Longest Possible Route in a Matrix with Hurdles - GeeksforGeeks

Longest Possible Route in a Matrix with Hurdles - GeeksforGeeks
Given an M x N matrix, with a few hurdles arbitrarily placed, calculate the length of longest possible route possible from source to destination within the matrix. We are allowed to move to only adjacent cells which are not hurdles. The route cannot contains any diagonal moves and a location once visited in a particular path cannot be visited again.
For example, longest path with no hurdles from source to destination is highlighted for below matrix. 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 leads to the solution or not. If destination is found, we update the value of longest path else if none of the above solutions work we return false from our function.

// 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];

 

    // initailize 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";

}

Read full article from Longest Possible Route in a Matrix with Hurdles - GeeksforGeeks