计算在矩阵中达到给定分数的方法数

给定一个N×N的矩阵垫[] []由非负整数的，任务是找到方法来达到给定的分数m距离细胞(0，0)开始次数和到达细胞(N – 1 , N – 1)仅向下(从 (i, j) 到 (i + 1, j))或向右(从 (i, j) 到 (i, j + 1))。每当到达单元格(i, j)时，总得分由currentScore + mat[i][j] 更新。
例子：

Input: mat[][] = {{1, 1, 1}, 
{1, 1, 1}, 
{1, 1, 1}}, M = 4 
Output: 0 
All the paths will result in a score of 5.
Input: mat[][] = {{1, 1, 1}, 
{1, 1, 1}, 
{1, 1, 1}}, M = 5 
Output: 6 
 

方法：这个问题可以用动态规划解决。首先，我们需要决定DP的状态。对于每个单元格(i, j)和数字0 ≤ X ≤ M ，我们将存储从单元格(0, 0)到达该单元格的方式数，总分为X 。因此，我们的解决方案将使用 3 维动态规划，两个用于单元格的坐标，一个用于所需的分数值。
所需的递归关系将是，

dp[i][j][M] = dp[i – 1][j][M – mat[i][j]] + dp[i][j-1][M – mat[i][j]] 
 

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define n 3
#define MAX 30
 
// To store the states of dp
int dp[n][n][MAX];
 
// To check whether a particular state
// of dp has been solved
bool v[n][n][MAX];
 
// Function to find the ways
// using memoization
int findCount(int mat[][n], int i, int j, int m)
{
    // Base cases
    if (i == 0 && j == 0) {
        if (m == mat[0][0])
            return 1;
        else
            return 0;
    }
 
    // If required score becomes negative
    if (m < 0)
        return 0;
 
    if (i < 0 || j < 0)
        return 0;
 
    // If current state has been reached before
    if (v[i][j][m])
        return dp[i][j][m];
 
    // Set current state to visited
    v[i][j][m] = true;
 
    dp[i][j][m] = findCount(mat, i - 1, j, m - mat[i][j])
                  + findCount(mat, i, j - 1, m - mat[i][j]);
    return dp[i][j][m];
}
 
// Driver code
int main()
{
    int mat[n][n] = { { 1, 1, 1 },
                      { 1, 1, 1 },
                      { 1, 1, 1 } };
    int m = 5;
    cout << findCount(mat, n - 1, n - 1, m);
 
    return 0;
}

Java

// Java implementation of the approach
class GFG
{
 
static int n = 3;
static int MAX =30;
 
// To store the states of dp
static int dp[][][] = new int[n][n][MAX];
 
// To check whether a particular state
// of dp has been solved
static boolean v[][][] = new boolean[n][n][MAX];
 
// Function to find the ways
// using memoization
static int findCount(int mat[][], int i, int j, int m)
{
    // Base cases
    if (i == 0 && j == 0)
    {
        if (m == mat[0][0])
            return 1;
        else
            return 0;
    }
 
    // If required score becomes negative
    if (m < 0)
        return 0;
 
    if (i < 0 || j < 0)
        return 0;
 
    // If current state has been reached before
    if (v[i][j][m])
        return dp[i][j][m];
 
    // Set current state to visited
    v[i][j][m] = true;
 
    dp[i][j][m] = findCount(mat, i - 1, j, m - mat[i][j])
                + findCount(mat, i, j - 1, m - mat[i][j]);
    return dp[i][j][m];
}
 
// Driver code
public static void main(String[] args)
{
    int mat[][] = { { 1, 1, 1 },
                    { 1, 1, 1 },
                    { 1, 1, 1 } };
    int m = 5;
    System.out.println(findCount(mat, n - 1, n - 1, m));
}
}
 
/* This code contributed by PrinciRaj1992 */

Python3

# Python3 implementation of the approach
n = 3
MAX = 60
 
# To store the states of dp
dp = [[[0 for i in range(30)]  
          for i in range(30)]
          for i in range(MAX + 1)]
 
# To check whether a particular state
# of dp has been solved
v = [[[0 for i in range(30)]
         for i in range(30)]
         for i in range(MAX + 1)]
 
# Function to find the ways
# using memoization
def findCount(mat, i, j, m):
     
    # Base cases
    if (i == 0 and j == 0):
        if (m == mat[0][0]):
            return 1
        else:
            return 0
 
    # If required score becomes negative
    if (m < 0):
        return 0
 
    if (i < 0 or j < 0):
        return 0
 
    # If current state has been reached before
    if (v[i][j][m] > 0):
        return dp[i][j][m]
 
    # Set current state to visited
    v[i][j][m] = True
 
    dp[i][j][m] = (findCount(mat, i - 1, j,
                             m - mat[i][j]) +
                   findCount(mat, i, j - 1,
                             m - mat[i][j]))
 
    return dp[i][j][m]
 
# Driver code
mat = [ [ 1, 1, 1 ],
        [ 1, 1, 1 ],
        [ 1, 1, 1 ] ]
m = 5
print(findCount(mat, n - 1, n - 1, m))
 
# This code is contributed by mohit kumar

C#

// C# implementation of the approach
using System;
 
class GFG
{
 
static int n = 3;
static int MAX = 30;
 
// To store the states of dp
static int [,,]dp = new int[n, n, MAX];
 
// To check whether a particular state
// of dp has been solved
static bool [,,]v = new bool[n, n, MAX];
 
// Function to find the ways
// using memoization
static int findCount(int [,]mat, int i,
                          int j, int m)
{
    // Base cases
    if (i == 0 && j == 0)
    {
        if (m == mat[0, 0])
            return 1;
        else
            return 0;
    }
 
    // If required score becomes negative
    if (m < 0)
        return 0;
 
    if (i < 0 || j < 0)
        return 0;
 
    // If current state has been reached before
    if (v[i, j, m])
        return dp[i, j, m];
 
    // Set current state to visited
    v[i, j, m] = true;
 
    dp[i, j, m] = findCount(mat, i - 1, j, m - mat[i, j]) +
                  findCount(mat, i, j - 1, m - mat[i, j]);
    return dp[i, j, m];
}
 
// Driver code
public static void Main()
{
    int [,]mat = {{ 1, 1, 1 },
                  { 1, 1, 1 },
                  { 1, 1, 1 }};
    int m = 5;
    Console.WriteLine(findCount(mat, n - 1, n - 1, m));
}
}
 
// This code is contributed by Ryuga

---

Javascript

 
// Javascript implementation of the approach
var n = 3;
var MAX = 30;
 
// To store the states of dp
var dp = Array(n);
for(var i =0; i<n; i++)
{
    dp[i] = Array(n);
    for(var j=0; j<n; j++)
    {
        dp[i][j]=Array(MAX);
    }
}
 
// To check whether a particular state
// of dp has been solved
var v = Array(n);
for(var i =0; i<n; i++)
{
    v[i] = Array(n);
    for(var j=0; j<n; j++)
    {
        v[i][j]=Array(MAX);
    }
}
 
// Function to find the ways
// using memoization
function findCount(mat, i, j, m)
{
    // Base cases
    if (i == 0 && j == 0) {
        if (m == mat[0][0])
            return 1;
        else
            return 0;
    }
 
    // If required score becomes negative
    if (m < 0)
        return 0;
 
    if (i < 0 || j < 0)
        return 0;
 
    // If current state has been reached before
    if (v[i][j][m])
        return dp[i][j][m];
 
    // Set current state to visited
    v[i][j][m] = true;
 
    dp[i][j][m] = findCount(mat, i - 1, j,
                            m - mat[i][j])
                  + findCount(mat, i, j - 1,
                              m - mat[i][j]);
    return dp[i][j][m];
}
 
// Driver code
var mat = [ [ 1, 1, 1 ],
                  [ 1, 1, 1 ],
                  [ 1, 1, 1 ] ];
var m = 5;
document.write( findCount(mat, n - 1, n - 1, m));
 

输出：

6

时间复杂度： O(N * N * M)

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