Find paths from corner cell to middle cell in maze in C++

C++Server Side ProgrammingProgramming

Suppose we have a square maze filled with numbers; we have to find all paths from a corner cell to the middle cell. Here, we will proceed exactly n steps from a cell in 4 directions Up, Down, Right and Left where n is the value of the cell. Thus, we can move to cell [i+n,j] to [i-n, j], [i, j+n], and [i, j-n] from a cell [i,j] where n is value of cell [i, j].

So, if the input is like

344473463
675662662
334325472
655123656
334301434
334321335
354326443
351375363
624345451

then the output will be

  • (0, 0)→(0, 3)→(0, 7)→(6, 7)→(6, 3)→(3, 3)→(3, 4)→(5, 4)→(5, 2)→(1, 2)→(1, 7)→(7, 7)→(7, 1)→(2, 1)→(5, 1)→(0, 1)→(4, 1)→(4, 4)→MIDDLE

  • (0, 0)→(0, 3)→(0, 7)→(6, 7)→(6, 3)→(3, 3)→(3, 4)→(5, 4)→(5, 2)→(1, 2)→(1, 7)→(7, 7)→(7, 1)→(2, 1)→(2, 4)→(4, 4→MIDDLE

  • (0, 0)→(0, 3)→(0, 7)→(0, 1)→(4, 1)→(7, 1)→(2, 1)→(2, 4)→(4, 4)→MIDDLE

  • (0, 0)→(0, 3)→(0, 7)→(0, 1)→(4, 1)→(4, 4)→MIDDLE

  • (8, 8)→(7, 8)→(4, 8)→(4, 4)→MIDDLE

To solve this, we will follow these steps −

  • N := 9

  • Define a function is_ok(), this will take one set of pairs called visited, one pair pt,

  • return true when first and second element of pt in range 0 to N and pt is not in visited

  • Define an array dir_row := { - 1, 1, 0, 0}

  • Define an array dir_col := { 0, 0, - 1, 1}

  • Define an array row := { 0, 0, N - 1, N - 1}

  • Define an array col := { 0, N - 1, 0, N - 1}

  • Define a function solve(), this will take maze, path, one set of pairs visited, one pair curr,

  • if first and second of curr is same as N / 2, then −

    • display the path

    • return

  • for initialize i := 0, when i < 4, update (increase i by 1), do −

    • n := maze[curr.first, curr.second]

    • x := curr.first + dir_row[i] * n

    • y := curr.second + dir_col[i] * n

    • n := a pair using x, y

    • if is_ok(visited, next), then −

      • insert next into visited

      • insert next at the end of path

      • solve(maze, path, visited, next)

      • delete last element from path

      • delete next from visited

  • From the main method, do the following −

  • Define one set of pairs called visited

  • for initialize i := 0, when i < 4, update (increase i by 1), do −

    • x := row[i]

    • y := col[i]

    • pt := make a pair using (x, y)

    • insert pt into visited

    • insert pt at the end of path

    • solve(maze, path, visited, pt)

    • delete last element from path

    • delete pt from visited

Example (C++)

Let us see the following implementation to get better understanding −

 Live Demo

#include <bits/stdc++.h>
using namespace std;
#define N 9
bool is_ok(set<pair<int, int> > visited, pair<int, int> pt) {
   return (pt.first >= 0) && (pt.first < N) && (pt.second >= 0) && (pt.second < N) && (visited.find(pt) == visited.end());
}
void display_path(list<pair<int, int> > path) {
   for (auto it = path.begin(); it != path.end(); it++)
   cout << "(" << it->first << ", " << it->second << ")->";
   cout << "MIDDLE" << endl << endl;
}
int dir_row[] = {-1, 1, 0, 0};
int dir_col[] = { 0, 0, -1, 1};
int row[] = { 0, 0, N-1, N-1};
int col[] = { 0, N-1, 0, N-1};
void solve(int maze[N][N], list<pair<int, int> > &path, set<pair<int, int> > &visited, pair<int, int> &curr) {
   if (curr.first == N / 2 && curr.second == N / 2) {
      display_path(path);
      return;
   }
   for (int i = 0; i < 4; ++i) {
      int n = maze[curr.first][curr.second];
      int x = curr.first + dir_row[i]*n;
      int y = curr.second + dir_col[i]*n;
      pair<int, int> next = make_pair(x, y);
      if (is_ok(visited, next)) {
         visited.insert(next);
         path.push_back(next);
         solve(maze, path, visited, next);
         path.pop_back();
         visited.erase(next);
      }
   }
}
void search_path(int maze[N][N]) {
   list<pair<int, int> > path;
   set<pair<int, int> > visited;
   for (int i = 0; i < 4; ++i) {
      int x = row[i];
      int y = col[i];
      pair<int, int> pt = make_pair(x, y);
      visited.insert(pt);
      path.push_back(pt);
      solve(maze, path, visited, pt);
      path.pop_back();
      visited.erase(pt);
   }
}
int main() {
   int maze[N][N] = {
      {3, 4, 4, 4, 7, 3, 4, 6, 3},
      {6, 7, 5, 6, 6, 2, 6, 6, 2},
      {3, 3, 4, 3, 2, 5, 4, 7, 2},
      {6, 5, 5, 1, 2, 3, 6, 5, 6},
      {3, 3, 4, 3, 0, 1, 4, 3, 4},
      {3, 5, 4, 3, 2, 1, 3, 3, 5},
      {3, 5, 4, 3, 2, 6, 4, 4, 3},
      {3, 5, 1, 3, 7, 5, 3, 6, 3},
      {6, 2, 4, 3, 4, 5, 4, 5, 1}
   };
   search_path(maze);
}

Input

{{3, 4, 4, 4, 7, 3, 4, 6, 3},
{6, 7, 5, 6, 6, 2, 6, 6, 2},
{3, 3, 4, 3, 2, 5, 4, 7, 2},
{6, 5, 5, 1, 2, 3, 6, 5, 6},
{3, 3, 4, 3, 0, 1, 4, 3, 4},
{3, 5, 4, 3, 2, 1, 3, 3, 5},
{3, 5, 4, 3, 2, 6, 4, 4, 3},
{3, 5, 1, 3, 7, 5, 3, 6, 3},
{6, 2, 4, 3, 4, 5, 4, 5, 1}}

Output

(0, 0)->(0, 3)->(0, 7)->(6, 7)->(6, 3)->(3, 3)->(3, 4)->(5, 4)->(5, 2)->(1, 2)->(1, 7)->(7, 7)->(7, 1)->(2, 1)->(5, 1)->(0, 1)->(4, 1)->(4, 4)->MIDDLE
(0, 0)->(0, 3)->(0, 7)->(6, 7)->(6, 3)->(3, 3)->(3, 4)->(5, 4)->(5, 2)->(1, 2)->(1, 7)->(7, 7)->(7, 1)->(2, 1)->(2, 4)->(4, 4)->MIDDLE
(0, 0)->(0, 3)->(0, 7)->(0, 1)->(4, 1)->(7, 1)->(2, 1)->(2, 4)->(4, 4)->MIDDLE
(0, 0)->(0, 3)->(0, 7)->(0, 1)->(4, 1)->(4, 4)->MIDDLE
(8, 8)->(7, 8)->(4, 8)->(4, 4)->MIDDLE
raja
Published on 19-Aug-2020 11:08:35
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