给定一个* n矩阵，其中所有数字都是不同的，请找到最大长度路径(从任何单元格开始)，以使路径上的所有单元格都以递增顺序排列，相差1。
我们可以从给定的像元(i，j)向4个方向移动，即我们可以移动到(i + 1，j)或(i，j + 1)或(i-1，j)或(i，j -1)，条件是相邻像元的差为1。
例子：
Input: mat[][] = {{1, 2, 9}
{5, 3, 8}
{4, 6, 7}}
Output: 4
The longest path is 6-7-8-9.
这个想法很简单，我们计算从每个单元格开始的最长路径。一旦计算出所有单元的最长，就返回所有最长路径的最大值。这种方法的一个重要发现是许多重叠的子问题。因此，可以使用动态编程来最佳地解决此问题。
以下是基于动态编程的实现，该实现使用查找表dp [] []来检查问题是否已解决。
C/C++
// C++ program to find the longest path in a matrix
// with given constraints
#include
#define n 3
using namespace std;
// Returns length of the longest path beginning with mat[i][j].
// This function mainly uses lookup table dp[n][n]
int findLongestFromACell(int i, int j, int mat[n][n], int dp[n][n])
{
if (i < 0 || i >= n || j < 0 || j >= n)
return 0;
// If this subproblem is already solved
if (dp[i][j] != -1)
return dp[i][j];
// To store the path lengths in all the four directions
int x = INT_MIN, y = INT_MIN, z = INT_MIN, w = INT_MIN;
// Since all numbers are unique and in range from 1 to n*n,
// there is atmost one possible direction from any cell
if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
x = 1 + findLongestFromACell(i, j + 1, mat, dp);
if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
y = 1 + findLongestFromACell(i, j - 1, mat, dp);
if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
z = 1 + findLongestFromACell(i - 1, j, mat, dp);
if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
w = 1 + findLongestFromACell(i + 1, j, mat, dp);
// If none of the adjacent fours is one greater we will take 1
// otherwise we will pick maximum from all the four directions
return dp[i][j] = max(x, max(y, max(z, max(w, 1))));
}
// Returns length of the longest path beginning with any cell
int finLongestOverAll(int mat[n][n])
{
int result = 1; // Initialize result
// Create a lookup table and fill all entries in it as -1
int dp[n][n];
memset(dp, -1, sizeof dp);
// Compute longest path beginning from all cells
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (dp[i][j] == -1)
findLongestFromACell(i, j, mat, dp);
// Update result if needed
result = max(result, dp[i][j]);
}
}
return result;
}
// Driver program
int main()
{
int mat[n][n] = { { 1, 2, 9 },
{ 5, 3, 8 },
{ 4, 6, 7 } };
cout << "Length of the longest path is "
<< finLongestOverAll(mat);
return 0;
} Java
// Java program to find the longest path in a matrix
// with given constraints
class GFG {
public static int n = 3;
// Function that returns length of the longest path
// beginning with mat[i][j]
// This function mainly uses lookup table dp[n][n]
static int findLongestFromACell(int i, int j, int mat[][], int dp[][])
{
// Base case
if (i < 0 || i >= n || j < 0 || j >= n)
return 0;
// If this subproblem is already solved
if (dp[i][j] != -1)
return dp[i][j];
// To store the path lengths in all the four directions
int x = Integer.MIN_VALUE, y = Integer.MIN_VALUE, z = Integer.MIN_VALUE, w = Integer.MIN_VALUE;
// Since all numbers are unique and in range from 1 to n*n,
// there is atmost one possible direction from any cell
if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1]))
x = dp[i][j] = 1 + findLongestFromACell(i, j + 1, mat, dp);
if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1]))
y = dp[i][j] = 1 + findLongestFromACell(i, j - 1, mat, dp);
if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j]))
z = dp[i][j] = 1 + findLongestFromACell(i - 1, j, mat, dp);
if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j]))
w = dp[i][j] = 1 + findLongestFromACell(i + 1, j, mat, dp);
// If none of the adjacent fours is one greater we will take 1
// otherwise we will pick maximum from all the four directions
return dp[i][j] = Math.max(x, Math.max(y, Math.max(z, Math.max(w, 1))));
}
// Function that returns length of the longest path
// beginning with any cell
static int finLongestOverAll(int mat[][])
{
// Initialize result
int result = 1;
// Create a lookup table and fill all entries in it as -1
int[][] dp = new int[n][n];
for (int i = 0; i < n; i++)
for (int j = 0; j < n; j++)
dp[i][j] = -1;
// Compute longest path beginning from all cells
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (dp[i][j] == -1)
findLongestFromACell(i, j, mat, dp);
// Update result if needed
result = Math.max(result, dp[i][j]);
}
}
return result;
}
// driver program
public static void main(String[] args)
{
int mat[][] = { { 1, 2, 9 },
{ 5, 3, 8 },
{ 4, 6, 7 } };
System.out.println("Length of the longest path is " + finLongestOverAll(mat));
}
}
// Contributed by Pramod KumarPython3
# Python3 program to find the longest path in a matrix
# with given constraints
n = 3
# Returns length of the longest path beginning with mat[i][j].
# This function mainly uses lookup table dp[n][n]
def findLongestFromACell(i, j, mat, dp):
# Base case
if (i<0 or i>= n or j<0 or j>= n):
return 0
# If this subproblem is already solved
if (dp[i][j] != -1):
return dp[i][j]
# To store the path lengths in all the four directions
x, y, z, w = -1, -1, -1, -1
# Since all numbers are unique and in range from 1 to n * n,
# there is atmost one possible direction from any cell
if (j0 and (mat[i][j] +1 == mat[i][j-1])):
y = 1 + findLongestFromACell(i, j-1, mat, dp)
if (i>0 and (mat[i][j] +1 == mat[i-1][j])):
z = 1 + findLongestFromACell(i-1, j, mat, dp)
if (i C#
// C# program to find the longest path
// in a matrix with given constraints
using System;
class GFG {
public static int n = 3;
// Function that returns length of
// the longest path beginning with mat[i][j]
// This function mainly uses lookup
// table dp[n][n]
public static int findLongestFromACell(int i, int j,
int[][] mat,
int[][] dp)
{
// Base case
if (i < 0 || i >= n || j < 0 || j >= n) {
return 0;
}
// If this subproblem is
// already solved
if (dp[i][j] != -1) {
return dp[i][j];
}
// To store the path lengths in all the four directions
int x = int.MinValue, y = int.MinValue, z = int.MinValue, w = int.MinValue;
// Since all numbers are unique and
// in range from 1 to n*n, there is
// atmost one possible direction
// from any cell
if (j < n - 1 && ((mat[i][j] + 1) == mat[i][j + 1])) {
x = dp[i][j] = 1 + findLongestFromACell(i, j + 1, mat, dp);
}
if (j > 0 && (mat[i][j] + 1 == mat[i][j - 1])) {
y = dp[i][j] = 1 + findLongestFromACell(i, j - 1, mat, dp);
}
if (i > 0 && (mat[i][j] + 1 == mat[i - 1][j])) {
z = dp[i][j] = 1 + findLongestFromACell(i - 1, j, mat, dp);
}
if (i < n - 1 && (mat[i][j] + 1 == mat[i + 1][j])) {
w = dp[i][j] = 1 + findLongestFromACell(i + 1, j, mat, dp);
}
// If none of the adjacent fours is one greater we will take 1
// otherwise we will pick maximum from all the four directions
dp[i][j] = Math.Max(x, Math.Max(y, Math.Max(z, Math.Max(w, 1))));
return dp[i][j];
}
// Function that returns length of the
// longest path beginning with any cell
public static int finLongestOverAll(int[][] mat)
{
// Initialize result
int result = 1;
// Create a lookup table and fill
// all entries in it as -1
int[][] dp = RectangularArrays.ReturnRectangularIntArray(n, n);
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
dp[i][j] = -1;
}
}
// Compute longest path beginning
// from all cells
for (int i = 0; i < n; i++) {
for (int j = 0; j < n; j++) {
if (dp[i][j] == -1) {
findLongestFromACell(i, j, mat, dp);
}
// Update result if needed
result = Math.Max(result, dp[i][j]);
}
}
return result;
}
public static class RectangularArrays {
public static int[][] ReturnRectangularIntArray(int size1,
int size2)
{
int[][] newArray = new int[size1][];
for (int array1 = 0;
array1 < size1; array1++) {
newArray[array1] = new int[size2];
}
return newArray;
}
}
// Driver Code
public static void Main(string[] args)
{
int[][] mat = new int[][] {
new int[] { 1, 2, 9 },
new int[] { 5, 3, 8 },
new int[] { 4, 6, 7 }
};
Console.WriteLine("Length of the longest path is " + finLongestOverAll(mat));
}
}
// This code is contributed by Shrikant13输出：
Length of the longest path is 4上述解决方案的时间复杂度为O(n 2 )。乍看起来似乎更多。如果仔细研究，我们会发现dp [i] [j]的所有值仅计算一次。