矩阵中的最长递增路径

给定一个N行M列的矩阵。从m[i][j]，我们可以移动到m[i+1][j]，如果m[i+1][j] > m[i][j]，或者可以移动到m[i] [j+1] 如果 m[i][j+1] > m[i][j]。如果我们从 (0, 0) 开始，则任务是打印最长路径长度。
例子：

Input : N = 4, M = 4
        m[][] = { { 1, 2, 3, 4 },
                  { 2, 2, 3, 4 },
                  { 3, 2, 3, 4 },
                  { 4, 5, 6, 7 } };
Output : 7
Longest path is 1 2 3 4 5 6 7.

Input : N = 2, M =2
        m[][] = { { 1, 2 },
                  { 3, 4 } };
Output :3
Longest path is either 1 2 4 or 
1 3 4.

这个想法是使用动态规划。维护二维矩阵 dp[][]，其中 dp[i][j] 存储从第 i 行第 j 列开始的子矩阵的最长递增序列的长度值。
让 m[i+1][j] 和 m[i][j+1] 的最长递增子序列值分别已知为 v1 和 v2。那么 m[i][j] 的值将是 max(v1, v2) + 1。
我们可以从 m[n-1][m-1] 开始，作为基本情况，最长递增子序列的长度为 1，向上和向左移动更新单元格的值。那么单元格 m[0][0] 的 LIP 值将是答案。
下面是这个方法的实现：

C++

// CPP program to find longest increasing
// path in a matrix.
#include 
#define MAX 10
using namespace std;
 
// Return the length of LIP in 2D matrix
int LIP(int dp[][MAX], int mat[][MAX], int n, int m, int x, int y)
{
    // If value not calculated yet.
    if (dp[x][y] < 0) {
        int result = 0;
 
        // If reach bottom left cell, return 1.
        if (x == n - 1 && y == m - 1)
            return dp[x][y] = 1;
 
        // If reach the corner of the matrix.
        if (x == n - 1 || y == m - 1)
            result = 1;
 
        // If value greater than below cell.
        if (mat[x][y] < mat[x + 1][y])
            result = 1 + LIP(dp, mat, n, m, x + 1, y);
 
        // If value greater than left cell.
        if (mat[x][y] < mat[x][y + 1])
            result = max(result, 1 + LIP(dp, mat, n, m, x, y + 1));
 
        dp[x][y] = result;
    }
 
    return dp[x][y];
}
 
// Wrapper function
int wrapper(int mat[][MAX], int n, int m)
{
    int dp[MAX][MAX];
    memset(dp, -1, sizeof dp);
 
    return LIP(dp, mat, n, m, 0, 0);
}
 
// Driven Program
int main()
{
    int mat[][MAX] = {
        { 1, 2, 3, 4 },
        { 2, 2, 3, 4 },
        { 3, 2, 3, 4 },
        { 4, 5, 6, 7 },
    };
    int n = 4, m = 4;
    cout << wrapper(mat, n, m) << endl;
 
    return 0;
}

Java

// Java program to find longest increasing
// path in a matrix.
import java.util.*;
 
class GFG {
 
    // Return the length of LIP in 2D matrix
    static int LIP(int dp[][], int mat[][], int n,
                   int m, int x, int y)
    {
        // If value not calculated yet.
        if (dp[x][y] < 0) {
            int result = 0;
 
            // If reach bottom left cell, return 1.
            if (x == n - 1 && y == m - 1)
                return dp[x][y] = 1;
 
            // If reach the corner of the matrix.
            if (x == n - 1 || y == m - 1)
                result = 1;
 
            // If value greater than below cell.
            if (x + 1 < n && mat[x][y] < mat[x + 1][y])
                result = 1 + LIP(dp, mat, n, m, x + 1, y);
 
            // If value greater than left cell.
            if (y + 1 < m && mat[x][y] < mat[x][y + 1])
                result = Math.max(result, 1 + LIP(dp, mat, n, m, x, y + 1));
 
            dp[x][y] = result;
        }
 
        return dp[x][y];
    }
 
    // Wrapper function
    static int wrapper(int mat[][], int n, int m)
    {
        int dp[][] = new int[10][10];
        for (int i = 0; i < 10; i++)
            Arrays.fill(dp[i], -1);
 
        return LIP(dp, mat, n, m, 0, 0);
    }
 
    /* Driver program to test above function */
    public static void main(String[] args)
    {
        int mat[][] = {
            { 1, 2, 3, 4 },
            { 2, 2, 3, 4 },
            { 3, 2, 3, 4 },
            { 4, 5, 6, 7 },
        };
        int n = 4, m = 4;
        System.out.println(wrapper(mat, n, m));
    }
}
 
// This code is contributed by Arnav Kr. Mandal.

Python3

# Python3 program to find longest
# increasing path in a matrix.
MAX = 20
 
# Return the length of
# LIP in 2D matrix
def LIP(dp, mat, n, m, x, y):
     
    # If value not calculated yet.
    if (dp[x][y] < 0):
        result = 0
         
        # If reach bottom left cell,
        # return 1.
        if (x == n - 1 and y == m - 1):
            dp[x][y] = 1
            return dp[x][y]
 
        # If reach the corner
        # of the matrix.
        if (x == n - 1 or y == m - 1):
            result = 1
 
        # If value greater than below cell.
        if (x + 1 < n and mat[x][y] < mat[x + 1][y]):
            result = 1 + LIP(dp, mat, n,
                            m, x + 1, y)
 
        # If value greater than left cell.
        if (y + 1 < m and mat[x][y] < mat[x][y + 1]):
            result = max(result, 1 + LIP(dp, mat, n,
                                        m, x, y + 1))
        dp[x][y] = result
    return dp[x][y]
 
# Wrapper function
def wrapper(mat, n, m):
    dp = [[-1 for i in range(MAX)]
            for i in range(MAX)]
    return LIP(dp, mat, n, m, 0, 0)
 
# Driver Code
mat = [[1, 2, 3, 4 ],
    [2, 2, 3, 4 ],
    [3, 2, 3, 4 ],
    [4, 5, 6, 7 ]]
n = 4
m = 4
print(wrapper(mat, n, m))
 
# This code is contributed
# by Sahil Shelangia

C#

// C# program to find longest increasing
// path in a matrix.
using System;
 
public class GFG {
 
    // Return the length of LIP in 2D matrix
    static int LIP(int[, ] dp, int[, ] mat, int n,
                   int m, int x, int y)
    {
        // If value not calculated yet.
        if (dp[x, y] < 0) {
            int result = 0;
 
            // If reach bottom left cell, return 1.
            if (x == n - 1 && y == m - 1)
                return dp[x, y] = 1;
 
            // If reach the corner of the matrix.
            if (x == n - 1 || y == m - 1)
                result = 1;
 
            // If value greater than below cell.
            if (x + 1 < n && mat[x, y] < mat[x + 1, y])
                result = 1 + LIP(dp, mat, n, m, x + 1, y);
 
            // If value greater than left cell.
            if (y + 1 < m && mat[x, y] < mat[x, y + 1])
                result = Math.Max(result, 1 + LIP(dp, mat, n, m, x, y + 1));
 
            dp[x, y] = result;
        }
 
        return dp[x, y];
    }
 
    // Wrapper function
    static int wrapper(int[, ] mat, int n, int m)
    {
        int[, ] dp = new int[10, 10];
        for (int i = 0; i < 10; i++) {
            for (int j = 0; j < 10; j++) {
                dp[i, j] = -1;
            }
        }
 
        return LIP(dp, mat, n, m, 0, 0);
    }
 
    /* Driver code */
    public static void Main()
    {
        int[, ] mat = {
            { 1, 2, 3, 4 },
            { 2, 2, 3, 4 },
            { 3, 2, 3, 4 },
            { 4, 5, 6, 7 },
        };
        int n = 4, m = 4;
        Console.WriteLine(wrapper(mat, n, m));
    }
}
 
/* This code contributed by PrinciRaj1992 */

Javascript

    // Javascript program to find longest increasing path in a matrix.
     
    // Return the length of LIP in 2D matrix
    function LIP(dp, mat, n, m, x, y)
    {
        // If value not calculated yet.
        if (dp[x][y] < 0) {
            let result = 0;
   
            // If reach bottom left cell, return 1.
            if (x == n - 1 && y == m - 1)
                return dp[x][y] = 1;
   
            // If reach the corner of the matrix.
            if (x == n - 1 || y == m - 1)
                result = 1;
   
            // If value greater than below cell.
            if (x + 1 < n && mat[x][y] < mat[x + 1][y])
                result = 1 + LIP(dp, mat, n, m, x + 1, y);
   
            // If value greater than left cell.
            if (y + 1 < m && mat[x][y] < mat[x][y + 1])
                result = Math.max(result, 1 + LIP(dp, mat, n, m, x, y + 1));
   
            dp[x][y] = result;
        }
   
        return dp[x][y];
    }
   
    // Wrapper function
    function wrapper(mat, n, m)
    {
        let dp = new Array(10);
        for (let i = 0; i < 10; i++)
        {
            dp[i] = new Array(10);
            for (let j = 0; j < 10; j++)
            {
                dp[i][j] = -1;
            }
        }
   
        return LIP(dp, mat, n, m, 0, 0);
    }
     
    let mat = [
            [ 1, 2, 3, 4 ],
            [ 2, 2, 3, 4 ],
            [ 3, 2, 3, 4 ],
            [ 4, 5, 6, 7 ],
        ];
    let n = 4, m = 4;
    document.write(wrapper(mat, n, m));
     

输出：

7

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