矩阵中的最小成本路径仅在 X 的值差上移动

给定一个矩阵mat[][]和一个整数X ，任务是找到达到
$(N, M)$从$(1, 1)$ .在每次移动中，我们可以在矩阵中向右或向下移动，但是要移动矩阵中的下一个单元格，该单元格的值必须是$mat[i][j] + X$ .在一次操作中，任何单元格的值都可以减 1。

例子：

Input: mat[][] = {{8, 10, 14}, {5, 41, 19}, {10, 2, 25}}, X = 3 
Output: 11 
Explanation: 
After performing the operations on the matrix: 
7 10 13 
5 41 16 
10 2 19
Here the minimum required operation is 11. 
8 => 7 = 1 
14 => 13 = 1 
19 => 16 = 3 
25 => 19 = 6
Path: 7 => 10 => 13 => 16 => 19

Input: mat[][] = {{15, 153}, {135, 17}}, X = 3 
Output: 125 
 

方法：思路是用动态规划来解决这个问题。一般的想法是迭代矩阵的每个可能的单元格，如果当前单元格的值在最终路径中没有改变，则找到所需的操作次数。如果当前单元格的值为$mat[i][j]$ , 那么单元格所需的值是$mat[i][j] - X * (i + j)$ .同样，我们可以递归计算所需的操作数。

下面是上述方法的实现：

C++

// C++ implementation to find the
// minimum number of operations
// required to move from
// (1, 1) to (N, M)
 
#include 
using namespace std;
 
const long long MAX = 1e18;
 
long long n, m;
vector v[151];
long long dp[151][151];
 
// Function to find the minimum
// operations required to move
// to bottom-right cell of matrix
long long min_operation(long long i,
    long long j, long long val, long long x) {
         
    // Condition to check if the
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1) {
        if (val > v[i][j]) {
            return dp[i][j] = MAX;
        }
        else {
            return dp[i][j] =
                v[i][j] - val;
        }
    }
     
    // Condition to check if the
    // current cell is out of
    // matrix
    if (i == n || j == m) {
        return dp[i][j] = MAX;
    }
     
    // Condition to check if the
    // current indices is already
    // computed
    if (dp[i][j] != -1) {
        return dp[i][j];
    }
     
    // Condition to check that the
    // movement with the current
    // value is not possible
    if (val > v[i][j]) {
        return dp[i][j] = MAX;
    }
    long long tmp = v[i][j] - val;
     
    // Recursive call to compute the
    // number of operation required
    // to compute the value
    tmp += min(min_operation(i + 1,
            j, val + x, x),
            min_operation(i,
            j + 1, val + x, x));
    return dp[i][j] = tmp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
long long solve(long long x)
{
    long long ans = INT64_MAX;
     
    // Loop to iterate over every
    // possible cell of the matrix
    for (long long i = 0;
        i < n; i++) {
 
        for (long long j = 0;
            j < m; j++) {
 
            long long val =
                v[i][j] - x * (i + j);
 
            memset(dp, -1,
                sizeof(dp));
 
            val = min_operation(
                0, 0, val, x);
 
            ans = min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
int main()
{
    n = 2, m = 2;
    long long x = 3;
     
    v[0] = { 15, 153 };
    v[1] = { 135, 17 };
     
    // Function Call
    cout << solve(x) << endl;
    return 0;
}

Java

// Java implementation to find the 
// minimum number of operations
// required to move from 
// (1, 1) to (N, M)
import java.lang.*;
import java.util.*;
 
class GFG{
     
static final long MAX = (long)1e18;
static long n, m;
 
static List> v = new ArrayList<>(151);
static long[][] dp = new long[151][151];
 
// Function to find the minimum
// operations required to move
// to bottom-right cell of matrix
static long min_operation(long i, long j,
                          long val, long x)
{
     
    // Condition to check if the
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1)
    {
        if (val > v.get((int)i).get((int)j))
        {
            return dp[(int)i][(int)j] = MAX;
        }
        else
        {
            return dp[(int)i][(int)j] =
                v.get((int)i).get((int)j) - val;
        }
    }
     
    // Condition to check if the
    // current cell is out of
    // matrix
    if (i == n || j == m)
    {
        return dp[(int)i][(int)j] = MAX;
    }
     
    // Condition to check if the
    // current indices is already
    // computed
    if (dp[(int)i][(int)j] != -1)
    {
        return dp[(int)i][(int)j];
    }
     
    // Condition to check that the
    // movement with the current
    // value is not possible
    if (val > v.get((int)i).get((int)j))
    {
        return dp[(int)i][(int)j] = MAX;
    }
    long tmp = v.get((int)i).get((int)j) - val;
     
    // Recursive call to compute the
    // number of operation required
    // to compute the value
    tmp += Math.min(min_operation(i + 1,
                             j, val + x, x),
                    min_operation(i, j + 1,
                                   val + x, x));
                                    
    return dp[(int)i][(int)j] = tmp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
static long solve(long x)
{
    long ans = Long.MAX_VALUE;
     
    // Loop to iterate over every
    // possible cell of the matrix
    for(long i = 0; i < n; i++)
    {
        for(long j = 0; j < m; j++)
        {
            long val = v.get((int)i).get((int)j) -
                       x * (i + j);
     
            for(int k = 0; k < dp.length; k++)
                for(int l = 0; l < dp[k].length; l++)
                    dp[k][l] = -1;
                     
            val = min_operation(0l, 0l, val, x);
            ans = Math.min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
public static void main (String[] args)
{
    n = 2; m = 2;
    long x = 3;
     
    v.add(Arrays.asList(15l, 153l));
    v.add(Arrays.asList(135l, 17l));
     
    // Function call
    System.out.println(solve(x));
}
}
 
// This code is contributed by offbeat

Python3

# Python3 implementation to find the
# minimum number of operations
# required to move from
# (1, 1) to (N, M)
MAX = 1e18
v = [[ 0, 0]] * (151)
dp = [[-1 for i in range(151)]
        for i in range(151)]
 
# Function to find the minimum
# operations required to move
# to bottom-right cell of matrix
def min_operation(i, j, val, x):
 
    # Condition to check if the
    # current cell is the bottom-right
    # cell of the matrix
    if (i == n - 1 and j == m - 1):
        if (val > v[i][j]):
            dp[i][j] = MAX
            return MAX
 
        else:
            dp[i][j] = v[i][j] - val
            return dp[i][j]
 
    # Condition to check if the
    # current cell is out of
    # matrix
    if (i == n or j == m):
        dp[i][j] = MAX
        return MAX
 
    # Condition to check if the
    # current indices is already
    # computed
    if (dp[i][j] != -1):
        return dp[i][j]
 
    # Condition to check that the
    # movement with the current
    # value is not possible
    if (val > v[i][j]):
        dp[i][j] = MAX
        return MAX
 
    tmp = v[i][j] - val
 
    # Recursive call to compute the
    # number of operation required
    # to compute the value
    tmp += min(min_operation(i + 1, j,
                        val + x, x),
            min_operation(i, j + 1,
                            val + x, x))
    dp[i][j] = tmp
 
    return tmp
 
# Function to find the minimum
# number of operations required
# to reach the bottom-right cell
def solve(x):
     
    ans = 10 ** 19
 
    # Loop to iterate over every
    # possible cell of the matrix
    for i in range(n):
        for j in range(m):
            val = v[i][j] - x * (i + j)
 
            for ii in range(151):
                for jj in range(151):
                    dp[ii][jj] = -1
 
            val = min_operation(0, 0, val, x)
            ans = min(ans, val)
    return ans
 
# Driver Code
if __name__ == '__main__':
     
    n = 2
    m = 2
    x = 3
 
    v[0] = [ 15, 153 ]
    v[1] = [ 135, 17 ]
 
    # Function Call
    print(solve(x))
 
# This code is contributed by mohit kumar 29

C#

// C# implementation to find the  
// minimum number of operations 
// required to move from  
// (1, 1) to (N, M) 
using System;
using System.Collections.Generic;
 
class GFG{
     
static long MAX = (long)1e18;
static long n, m;
static List> v = new List>();
 
static long[,] dp = new long[151, 151];
 
// Function to find the minimum 
// operations required to move
// to bottom-right cell of matrix
static long min_operation(long i, long j,
                          long val, long x)
{
     
    // Condition to check if the 
    // current cell is the bottom-right
    // cell of the matrix
    if (i == n - 1 && j == m - 1)
    {
        if (val > v[(int)i][(int)j])
        {
            return dp[(int)i, (int)j] = MAX;
        }
        else
        {
            return dp[(int)i, (int)j] = v[(int)i][(int)j] - val;
        }
    }
     
    // Condition to check if the
    // current cell is out of 
    // matrix
    if (i == n || j == m) 
    {
        return dp[(int)i, (int)j] = MAX;
    }
     
    // Condition to check if the
    // current indices is already 
    // computed 
    if (dp[(int)i, (int)j] != -1)
    {
        return dp[(int)i, (int)j];
    }
     
    // Condition to check that the 
    // movement with the current 
    // value is not possible 
    if (val > v[(int)i][(int)j])
    {
        return dp[(int)i, (int)j] = MAX;
         
    }
     
    long temp = v[(int)i][(int)j] - val;
     
    // Recursive call to compute the 
    // number of operation required
    // to compute the value
    temp += Math.Min(min_operation(i + 1, j, val + x, x),
                     min_operation(i, j + 1, val + x, x));
    return dp[(int)i, (int)j] = temp;
}
 
// Function to find the minimum
// number of operations required
// to reach the bottom-right cell
static long solve(long x)
{
     
    long ans = Int64.MaxValue;
     
    // Loop to iterate over every 
    // possible cell of the matrix
    for(long i = 0; i < n; i++)
    {
        for(long j = 0; j < m; j++)
        {
            long val = v[(int)i][(int)j] - x * (i + j);
            for(long k = 0; k < dp.GetLength(0); k++)
            {
                for(long l = 0; l < dp.GetLength(1); l++)
                {
                    dp[k, l] = -1;
                }
            }
            val = min_operation(0, 0, val, x);
            ans = Math.Min(ans, val);
        }
    }
    return ans;
}
 
// Driver Code
static public void Main()
{
    for(int i = 0; i < 151; i++)
    {
        v.Add(new List());
        v[i].Add(0);
        v[i].Add(0);
    }
    v[0][0] = 15;
    v[0][1] = 153;
    v[1][0] = 135;
    v[1][1] = 17;
    n = 2; m = 2;
    long x = 3;
     
    // Function call
    Console.WriteLine(solve(x));
}
}
 
// This code is contributed by avanitrachhadiya2155

输出：

125

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