Domino 和 Tromino 平铺问题

给定一个正整数N ，任务是找到用尺寸为2 × 1 , 1 × 2的瓷砖(也称为多米诺骨牌)和一个“ L ”形瓷砖填充尺寸为2*N的板的方法数(也称为tromino )如下所示，可以旋转90度。

The L shape tile:

XX
X
After rotating L shape tile by 90:

XX
 X
or
X
XX

例子：

Input: N = 3
Output: 5
Explanation:
Below is the image to illustrate all the combinations:

Input: N = 1
Output: 1

方法：可以根据以下观察结果解决给定的问题：

让我们定义一个2状态，动态规划说dp[i, j]表示列索引i中的以下安排之一。

• 当前列可以在状态0下被填充1，2×1张多米诺骨牌，如果前一列有状态0。
• 当前列可以填充有2，1×2在状态0水平骨牌中，如果i – 2列具有状态0。
• 如果前一列的状态为0 ，则当前列可以用状态1和状态2的“ L ”形多米诺骨牌填充。
• 如果前一列具有状态2或在状态2 中(如果前一列具有状态1 ) ，则当前列可以在状态1 中填充1 × 2形状的多米诺骨牌。
• 因此，状态的转移可以定义如下：
  1. dp[i][0] = (dp[i – 1][0] + dp[i – 2][0]+ dp[i – 2][1] + dp[i – 2][2]) 。
  2. dp[i][1] = dp[i – 1][0] + dp[i – 1][2] 。
  3. dp[i][2] = dp[i – 1][0] + dp[i – 1][1] 。

根据以上观察，请按照以下步骤解决问题：

• 如果N的值小于3 ，则打印N作为总路数。
• 初始化一个二维数组，比如说dp[][3] ，它存储了 dp 的所有状态。
• 考虑基本情况： dp[0][0] = dp[1][0] = dp[1][1] = dp[1][2] = 1 。
• 迭代给定范围[2, N]并使用变量i并在 dp 中执行以下转换：
  • dp[i][0]等于(dp[i – 1][0] + dp[i – 2][0]+ dp[i – 2][1] + dp[i – 2][2]) 。
  • dp[i][1]等于dp[i – 1][0] + dp[i – 1][2] 。
  • dp[i][2]等于dp[i – 1][0] + dp[i – 1][1] 。
• 完成上述步骤后，打印dp[N][0] 中存储的总路数。

下面是上述方法的实现：

C++

// C++ program for the above approach
 
#include 
using namespace std;
const long long MOD = 1e9 + 7;
 
// Function to find the total number
// of ways to tile a 2*N board using
// the given types of tile
int numTilings(int N)
{
    // If N is less than 3
    if (N < 3) {
        return N;
    }
 
    // Store all dp-states
    vector > dp(
        N + 1, vector(3, 0));
 
    // Base Case
    dp[0][0] = dp[1][0] = 1;
    dp[1][1] = dp[1][2] = 1;
 
    // Traverse the range [2, N]
    for (int i = 2; i <= N; i++) {
 
        // Update the value of dp[i][0]
        dp[i][0] = (dp[i - 1][0]
                    + dp[i - 2][0]
                    + dp[i - 2][1]
                    + dp[i - 2][2])
                   % MOD;
 
        // Update the value of dp[i][1]
        dp[i][1] = (dp[i - 1][0]
                    + dp[i - 1][2])
                   % MOD;
 
        // Update the value of dp[i][2]
        dp[i][2] = (dp[i - 1][0]
                    + dp[i - 1][1])
                   % MOD;
    }
 
    // Return the number of ways as
    // the value of dp[N][0]
    return dp[N][0];
}
 
// Driver Code
int main()
{
    int N = 3;
    cout << numTilings(N);
 
    return 0;
}

Java

// Java program for the above approach
import java.util.Arrays;
 
class GFG{
 
public static long MOD = 1000000007l;
 
// Function to find the total number
// of ways to tile a 2*N board using
// the given types of tile
public static long numTilings(int N)
{
     
    // If N is less than 3
    if (N < 3)
    {
        return N;
    }
 
    // Store all dp-states
    long[][] dp = new long[N + 1][3];
 
    for(long[] row : dp)
    {
        Arrays.fill(row, 0);
    }
     
    // Base Case
    dp[0][0] = dp[1][0] = 1;
    dp[1][1] = dp[1][2] = 1;
 
    // Traverse the range [2, N]
    for(int i = 2; i <= N; i++)
    {
         
        // Update the value of dp[i][0]
        dp[i][0] = (dp[i - 1][0] + dp[i - 2][0] +
                    dp[i - 2][1] + dp[i - 2][2]) % MOD;
 
        // Update the value of dp[i][1]
        dp[i][1] = (dp[i - 1][0] + dp[i - 1][2]) % MOD;
 
        // Update the value of dp[i][2]
        dp[i][2] = (dp[i - 1][0] + dp[i - 1][1]) % MOD;
    }
 
    // Return the number of ways as
    // the value of dp[N][0]
    return dp[N][0];
}
 
// Driver Code
public static void main(String args[])
{
    int N = 3;
     
    System.out.println(numTilings(N));
}
}
 
// This code is contributed by gfgking

Python3

# Python3 program for the above approache9 + 7;
 
# Function to find the total number
# of ways to tile a 2*N board using
# the given types of tile
MOD = 1e9 + 7
 
def numTilings(N):
     
    # If N is less than 3
    if (N < 3):
        return N
 
    # Store all dp-states
    dp = [[0] * 3 for i in range(N + 1)]
 
    # Base Case
    dp[0][0] = dp[1][0] = 1
    dp[1][1] = dp[1][2] = 1
 
    # Traverse the range [2, N]
    for i in range(2, N + 1):
 
        # Update the value of dp[i][0]
        dp[i][0] = (dp[i - 1][0] +
                    dp[i - 2][0] +
                    dp[i - 2][1] +
                    dp[i - 2][2]) % MOD
 
        # Update the value of dp[i][1]
        dp[i][1] = (dp[i - 1][0] +
                    dp[i - 1][2]) % MOD
 
        # Update the value of dp[i][2]
        dp[i][2] = (dp[i - 1][0] +
                    dp[i - 1][1]) % MOD
 
    # Return the number of ways as
    # the value of dp[N][0]
    return int(dp[N][0])
 
# Driver Code
N = 3
 
print(numTilings(N))
 
# This code is contributed by gfgking

C#

// C# program for the above approach
using System;
using System.Collections.Generic;
 
class GFG{
 
static int MOD = 1000000007;
 
// Function to find the total number
// of ways to tile a 2*N board using
// the given types of tile
static int numTilings(int N)
{
    // If N is less than 3
    if (N < 3) {
        return N;
    }
 
    // Store all dp-states
    int [,]dp = new int[N+1,3];
 
    // Base Case
    dp[0,0] = dp[1,0] = 1;
    dp[1,1] = dp[1,2] = 1;
 
    // Traverse the range [2, N]
    for (int i = 2; i <= N; i++) {
 
        // Update the value of dp[i,0]
        dp[i,0] = (dp[i - 1,0]
                    + dp[i - 2,0]
                    + dp[i - 2,1]
                    + dp[i - 2,2])
                   % MOD;
 
        // Update the value of dp[i,1]
        dp[i,1] = (dp[i - 1,0]
                    + dp[i - 1,2])
                   % MOD;
 
        // Update the value of dp[i,2]
        dp[i,2] = (dp[i - 1,0]
                    + dp[i - 1,1])
                   % MOD;
    }
 
    // Return the number of ways as
    // the value of dp[N,0]
    return dp[N,0];
}
 
// Driver Code
public static void Main()
{
    int N = 3;
    Console.Write(numTilings(N));
}
}
 
// This code is contributed by SURENDRA_GANGWAR.

Javascript

 
        // JavaScript program for the above approache9 + 7;
 
        // Function to find the total number
        // of ways to tile a 2*N board using
        // the given types of tile
        const MOD = 1e9 + 7;
        function numTilings(N) {
            // If N is less than 3
            if (N < 3) {
                return N;
            }
 
            // Store all dp-states
            let dp = Array(N + 1).fill().map(() => Array(3).fill(0))
 
            // Base Case
            dp[0][0] = dp[1][0] = 1;
            dp[1][1] = dp[1][2] = 1;
 
            // Traverse the range [2, N]
            for (let i = 2; i <= N; i++) {
 
                // Update the value of dp[i][0]
                dp[i][0] = (dp[i - 1][0]
                    + dp[i - 2][0]
                    + dp[i - 2][1]
                    + dp[i - 2][2])
                    % MOD;
 
                // Update the value of dp[i][1]
                dp[i][1] = (dp[i - 1][0]
                    + dp[i - 1][2])
                    % MOD;
 
                // Update the value of dp[i][2]
                dp[i][2] = (dp[i - 1][0]
                    + dp[i - 1][1])
                    % MOD;
            }
 
            // Return the number of ways as
            // the value of dp[N][0]
            return dp[N][0];
        }
 
        // Driver Code
 
        let N = 3;
        document.write(numTilings(N));
 
 
 
// This code is contributed by Potta Lokesh
    

输出：

5

时间复杂度： O(N)
辅助空间： O(N)

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