package class43;

public class Code03_PavingTile {

	/*
	 * 2*M铺地的问题非常简单，这个是解决N*M铺地的问题
	 */

	public static int ways1(int N, int M) {
		if (N < 1 || M < 1 || ((N * M) & 1) != 0) {
			return 0;
		}
		if (N == 1 || M == 1) {
			return 1;
		}
		int[] pre = new int[M]; // pre代表-1行的状况
		for (int i = 0; i < pre.length; i++) {
			pre[i] = 1;
		}
		return process(pre, 0, N);
	}

	// pre 表示level-1行的状态
	// level表示，正在level行做决定
	// N 表示一共有多少行 固定的
	// level-2行及其之上所有行，都摆满砖了
	// level做决定，让所有区域都满，方法数返回
	public static int process(int[] pre, int level, int N) {
		if (level == N) { // base case
			for (int i = 0; i < pre.length; i++) {
				if (pre[i] == 0) {
					return 0;
				}
			}
			return 1;
		}

		// 没到终止行，可以选择在当前的level行摆瓷砖
		int[] op = getOp(pre);
		return dfs(op, 0, level, N);
	}

	// op[i] == 0 可以考虑摆砖
	// op[i] == 1 只能竖着向上
	public static int dfs(int[] op, int col, int level, int N) {
		// 在列上自由发挥，玩深度优先遍历，当col来到终止列，i行的决定做完了
		// 轮到i+1行，做决定
		if (col == op.length) {
			return process(op, level + 1, N);
		}
		int ans = 0;
		// col位置不横摆
		ans += dfs(op, col + 1, level, N); // col位置上不摆横转
		// col位置横摆, 向右
		if (col + 1 < op.length && op[col] == 0 && op[col + 1] == 0) {
			op[col] = 1;
			op[col + 1] = 1;
			ans += dfs(op, col + 2, level, N);
			op[col] = 0;
			op[col + 1] = 0;
		}
		return ans;
	}

	public static int[] getOp(int[] pre) {
		int[] cur = new int[pre.length];
		for (int i = 0; i < pre.length; i++) {
			cur[i] = pre[i] ^ 1;
		}
		return cur;
	}

	// Min (N,M) 不超过 32
	public static int ways2(int N, int M) {
		if (N < 1 || M < 1 || ((N * M) & 1) != 0) {
			return 0;
		}
		if (N == 1 || M == 1) {
			return 1;
		}
		int max = Math.max(N, M);
		int min = Math.min(N, M);
		int pre = (1 << min) - 1;
		return process2(pre, 0, max, min);
	}

	// 上一行的状态，是pre，limit是用来对齐的，固定参数不用管
	// 当前来到i行，一共N行，返回填满的方法数
	public static int process2(int pre, int i, int N, int M) {
		if (i == N) { // base case
			return pre == ((1 << M) - 1) ? 1 : 0;
		}
		int op = ((~pre) & ((1 << M) - 1));
		return dfs2(op, M - 1, i, N, M);
	}

	public static int dfs2(int op, int col, int level, int N, int M) {
		if (col == -1) {
			return process2(op, level + 1, N, M);
		}
		int ans = 0;
		ans += dfs2(op, col - 1, level, N, M);
		if ((op & (1 << col)) == 0 && col - 1 >= 0 && (op & (1 << (col - 1))) == 0) {
			ans += dfs2((op | (3 << (col - 1))), col - 2, level, N, M);
		}
		return ans;
	}

	// 记忆化搜索的解
	// Min(N,M) 不超过 32
	public static int ways3(int N, int M) {
		if (N < 1 || M < 1 || ((N * M) & 1) != 0) {
			return 0;
		}
		if (N == 1 || M == 1) {
			return 1;
		}
		int max = Math.max(N, M);
		int min = Math.min(N, M);
		int pre = (1 << min) - 1;
		int[][] dp = new int[1 << min][max + 1];
		for (int i = 0; i < dp.length; i++) {
			for (int j = 0; j < dp[0].length; j++) {
				dp[i][j] = -1;
			}
		}
		return process3(pre, 0, max, min, dp);
	}

	public static int process3(int pre, int i, int N, int M, int[][] dp) {
		if (dp[pre][i] != -1) {
			return dp[pre][i];
		}
		int ans = 0;
		if (i == N) {
			ans = pre == ((1 << M) - 1) ? 1 : 0;
		} else {
			int op = ((~pre) & ((1 << M) - 1));
			ans = dfs3(op, M - 1, i, N, M, dp);
		}
		dp[pre][i] = ans;
		return ans;
	}

	public static int dfs3(int op, int col, int level, int N, int M, int[][] dp) {
		if (col == -1) {
			return process3(op, level + 1, N, M, dp);
		}
		int ans = 0;
		ans += dfs3(op, col - 1, level, N, M, dp);
		if (col > 0 && (op & (3 << (col - 1))) == 0) {
			ans += dfs3((op | (3 << (col - 1))), col - 2, level, N, M, dp);
		}
		return ans;
	}

	// 严格位置依赖的动态规划解
	public static int ways4(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;
		int[] dp = new int[sn];
		dp[limit] = 1;
		int[] cur = new int[sn];
		for (int level = 0; level < big; level++) {
			for (int status = 0; status < sn; status++) {
				if (dp[status] != 0) {
					int op = (~status) & limit;
					dfs4(dp[status], op, 0, small - 1, cur);
				}
			}
			for (int i = 0; i < sn; i++) {
				dp[i] = 0;
			}
			int[] tmp = dp;
			dp = cur;
			cur = tmp;
		}
		return dp[limit];
	}

	public static void dfs4(int way, int op, int index, int end, int[] cur) {
		if (index == end) {
			cur[op] += way;
		} else {
			dfs4(way, op, index + 1, end, cur);
			if (((3 << index) & op) == 0) { // 11 << index 可以放砖
				dfs4(way, op | (3 << index), index + 1, end, cur);
			}
		}
	}

	public static void main(String[] args) {
		int N = 8;
		int M = 6;
		System.out.println(ways1(N, M));
		System.out.println(ways2(N, M));
		System.out.println(ways3(N, M));
		System.out.println(ways4(N, M));

		N = 10;
		M = 10;
		System.out.println("=========");
		System.out.println(ways3(N, M));
		System.out.println(ways4(N, M));
	}

}