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体系学习班/class43/Code03_PavingTile1.java

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+package class43;
+
+// 找到了贴瓷砖问题在线测试
+// 测试链接 : http://poj.org/problem?id=2411
+// 注册一个北京大学评测平台的号提交以下
+// 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
+// 提交以下的code，提交时请把类名改成"Main"
+// 本文件是状态压缩的动态规划版本，也就是课上讲的版本
+import java.io.BufferedReader;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.io.OutputStreamWriter;
+import java.io.PrintWriter;
+import java.io.StreamTokenizer;
+
+public class Code03_PavingTile1 {
+
+	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;
+			if (n != 0 || m != 0) {
+				long ans = ways(n, m);
+				out.println(ans);
+				out.flush();
+			}
+		}
+	}
+
+	// 状态压缩动态规划，最后一个版本
+	// 其实其他版本也能通过
+	public static long ways(int N, int M) {
+		if (N < 1 || M < 1 || ((N * M) & 1) != 0) {
+			return 0;
+		}
+		if (N == 1 || M == 1) {
+			return 1;
+		}
+		int big = N > M ? N : M;
+		int small = big == N ? M : N;
+		int sn = 1 << small;
+		int limit = sn - 1;
+		long[] dp = new long[sn];
+		dp[limit] = 1;
+		long[] cur = new long[sn];
+		for (int level = 0; level < big; level++) {
+			for (int status = 0; status < sn; status++) {
+				if (dp[status] != 0) {
+					int op = (~status) & limit;
+					dfs(dp[status], op, 0, small - 1, cur);
+				}
+			}
+			for (int i = 0; i < sn; i++) {
+				dp[i] = 0;
+			}
+			long[] tmp = dp;
+			dp = cur;
+			cur = tmp;
+		}
+		return dp[limit];
+	}
+
+	public static void dfs(long way, int op, int index, int end, long[] cur) {
+		if (index == end) {
+			cur[op] += way;
+		} else {
+			dfs(way, op, index + 1, end, cur);
+			if (((3 << index) & op) == 0) {
+				dfs(way, op | (3 << index), index + 1, end, cur);
+			}
+		}
+	}
+
+}

体系学习班/class43/Code03_PavingTile2.java

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+package class43;
+
+// 找到了贴瓷砖问题在线测试
+// 测试链接 : http://poj.org/problem?id=2411
+// 注册一个北京大学评测平台的号提交以下
+// 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
+// 提交以下的code，提交时请把类名改成"Main"
+// 本文件是轮廓线dp的版本，轮廓线dp的讲解在每周直播课 : 
+// 2022年9月第4周的课，看了就能懂
+// 这是很难的一类题型，不做要求，大厂几乎不考
+import java.io.BufferedReader;
+import java.io.IOException;
+import java.io.InputStreamReader;
+import java.io.OutputStreamWriter;
+import java.io.PrintWriter;
+import java.io.StreamTokenizer;
+
+public class Code03_PavingTile2 {
+
+	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;
+			if (n != 0 || m != 0) {
+				long ans = ways(n, m);
+				out.println(ans);
+				out.flush();
+			}
+		}
+	}
+
+	// 轮廓线dp的版本
+	// 看课，上面说了看哪的
+	public static int MAXN = 12;
+
+	public static int MAXM = (1 << MAXN);
+
+	public static long[][][] dp = new long[MAXN][MAXN][MAXM];
+
+	public static long ways(int n, int m) {
+		for (int i = 0; i <= n; i++) {
+			for (int j = 0; j <= m; j++) {
+				for (int k = 0; k <= (1 << m); k++) {
+					dp[i][j][k] = -1;
+				}
+			}
+		}
+		return process(0, 0, 0, n, m);
+	}
+
+	public static long process(int r, int c, int s, int n, int m) {
+		if (r == n) {
+			return s == 0 ? 1 : 0;
+		}
+		if (c == m) {
+			return process(r + 1, 0, s, n, m);
+		}
+		if (dp[r][c][s] != -1) {
+			return dp[r][c][s];
+		}
+		long ans;
+		int cur = (s & (1 << c)) == 0 ? 0 : 1;
+		if (cur == 1) {
+			ans = process(r, c + 1, s ^ (1 << c), n, m);
+		} else {
+			long p1 = process(r, c + 1, s | (1 << c), n, m);
+			long next = c + 1 < m && (s & (1 << (c + 1))) == 0 ? 0 : 1;
+			long p2 = 0;
+			if (next == 0) {
+				p2 = process(r, c + 2, s, n, m);
+			}
+			ans = p1 + p2;
+		}
+		dp[r][c][s] = ans;
+		return ans;
+	}
+
+}