
+				if (process(0, rest - pick, jia, yi + pick).equals("乙")) {
+					return "乙";
+				}
+			}
+			return "甲";
+		}
+	}
+
+	// 我们补充一下设定，假设一开始的球数量不是21，是任意的正数
+	// 如果最终两个人拿的都是偶数，认为无人获胜，平局
+	// 如果最终两个人拿的都是奇数，认为无人获胜，平局
+	// rest代表目前剩下多少球
+	// cur == 0 代表目前是甲行动
+	// cur == 1 代表目前是乙行动
+	// first == 0 代表目前甲所选的球数，加起来是偶数
+	// first == 1 代表目前甲所选的球数，加起来是奇数
+	// second == 0 代表目前乙所选的球数，加起来是偶数
+	// second == 1 代表目前乙所选的球数，加起来是奇数
+	// 返回选完了rest个球，谁会赢，只会返回"甲"、"乙"、"平"
+	// win1方法，就是彻底暴力的做所有尝试，并且返回最终的胜利者
+	// 在甲的回合，甲会尝试所有的可能，以保证自己会赢，如果自己怎么都不会赢，那也要尽量平局，如果这个也不行，只能对方赢
+	// 在乙的回合，乙会尝试所有的可能，以保证自己会赢，如果自己怎么都不会赢，那也要尽量平局，如果这个也不行，只能对方赢
+	// 算法和数据结构体系学习班，视频39章节，牛羊吃草问题，就是类似这种递归
+	public static String win1(int rest, int cur, int first, int second) {
+		if (rest == 0) {
+			if (first == 0 && second == 1) {
+				return "甲";
+			}
+			if (first == 1 && second == 0) {
+				return "乙";
+			}
+			return "平";
+		}
+		if (cur == 0) { // 甲行动
+			String bestAns = "乙";
+			for (int pick = 1; pick <= Math.min(3, rest); pick++) {
+				String curAns = win1(rest - pick, 1, first ^ (pick & 1), second);
+				if (curAns.equals("甲")) {
+					bestAns = "甲";
+					break;
+				}
+				if (curAns.equals("平")) {
+					bestAns = "平";
+				}
+			}
+			return bestAns;
+		} else { // 乙行动
+			String bestAns = "甲";
+			for (int pick = 1; pick <= Math.min(3, rest); pick++) {
+				String curAns = win1(rest - pick, 0, first, second ^ (pick & 1));
+				if (curAns.equals("乙")) {
+					bestAns = "乙";
+					break;
+				}
+				if (curAns.equals("平")) {
+					bestAns = "平";
+				}
+			}
+			return bestAns;
+		}
+	}
+
+	// 下面的win2方法，仅仅是把win1方法，做了记忆化搜索
+	// 变成了动态规划
+	public static String[][][][] dp = new String[5000][2][2][2];
+
+	public static String win2(int rest, int cur, int first, int second) {
+		if (rest == 0) {
+			if (first == 0 && second == 1) {
+				return "甲";
+			}
+			if (first == 1 && second == 0) {
+				return "乙";
+			}
+			return "平";
+		}
+		if (dp[rest][cur][first][second] != null) {
+			return dp[rest][cur][first][second];
+		}
+		if (cur == 0) { // 甲行动
+			String bestAns = "乙";
+			for (int pick = 1; pick <= Math.min(3, rest); pick++) {
+				String curAns = win2(rest - pick, 1, first ^ (pick & 1), second);
+				if (curAns.equals("甲")) {
+					bestAns = "甲";
+					break;
+				}
+				if (curAns.equals("平")) {
+					bestAns = "平";
+				}
+			}
+			dp[rest][cur][first][second] = bestAns;
+			return bestAns;
+		} else { // 乙行动
+			String bestAns = "甲";
+			for (int pick = 1; pick <= Math.min(3, rest); pick++) {
+				String curAns = win2(rest - pick, 0, first, second ^ (pick & 1));
+				if (curAns.equals("乙")) {
+					bestAns = "乙";
+					break;
+				}
+				if (curAns.equals("平")) {
+					bestAns = "平";
+				}
+			}
+			dp[rest][cur][first][second] = bestAns;
+			return bestAns;
+		}
+	}
+
+	// 为了测试
+	public static void main(String[] args) {
+		for (int balls = 1; balls <= 500; balls += 2) {
+			System.out.println("球数为 " + balls + " 时 , 赢的是 " + win(balls));
+		}
+	}
+
+}
\ No newline at end of file
diff --git a/MCA算法突击课/第02期/mca_02/Code07_RedAndWhiteSquares.java b/MCA算法突击课/第02期/mca_02/Code07_RedAndWhiteSquares.java
new file mode 100644
index 0000000..f1bfded
--- /dev/null
+++ b/MCA算法突击课/第02期/mca_02/Code07_RedAndWhiteSquares.java
@@ -0,0 +1,213 @@
+package 第02期.mca_02;
+
+// 每周有营养的大厂算法面试题
+// 课时179
+// 来自网易
+// 小红拿到了一个大立方体，该大立方体由1*1*1的小方块拼成，初始每个小方块都是白色。
+// 小红可以每次选择一个小方块染成红色
+// 每次小红可能选择同一个小方块重复染色
+// 每次染色以后，你需要帮小红回答出当前的白色连通块数
+// 如果两个小方块共用同一个面，且颜色相同，则它们是连通的
+// 给定n、m、h，表示大立方体的长、宽、高
+// 给定k次操作，每一次操作用(a, b, c)表示在大立方体的该位置进行染色
+// 返回长度为k的数组，表示每一次操作后，白色方块的连通块数
+// n * m * h <= 10 ^ 5，k <= 10 ^ 5
+public class Code07_RedAndWhiteSquares {
+
+	// 暴力方法
+	// 时间复杂度(k * n * m * h);
+	public static int[] blocks1(int n, int m, int h, int[][] ops) {
+		int k = ops.length;
+		int[][][] cube = new int[n][m][h];
+		int value = 1;
+		int[] ans = new int[k];
+		for (int i = 0; i < k; i++) {
+			cube[ops[i][0]][ops[i][1]][ops[i][2]] = -1;
+			for (int x = 0; x < n; x++) {
+				for (int y = 0; y < m; y++) {
+					for (int z = 0; z < h; z++) {
+						if (cube[x][y][z] != -1 && cube[x][y][z] != value) {
+							ans[i]++;
+							infect(cube, x, y, z, value);
+						}
+					}
+				}
+			}
+			value++;
+		}
+		return ans;
+	}
+
+	public static void infect(int[][][] cube, int a, int b, int c, int change) {
+		if (a < 0 || a == cube.length || b < 0 || b == cube[0].length || c < 0 || c == cube[0][0].length
+				|| cube[a][b][c] == -1 || cube[a][b][c] == change) {
+			return;
+		}
+		cube[a][b][c] = change;
+		infect(cube, a - 1, b, c, change);
+		infect(cube, a + 1, b, c, change);
+		infect(cube, a, b - 1, c, change);
+		infect(cube, a, b + 1, c, change);
+		infect(cube, a, b, c - 1, change);
+		infect(cube, a, b, c + 1, change);
+	}
+
+	// 最优解
+	// O(k + n * m * h)
+	public static int[] blocks2(int n, int m, int h, int[][] ops) {
+		int k = ops.length;
+		int[][][] red = new int[n][m][h];
+		for (int[] op : ops) {
+			red[op[0]][op[1]][op[2]]++;
+		}
+		UnionFind uf = new UnionFind(n, m, h, red);
+		int[] ans = new int[k];
+		for (int i = k - 1; i >= 0; i--) {
+			ans[i] = uf.sets;
+			int x = ops[i][0];
+			int y = ops[i][1];
+			int z = ops[i][2];
+			if (--red[x][y][z] == 0) {
+				// x, y ,z 这个格子，变白，建立自己的小集合
+				// 然后6个方向，集合该合并合并
+				uf.finger(x, y, z);
+			}
+		}
+		return ans;
+	}
+
+	public static class UnionFind {
+		public int n;
+		public int m;
+		public int h;
+		public int[] father;
+		public int[] size;
+		public int[] help;
+		public int sets;
+
+		public UnionFind(int a, int b, int c, int[][][] red) {
+			n = a;
+			m = b;
+			h = c;
+			int len = n * m * h;
+			father = new int[len];
+			size = new int[len];
+			help = new int[len];
+			for (int x = 0; x < n; x++) {
+				for (int y = 0; y < m; y++) {
+					for (int z = 0; z < h; z++) {
+						if (red[x][y][z] == 0) {
+							finger(x, y, z);
+						}
+					}
+				}
+			}
+		}
+
+		public void finger(int x, int y, int z) {
+			// x，y，z
+			// 一维数值
+			int i = index(x, y, z);
+			father[i] = i;
+			size[i] = 1;
+			sets++;
+			union(i, x - 1, y, z);
+			union(i, x + 1, y, z);
+			union(i, x, y - 1, z);
+			union(i, x, y + 1, z);
+			union(i, x, y, z - 1);
+			union(i, x, y, z + 1);
+		}
+
+		private int index(int x, int y, int z) {
+			return z * n * m + y * n + x;
+		}
+
+		private void union(int i, int x, int y, int z) {
+			if (x < 0 || x == n || y < 0 || y == m || z < 0 || z == h) {
+				return;
+			}
+			int j = index(x, y, z);
+			if (size[j] == 0) {
+				return;
+			}
+			i = find(i);
+			j = find(j);
+			if (i != j) {
+				if (size[i] >= size[j]) {
+					father[j] = i;
+					size[i] += size[j];
+				} else {
+					father[i] = j;
+					size[j] += size[i];
+				}
+				sets--;
+			}
+		}
+
+		private int find(int i) {
+			int s = 0;
+			while (i != father[i]) {
+				help[s++] = i;
+				i = father[i];
+			}
+			while (s > 0) {
+				father[help[--s]] = i;
+			}
+			return i;
+		}
+
+	}
+
+	// 为了测试
+	public static int[][] randomOps(int n, int m, int h) {
+		int size = (int) (Math.random() * (n * m * h)) + 1;
+		int[][] ans = new int[size][3];
+		for (int i = 0; i < size; i++) {
+			ans[i][0] = (int) (Math.random() * n);
+			ans[i][1] = (int) (Math.random() * m);
+			ans[i][2] = (int) (Math.random() * h);
+		}
+		return ans;
+	}
+
+	// 为了测试
+	public static void main(String[] args) {
+		int size = 10;
+		int testTime = 5000;
+		System.out.println("测试开始");
+		for (int i = 0; i < testTime; i++) {
+			int n = (int) (Math.random() * size) + 1;
+			int m = (int) (Math.random() * size) + 1;
+			int h = (int) (Math.random() * size) + 1;
+			int[][] ops = randomOps(n, m, h);
+			int[] ans1 = blocks1(n, m, h, ops);
+			int[] ans2 = blocks2(n, m, h, ops);
+			for (int j = 0; j < ops.length; j++) {
+				if (ans1[j] != ans2[j]) {
+					System.out.println("出错了!");
+				}
+			}
+		}
+		System.out.println("测试结束");
+
+		// 立方体达到10^6规模
+		int n = 100;
+		int m = 100;
+		int h = 100;
+		int len = n * m * h;
+		// 操作条数达到10^6规模
+		int[][] ops = new int[len][3];
+		for (int i = 0; i < len; i++) {
+			ops[i][0] = (int) (Math.random() * n);
+			ops[i][1] = (int) (Math.random() * m);
+			ops[i][2] = (int) (Math.random() * h);
+		}
+		long start = System.currentTimeMillis();
+		blocks2(n, m, h, ops);
+		long end = System.currentTimeMillis();
+		System.out.println("运行时间(毫秒) : " + (end - start));
+
+	}
+
+}