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81 changes: 81 additions & 0 deletions dynamic_programming/ways_maze.cpp
Original file line number Diff line number Diff line change
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// C++ program to count number of paths in a maze
// with obstacles.
#include<bits/stdc++.h>
using namespace std;
#define R 4
#define C 4

// Returns count of possible paths in a maze[R][C]
// from (0,0) to (R-1,C-1)
int countPaths(int maze[][C])
{
// If the initial cell is blocked, there is no
// way of moving anywhere
if (maze[0][0]==-1)
return 0;

// Initializing the leftmost column
for (int i=0; i<R; i++)
{
if (maze[i][0] == 0)
maze[i][0] = 1;

// If we encounter a blocked cell in leftmost
// row, there is no way of visiting any cell
// directly below it.
else
break;
}

// Similarly initialize the topmost row
for (int i=1; i<C; i++)
{
if (maze[0][i] == 0)
maze[0][i] = 1;

// If we encounter a blocked cell in bottommost
// row, there is no way of visiting any cell
// directly below it.
else
break;
}

// The only difference is that if a cell is -1,
// simply ignore it else recursively compute
// count value maze[i][j]
for (int i=1; i<R; i++)
{
for (int j=1; j<C; j++)
{
// If blockage is found, ignore this cell
if (maze[i][j] == -1)
continue;

// If we can reach maze[i][j] from maze[i-1][j]
// then increment count.
if (maze[i-1][j] > 0)
maze[i][j] = (maze[i][j] + maze[i-1][j]);

// If we can reach maze[i][j] from maze[i][j-1]
// then increment count.
if (maze[i][j-1] > 0)
maze[i][j] = (maze[i][j] + maze[i][j-1]);
}
}

// If the final cell is blocked, output 0, otherwise
// the answer
return (maze[R-1][C-1] > 0)? maze[R-1][C-1] : 0;
}

// Driver code
int main()
{
int maze[R][C] = {{0, 0, 0, 0},
{0, -1, 0, 0},
{-1, 0, 0, 0},
{0, 0, 0, 0}};
cout << countPaths(maze);
return 0;
}