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.
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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 pathstruct 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";}