Merge branch 'algorithmzuo:main' into main

Author: moleoxa

src/class119/Code04_EmergencyAssembly1.java renamed to src/class119/Code01_EmergencyAssembly1.java

@@ -10,7 +10,7 @@
 // 1 <= m <= 5 * 10^5
 // 测试链接 : https://www.luogu.com.cn/problem/P4281
 // 如下实现是正确的，但是洛谷平台对空间卡的很严，只有使用C++能全部通过
-// C++版本就是本节代码中的Code04_EmergencyAssembly2文件
+// C++版本就是本节代码中的Code01_EmergencyAssembly2文件
 // C++版本和java版本逻辑完全一样，但只有C++版本可以通过所有测试用例
 // 这是洛谷平台没有照顾各种语言的实现所导致的
 // 在真正笔试、比赛时，一定是兼顾各种语言的，该实现是一定正确的
@@ -24,7 +24,7 @@
 import java.io.StreamTokenizer;
 import java.util.Arrays;
 
-public class Code04_EmergencyAssembly1 {
+public class Code01_EmergencyAssembly1 {
 
 	public static int MAXN = 500001;
 
@@ -40,10 +40,10 @@ public class Code04_EmergencyAssembly1 {
 
 	public static int cnt;
 
-	public static int[][] stjump = new int[MAXN][LIMIT];
-
 	public static int[] deep = new int[MAXN];
 
+	public static int[][] stjump = new int[MAXN][LIMIT];
+
 	public static int togather;
 
 	public static long cost;
@@ -138,9 +138,12 @@ public static void main(String[] args) throws IOException {
 	}
 
 	public static void compute(int a, int b, int c) {
+		// 来自对结构关系的深入分析，课上重点解释
 		int h1 = lca(a, b), h2 = lca(a, c), h3 = lca(b, c);
-		togather = h1 == h2 ? h3 : (h1 == h3 ? h2 : h1);
-		cost = (long) deep[a] + deep[b] + deep[c] - deep[h1] - deep[h2] - deep[h3];
+		int high = h1 != h2 ? (deep[h1] < deep[h2] ? h1 : h2) : h1;
+		int low = h1 != h2 ? (deep[h1] > deep[h2] ? h1 : h2) : h3;
+		togather = low;
+		cost = (long) deep[a] + deep[b] + deep[c] - deep[high] * 2 - deep[low];
 	}
 
 }

src/class119/Code04_EmergencyAssembly2.java renamed to src/class119/Code01_EmergencyAssembly2.java

@@ -27,8 +27,8 @@
 //int edgeNext[MAXN << 1];
 //int edgeTo[MAXN << 1];
 //int cnt;
-//int stjump[MAXN][LIMIT];
 //int deep[MAXN];
+//int stjump[MAXN][LIMIT];
 //int togather;
 //long long cost;
 //
@@ -83,9 +83,11 @@
 //}
 //
 //void compute(int a, int b, int c) {
-//    int h1 = lca(a, b), h2 = lca(a, c), h3 = lca(b, c);
-//    togather = h1 == h2 ? h3 : (h1 == h3 ? h2 : h1);
-//    cost = deep[a] + deep[b] + deep[c] - deep[h1] - deep[h2] - deep[h3];
+//	int h1 = lca(a, b), h2 = lca(a, c), h3 = lca(b, c);
+//	int high = h1 != h2 ? (deep[h1] < deep[h2] ? h1 : h2) : h1;
+//	int low = h1 != h2 ? (deep[h1] > deep[h2] ? h1 : h2) : h3;
+//	togather = low;
+//	cost = (long) deep[a] + deep[b] + deep[c] - deep[high] * 2 - deep[low];
 //}
 //
 //int main() {

src/class119/Code03_Trucking.java renamed to src/class119/Code02_Trucking.java

@@ -23,20 +23,21 @@
 import java.io.StreamTokenizer;
 import java.util.Arrays;
 
-public class Code03_Trucking {
+public class Code02_Trucking {
 
 	public static int MAXN = 10001;
 
 	public static int MAXM = 50001;
 
 	public static int LIMIT = 21;
 
+	public static int power;
+
 	public static int[][] edges = new int[MAXM][3];
 
+	// 并查集
 	public static int[] father = new int[MAXN];
 
-	public static int power;
-
 	public static int cnt;
 
 	public static int[] head = new int[MAXN];

src/class119/Code02_QueryPathMinimumChangesToSame.java renamed to src/class119/Code03_QueryPathMinimumChangesToSame.java

@@ -11,7 +11,7 @@
 // 0 <= u、v、a、b < n
 // 1 <= w <= 26
 // 测试链接 : https://leetcode.cn/problems/minimum-edge-weight-equilibrium-queries-in-a-tree/
-public class Code02_QueryPathMinimumChangesToSame {
+public class Code03_QueryPathMinimumChangesToSame {
 
 	public static int MAXN = 10001;
 
@@ -39,7 +39,7 @@ public class Code02_QueryPathMinimumChangesToSame {
 
 	public static int qcnt;
 
-	public static int[][] cnts = new int[MAXN][MAXW + 1];
+	public static int[][] stcnt = new int[MAXN][MAXW + 1];
 
 	public static boolean[] visited = new boolean[MAXN];
 
@@ -65,7 +65,7 @@ public int[] minOperationsQueries(int n, int[][] edges, int[][] queries) {
 			for (int j = 1; j <= MAXW; j++) {
 				// 特别重要的结论
 				// 很多题可以用到
-				pathCnt = cnts[queries[i][0]][j] + cnts[queries[i][1]][j] - 2 * cnts[lca[i]][j];
+				pathCnt = stcnt[queries[i][0]][j] + stcnt[queries[i][1]][j] - 2 * stcnt[lca[i]][j];
 				maxCnt = Math.max(maxCnt, pathCnt);
 				allCnt += pathCnt;
 			}
@@ -108,12 +108,12 @@ public static int find(int i) {
 	public void tarjan(int u, int w, int f) {
 		visited[u] = true;
 		if (u == 0) {
-			Arrays.fill(cnts[u], 0);
+			Arrays.fill(stcnt[u], 0);
 		} else {
 			for (int i = 1; i <= MAXW; i++) {
-				cnts[u][i] = cnts[f][i];
+				stcnt[u][i] = stcnt[f][i];
 			}
-			cnts[u][w]++;
+			stcnt[u][w]++;
 		}
 		for (int e = headEdge[u]; e != 0; e = edgeNext[e]) {
 			if (edgeTo[e] != f) {

src/class119/Code01_PassingBallMaximizeValue.java renamed to src/class119/Code04_PassingBallMaximizeValue.java

@@ -14,7 +14,7 @@
 // f(x) = x + receiver[x] + receiver[receiver[x]] + ... 
 // 你的任务时选择开始玩家x，目的是最大化f(x)，返回函数的最大值
 // 测试链接 : https://leetcode.cn/problems/maximize-value-of-function-in-a-ball-passing-game/
-public class Code01_PassingBallMaximizeValue {
+public class Code04_PassingBallMaximizeValue {
 
 	public static int MAXN = 100001;
 
@@ -28,8 +28,26 @@ public class Code01_PassingBallMaximizeValue {
 
 	public static long[][] stsum = new long[MAXN][LIMIT];
 
-	// 该方法实现了树上倍增的解法，打败比例很一般，但是非常好想
-	// 这个题最优解来自基环树分析，后面的课会安排这部分内容
+	public static void build(long k) {
+		power = 0;
+		while ((1L << power) <= (k >> 1)) {
+			power++;
+		}
+		m = 0;
+		// 数字k在哪些位有1
+		// 都收集起来
+		for (int p = power; p >= 0; p--) {
+			if ((1L << p) <= k) {
+				kbits[m++] = p;
+				k -= 1L << p;
+			}
+		}
+	}
+
+	// 该方法是树上倍增的解法
+	// 打败比例很一般但是非常好想
+	// 这个题最优解来自对基环树的分析
+	// 后面的课会安排基环树的内容
 	public static long getMaxFunctionValue(List<Integer> receiver, long k) {
 		build(k);
 		int n = receiver.size();
@@ -57,20 +75,4 @@ public static long getMaxFunctionValue(List<Integer> receiver, long k) {
 		return ans;
 	}
 
-	public static void build(long k) {
-		power = 0;
-		while ((1L << power) <= (k >> 1)) {
-			power++;
-		}
-		m = 0;
-		// 数字k在哪些位有1
-		// 都收集起来
-		for (int p = power; p >= 0; p--) {
-			if ((1L << p) <= k) {
-				kbits[m++] = p;
-				k -= 1L << p;
-			}
-		}
-	}
-
 }