检查从给定像元到矩阵的任何边界元素的路径是否存在，且元素的总和不超过K

给定尺寸为M * N的矩阵grid [] [] ，三个整数X，Y和K ，任务是检查是否存在从单元(X，Y)到矩阵的任何边界单元的路径，从而路径中存在的矩阵元素之和最多为K。如果不存在这样的路径，请打印“ No” 。否则，打印“是” 。任何单元格都可能向上，向下，向左或向右移动。

例子：

Input: grid[][] = {{25, 5, 25, 25, 25, 25},  {25, 1, 1, 5, 12, 25}, {25, 1, 12, 0, 15, 25}, {22, 1, 11, 2, 19, 15}, {25, 2, 2, 1, 12, 15}, {25, 9, 10, 1, 11, 25}, {25, 25, 25, 25, 25, 25}}, X = 2, Y = 3, K = 17
Output: Yes
Explanation:
Below image illustrates one of the possible ways to reach a boundary element of the given matrix from the cell (2, 3):
 

The concerned path is (2, 3) -> (3, 3) -> (4, 3) -> (4, 2) -> (4, 1) -> (3, 1) -> (2, 1) -> (1, 1).
The cost of the path = 0 + 2 + 1 + 2 + 2 + 1 + 1 + 1 + 0 = 15(  < K).

Input: grid[][] = {{1, 2, 3}, {1, 2, 3}, {3, 4, 5}}, X = 1, Y = 1, K = 0
Output: -1

方法：可以通过使用回溯法考虑从给定起始单元格开始的所有可能路径，并检查到给定矩阵的边界单元是否存在其组成元素之和等于K的路径，从而解决给定问题。

请按照以下步骤解决问题：

• 定义一个递归函数，例如findPath(X，Y，K，board) ，以检查是否存在从起始像元(X，Y)到任何边界元素的路径。
  • 基本情况：如果达到X <0或Y <0或X> = M或Y> = N ，则返回true 。
  • 现在，检查grid [X] [Y]> =K 。如果发现为真，则标记当前访问的单元格。将K减小grid [X] [Y]的值，然后访问未访问的相邻像元，即递归调用findPath(X + 1，Y，K)，findPath(X，Y + 1，K)，findPath(X – 1，Y，K)和findPath(X，Y – 1，K) 。
    • 如果上述任何递归调用返回true ，则从当前递归调用返回true 。
    • 否则，将当前单元标记为未访问。
  • 如果以上条件均不满足，则从当前递归调用中返回false 。
• 现在，如果函数findPath(X，Y，K)返回true，则输出“ Yes” 。否则，打印“否” 。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to check if it is valid
// to move to the given index or not
bool isValid(vector >& board,
             int i, int j, int K)
{
    if (board[i][j] <= K) {
        return true;
    }
 
    // Otherwise return false
    return false;
}
 
// Function to check if there exists a
// path from the cell (X, Y) of the
// matrix to any boundary cell with
// sum of elements at most K or not
bool findPath(vector >& board,
              int X, int Y,
              int M, int N, int K)
{
 
    // Base Case
    if (X < 0 || X == M
        || Y < 0 || Y == N) {
        return true;
    }
 
    // If K >= board[X][Y]
    if (isValid(board, X, Y, K)) {
 
        // Stores the current element
        int board_XY = board[X][Y];
 
        // Mark the current cell as visited
        board[X][Y] = INT_MAX;
 
        // Visit all the adjacent cells
        if (findPath(board, X + 1, Y, M,
                     N, K - board_XY)
            || findPath(board, X - 1, Y, M,
                        N, K - board_XY)
            || findPath(board, X, Y + 1, M,
                        N, K - board_XY)
            || findPath(board, X, Y - 1, M,
                        N, K - board_XY)) {
            return true;
        }
 
        // Mark the cell unvisited
        board[X][Y] = board_XY;
    }
 
    // Return false
    return false;
}
 
// Driver Code
int main()
{
    vector > grid = {
        { 25, 5, 25, 25, 25, 25 },
        { 25, 1, 1, 5, 12, 25 },
        { 25, 1, 12, 0, 15, 25 },
        { 22, 1, 11, 2, 19, 15 },
        { 25, 2, 2, 1, 12, 15 },
        { 25, 9, 10, 1, 11, 25 },
        { 25, 25, 25, 25, 25, 25 }
    };
 
    int K = 17;
    int M = grid.size();
    int N = grid[0].size();
    int X = 2, Y = 3;
 
    if (findPath(grid, X, Y, M, N, K))
        cout << "Yes";
    else
        cout << "No";
 
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
 
class GFG{
     
// Function to check if it is valid
// to move to the given index or not
static boolean isValid(int[][] board, int i,
                       int j, int K)
{
    if (board[i][j] <= K)
    {
        return true;
    }
 
    // Otherwise return false
    return false;
}
 
// Function to check if there exists a
// path from the cell (X, Y) of the
// matrix to any boundary cell with
// sum of elements at most K or not
static boolean findPath(int[][] board, int X, int Y,
                        int M, int N, int K)
{
     
    // Base Case
    if (X < 0 || X == M || Y < 0 || Y == N)
    {
        return true;
    }
 
    // If K >= board[X][Y]
    if (isValid(board, X, Y, K))
    {
         
        // Stores the current element
        int board_XY = board[X][Y];
 
        // Mark the current cell as visited
        board[X][Y] = Integer.MAX_VALUE;
 
        // Visit all the adjacent cells
        if (findPath(board, X + 1, Y, M, N,
                            K - board_XY) ||
            findPath(board, X - 1, Y, M, N,
                            K - board_XY) ||
            findPath(board, X, Y + 1, M, N,
                               K - board_XY) ||
            findPath(board, X, Y - 1, M, N,
                               K - board_XY))
        {
            return true;
        }
 
        // Mark the cell unvisited
        board[X][Y] = board_XY;
    }
 
    // Return false
    return false;
}
 
// Driver Code
public static void main(String[] args)
{
    int[][] grid = { { 25, 5, 25, 25, 25, 25 },
                     { 25, 1, 1, 5, 12, 25 },
                     { 25, 1, 12, 0, 15, 25 },
                     { 22, 1, 11, 2, 19, 15 },
                     { 25, 2, 2, 1, 12, 15 },
                     { 25, 9, 10, 1, 11, 25 },
                     { 25, 25, 25, 25, 25, 25 } };
 
    int K = 17;
    int M = grid.length;
    int N = grid[0].length;
    int X = 2, Y = 3;
 
    if (findPath(grid, X, Y, M, N, K))
        System.out.println("Yes");
    else
        System.out.println("No");
}
}
 
// This code is contributed by ukasp

Python3

# Python3 program for the above approach
import sys
 
INT_MAX = sys.maxsize
 
# Function to check if it is valid
# to move to the given index or not
def isValid(board, i, j, K):
 
    if (board[i][j] <= K):
        return True
 
    # Otherwise return false
    return False
 
# Function to check if there exists a
# path from the cell (X, Y) of the
# matrix to any boundary cell with
# sum of elements at most K or not
def findPath(board, X, Y, M, N, K):
 
    # Base Case
    if (X < 0 or X == M or
        Y < 0 or Y == N):
        return True
 
    # If K >= board[X][Y]
    if (isValid(board, X, Y, K)):
 
        # Stores the current element
        board_XY = board[X][Y]
 
        # Mark the current cell as visited
        board[X][Y] = INT_MAX
 
        # Visit all the adjacent cells
        if (findPath(board, X + 1, Y, M,
                     N, K - board_XY) or
            findPath(board, X - 1, Y, M,
                     N, K - board_XY) or
            findPath(board, X, Y + 1, M,
                     N, K - board_XY) or
            findPath(board, X, Y - 1, M,
                     N, K - board_XY)):
            return True;
     
        # Mark the cell unvisited
        board[X][Y] = board_XY
 
    # Return false
    return False
 
# Driver Code
if __name__ ==  "__main__":
     
    grid = [
        [ 25, 5, 25, 25, 25, 25 ],
        [ 25, 1, 1, 5, 12, 25 ],
        [ 25, 1, 12, 0, 15, 25 ],
        [ 22, 1, 11, 2, 19, 15 ],
        [ 25, 2, 2, 1, 12, 15 ],
        [ 25, 9, 10, 1, 11, 25 ],
        [ 25, 25, 25, 25, 25, 25 ] ]
 
    K = 17
    M = len(grid)
    N = len(grid[0])
    X = 2
    Y = 3
 
    if (findPath(grid, X, Y, M, N, K)):
        print("Yes")
    else:
        print("No")
 
# This code is contributed by AnkThon

C#

// C# program for the above approach
using System;
 
class GFG{
     
// Function to check if it is valid
// to move to the given index or not
static bool isValid(int[,] board, int i,
                    int j, int K)
{
    if (board[i, j] <= K)
    {
        return true;
    }
 
    // Otherwise return false
    return false;
}
 
// Function to check if there exists a
// path from the cell (X, Y) of the
// matrix to any boundary cell with
// sum of elements at most K or not
static bool findPath(int[,] board, int X, int Y,
                     int M, int N, int K)
{
     
    // Base Case
    if (X < 0 || X == M ||
        Y < 0 || Y == N)
    {
        return true;
    }
 
    // If K >= board[X][Y]
    if (isValid(board, X, Y, K))
    {
         
        // Stores the current element
        int board_XY = board[X, Y];
 
        // Mark the current cell as visited
        board[X, Y] = int.MaxValue;
 
        // Visit all the adjacent cells
        if (findPath(board, X + 1, Y, M, N,
                            K - board_XY) ||
            findPath(board, X - 1, Y, M, N,
                            K - board_XY) ||
            findPath(board, X, Y + 1, M, N,
                               K - board_XY) ||
            findPath(board, X, Y - 1, M, N,
                               K - board_XY))
        {
            return true;
        }
 
        // Mark the cell unvisited
        board[X, Y] = board_XY;
    }
 
    // Return false
    return false;
}
 
// Driver Code
public static void Main(string[] args)
{
    int[,] grid = { { 25, 5, 25, 25, 25, 25 },
                    { 25, 1, 1, 5, 12, 25 },
                    { 25, 1, 12, 0, 15, 25 },
                    { 22, 1, 11, 2, 19, 15 },
                    { 25, 2, 2, 1, 12, 15 },
                    { 25, 9, 10, 1, 11, 25 },
                    { 25, 25, 25, 25, 25, 25 } };
 
    int K = 17;
    int M = grid.Length;
    int N = grid.GetLength(0);
    int X = 2, Y = 3;
 
    if (findPath(grid, X, Y, M, N, K))
        Console.WriteLine("Yes");
    else
        Console.WriteLine("No");
}
}
 
// This code is contributed by AnkThon

输出：

Yes

时间复杂度： O(3 ^{N * M} )
辅助空间： O(N * M)