Given an M x N matrix, with a few hurdles arbitrarily placed, calculate the length of the longest possible route possible from source to a destination within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contain any diagonal moves and a location once visited in a particular path cannot be visited again.
For example, the longest path with no hurdles from source to destination is highlighted below. The length of the path is 24.
The idea is to use Backtracking. We start from the source cell of the matrix, move forward in all four allowed directions, and recursively checks if they lead to the solution or not. If the destination is found, we update the value of the longest path else if none of the above solutions work we return false from our function.
// C++ program to find Longest Possible Route in a
// matrix with hurdles
#include <bits/stdc++.h>
using namespace std;
#define R 3
#define C 10
// A Pair to store status of a cell. found is set to
// true of destination is reachable and value stores
// distance of longest path
struct Pair {
// true if destination is found
bool found;
// stores cost of longest path from current cell to
// destination cell
int value;
};
// Function to find Longest Possible Route in the
// matrix with hurdles. If the destination is not reachable
// the function returns false with cost INT_MAX.
// (i, j) is source cell and (x, y) is destination cell.
Pair findLongestPathUtil(int mat[R][C], int i, int j, int x,
int y, bool visited[R][C])
{
// if (i, j) itself is destination, return true
if (i == x && j == y) {
Pair p = { true, 0 };
return p;
}
// if not a valid cell, return false
if (i < 0 || i >= R || j < 0 || j >= C || mat[i][j] == 0
|| visited[i][j]) {
Pair p = { false, INT_MAX };
return p;
}
// include (i, j) in current path i.e.
// set visited(i, j) to true
visited[i][j] = true;
// res stores longest path from current cell (i, j) to
// destination cell (x, y)
int res = INT_MIN;
// go left from current cell
Pair sol
= findLongestPathUtil(mat, i, j - 1, x, y, visited);
// if destination can be reached on going left from
// current cell, update res
if (sol.found)
res = max(res, sol.value);
// go right from current cell
sol = findLongestPathUtil(mat, i, j + 1, x, y, visited);
// if destination can be reached on going right from
// current cell, update res
if (sol.found)
res = max(res, sol.value);
// go up from current cell
sol = findLongestPathUtil(mat, i - 1, j, x, y, visited);
// if destination can be reached on going up from
// current cell, update res
if (sol.found)
res = max(res, sol.value);
// go down from current cell
sol = findLongestPathUtil(mat, i + 1, j, x, y, visited);
// if destination can be reached on going down from
// current cell, update res
if (sol.found)
res = max(res, sol.value);
// Backtrack
visited[i][j] = false;
// if destination can be reached from current cell,
// return true
if (res != INT_MIN) {
Pair p = { true, 1 + res };
return p;
}
// if destination can't be reached from current cell,
// return false
else {
Pair p = { false, INT_MAX };
return p;
}
}
// A wrapper function over findLongestPathUtil()
void findLongestPath(int mat[R][C], int i, int j, int x,
int y)
{
// create a boolean matrix to store info about
// cells already visited in current route
bool visited[R][C];
// initialize visited to false
memset(visited, false, sizeof visited);
// find longest route from (i, j) to (x, y) and
// print its maximum cost
Pair p = findLongestPathUtil(mat, i, j, x, y, visited);
if (p.found)
cout << "Length of longest possible route is "
<< p.value;
// If the destination is not reachable
else
cout << "Destination not reachable from given "
"source";
}
// Driver code
int main()
{
// input matrix with hurdles shown with number 0
int mat[R][C] = { { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 },
{ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 } };
// find longest path with source (0, 0) and
// destination (1, 7)
findLongestPath(mat, 0, 0, 1, 7);
return 0;
}
import java.io.*;
class solution {
static int R = 3;
static int C = 10;
// A Pair to store status of a cell. found is set to
// true of destination is reachable and value stores
// distance of longest path
static class Pair {
// true if destination is found
boolean found;
// stores cost of longest path from current cell to
// destination cell
int val;
Pair (boolean x, int y){
found = x;
val = y;
}
}
// Function to find Longest Possible Route in the
// matrix with hurdles. If the destination is not reachable
// the function returns false with cost Integer.MAX_VALUE.
// (i, j) is source cell and (x, y) is destination cell.
static Pair findLongestPathUtil (int mat[][], int i, int j, int x, int y, boolean visited[][]) {
// if (i, j) itself is destination, return true
if(i == x && j == y)
return new Pair(true, 0);
// if not a valid cell, return false
if(i < 0 || i >= R || j < 0 || j >= C || mat[i][j] == 0 || visited[i][j] )
return new Pair(false, Integer.MAX_VALUE);
// include (i, j) in current path i.e.
// set visited(i, j) to true
visited[i][j] = true;
// res stores longest path from current cell (i, j) to
// destination cell (x, y)
int res = Integer.MIN_VALUE;
// go left from current cell
Pair sol = findLongestPathUtil(mat, i, j-1, x, y, visited);
// if destination can be reached on going left from current
// cell, update res
if(sol.found)
res = Math.max(sol.val, res);
// go right from current cell
sol = findLongestPathUtil(mat, i, j+1, x, y, visited);
// if destination can be reached on going right from current
// cell, update res
if(sol.found)
res = Math.max(sol.val, res);
// go up from current cell
sol = findLongestPathUtil(mat, i-1, j, x, y, visited);
// if destination can be reached on going up from current
// cell, update res
if(sol.found)
res = Math.max(sol.val, res);
// go down from current cell
sol = findLongestPathUtil(mat, i+1, j, x, y, visited);
// if destination can be reached on going down from current
// cell, update res
if(sol.found)
res = Math.max(sol.val, res);
// Backtrack
visited[i][j] = false;
// if destination can be reached from current cell,
// return true
if(res != Integer.MIN_VALUE)
return new Pair(true, res+1);
// if destination can't be reached from current cell,
// return false
else
return new Pair(false, Integer.MAX_VALUE);
}
// A wrapper function over findLongestPathUtil()
static void findLongestPath (int mat[][], int i, int j, int x, int y) {
// create a boolean matrix to store info about
// cells already visited in current route
boolean visited[][] = new boolean[R][C];
// find longest route from (i, j) to (x, y) and
// print its maximum cost
Pair p = findLongestPathUtil(mat, i, j, x, y, visited);
if(p.found)
System.out.println("Length of longest possible route is " + p.val);
// If the destination is not reachable
else
System.out.println("Destination not reachable from given source");
}
// Driver Code
public static void main (String[] args) {
// input matrix with hurdles shown with number 0
int mat[][] = { { 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1, 1, 0, 1 },
{ 1, 1, 1, 1, 1, 1, 1, 1, 1, 1 } };
// find longest path with source (0, 0) and
// destination (1, 7)
findLongestPath(mat, 0, 0, 1, 7);
}
}
## Output
Length of longest possible route is 24