以K步计算遍历矩阵并返回原点的方法数量

给定三个整数N ， M和K ，其中N和M是矩阵的维数，而K是最大可能步长，任务是计算从(0，0)开始并通过以下方式返回遍历矩阵的数量方法仅K步。
注意：在每一步中，可以向上，向下，向左，向右移动或停留在当前位置。答案可能很大，因此将答案模数为10 ⁹ + 7
例子：

Input: N = 2, M = 2, K = 2 
Output: 3 
Explanation: 
Three ways are: 
1)Stay, Stay. 
2)Up, Down 
3)Right, Left

Input: N = 3, M = 3, K = 4 
Output: 23 
 

方法：也可以使用记忆化技术解决问题。请按照以下步骤解决问题：

• 从位置(0，0)开始，递归调用每个可能的位置并将step的值减小1。
• 创建大小为N * M * K的3D数组( DP [N] [M] [S] )

  Possible ways for each step
  can be calculated by:
  
   DP[i][j][k]
    =  Recursion(i+1, j, k-1)  +
       Recursion(i-1, j, k-1)  +  
       Recursion(i, j-1, k-1)  +  
       Recursion(i, j+1, k-1)  +  
       Recursion(i, j, k-1).
  
  With base condition returning 0,
  whenever
  i >= l or i = b or j < 0 or k < 0.
  

下面是上述方法的实现：

C++

// C++ program to count total
// number of ways to return
// to origin after completing
// given number of steps.
  
#include 
using namespace std;
#define MOD 1000000007
  
long long dp[101][101][101];
int N, M, K;
  
// Function Initialize dp[][][]
// array with -1
void Initialize()
{
  
    for (int i = 0; i <= 100; i++)
        for (int j = 0; j <= 100; j++)
            for (int z = 0; z <= 100; z++)
                dp[i][j][z] = -1;
}
  
// Function returns the total count
int CountWays(int i, int j, int k)
{
    if (i >= N || i < 0
        || j >= M || j < 0 || k < 0)
        return 0;
  
    if (i == 0 && j == 0
        && k == 0)
        return 1;
  
    if (dp[i][j][k] != -1)
        return dp[i][j][k];
    else
        dp[i][j][k]
            = (CountWays(i + 1, j, k - 1) % MOD
               + CountWays(i - 1, j, k - 1) % MOD
               + CountWays(i, j - 1, k - 1) % MOD
               + CountWays(i, j + 1, k - 1) % MOD
               + CountWays(i, j, k - 1) % MOD)
              % MOD;
  
    return dp[i][j][k];
}
  
// Driver Program
int main()
{
    N = 3;
    M = 3;
    K = 4;
  
    Initialize();
    cout << CountWays(0, 0, K)
         << "\n";
  
    return 0;
}

Java

// Java program to count total 
// number of ways to return 
// to origin after completing 
// given number of steps. 
class GFG{
      
static int [][][] dp = new int[101][101][101]; 
static int N, M, K;
static int MOD = 1000000007; 
      
// Function Initialize dp[][][] 
// array with -1 
public static void Initialize() 
{ 
    for(int i = 0; i <= 100; i++) 
       for(int j = 0; j <= 100; j++) 
          for(int z = 0; z <= 100; z++) 
             dp[i][j][z] = -1; 
} 
      
// Function returns the total count 
public static int CountWays(int i, int j, int k) 
{ 
    if (i >= N || i < 0 ||
        j >= M || j < 0 || k < 0) 
        return 0; 
      
    if (i == 0 && j == 0 && k == 0) 
        return 1; 
      
    if (dp[i][j][k] != -1) 
        return dp[i][j][k]; 
      
    else
        dp[i][j][k] = (CountWays(i + 1, j, k - 1) % MOD +
                       CountWays(i - 1, j, k - 1) % MOD +
                       CountWays(i, j - 1, k - 1) % MOD +
                       CountWays(i, j + 1, k - 1) % MOD +
                       CountWays(i, j, k - 1) % MOD) % MOD; 
      
    return dp[i][j][k]; 
} 
  
// Driver code
public static void main(String[] args) 
{ 
    N = 3; 
    M = 3; 
    K = 4; 
      
    Initialize(); 
    System.out.println(CountWays(0, 0, K));
} 
} 
  
// This code is contributed by grand_master

Python3

# Python3 program to count total
# number of ways to return
# to origin after completing
# given number of steps.
MOD = 1000000007
  
dp = [[[0 for i in range(101)] 
          for i in range(101)] 
          for i in range(101)]
N, M, K = 0, 0, 0
  
# Function Initialize dp[][][]
# array with -1
def Initialize():
  
    for i in range(101):
        for j in range(101):
            for z in range(101):
                dp[i][j][z] = -1
  
# Function returns the total count
def CountWays(i, j, k):
  
    if (i >= N or i < 0 or
        j >= M or j < 0 or k < 0):
        return 0
  
    if (i == 0 and j == 0 and k == 0):
        return 1
  
    if (dp[i][j][k] != -1):
        return dp[i][j][k]
    else:
        dp[i][j][k] = (CountWays(i + 1, j, k - 1) % MOD +
                       CountWays(i - 1, j, k - 1) % MOD +
                       CountWays(i, j - 1, k - 1) % MOD +
                       CountWays(i, j + 1, k - 1) % MOD +
                       CountWays(i, j, k - 1) % MOD) % MOD
    return dp[i][j][k]
  
# Driver code
if __name__ == '__main__':
      
    N = 3
    M = 3
    K = 4
  
    Initialize()
    print(CountWays(0, 0, K))
  
# This code is contributed by mohit kumar 29

C#

// C# program to count total 
// number of ways to return 
// to origin after completing 
// given number of steps. 
using System;
  
class GFG{ 
      
static int [,,] dp = new int[101, 101, 101]; 
static int N, M, K; 
static int MOD = 1000000007; 
      
// Function Initialize [,]dp[] 
// array with -1 
public static void Initialize() 
{ 
    for(int i = 0; i <= 100; i++) 
        for(int j = 0; j <= 100; j++) 
            for(int z = 0; z <= 100; z++) 
                dp[i, j, z] = -1; 
} 
      
// Function returns the total count 
public static int CountWays(int i, int j, int k) 
{ 
    if (i >= N || i < 0 || 
        j >= M || j < 0 || k < 0) 
        return 0; 
      
    if (i == 0 && j == 0 && k == 0) 
        return 1; 
      
    if (dp[i, j, k] != -1) 
        return dp[i, j, k]; 
      
    else
        dp[i, j, k] = (CountWays(i + 1, j, k - 1) % MOD + 
                       CountWays(i - 1, j, k - 1) % MOD + 
                       CountWays(i, j - 1, k - 1) % MOD + 
                       CountWays(i, j + 1, k - 1) % MOD + 
                       CountWays(i, j, k - 1) % MOD) % MOD; 
      
    return dp[i, j, k]; 
} 
  
// Driver code 
public static void Main(String[] args) 
{ 
    N = 3; 
    M = 3; 
    K = 4; 
      
    Initialize(); 
    Console.WriteLine(CountWays(0, 0, K)); 
} 
} 
  
// This code is contributed by Amit Katiyar

输出：

23

如果您希望与行业专家一起参加现场课程，请参阅《 Geeks现场课程》和《 Geeks现场课程美国》。