On a 2-dimensional grid, there are 4 types of squares:

1 represents the starting square. There is exactly one starting square.
2 represents the ending square. There is exactly one ending square.
0 represents empty squares we can walk over.
-1 represents obstacles that we cannot walk over.

Return the number of 4-directional walks from the starting square to the ending square, that walk over every non-obstacle square exactly once.

Example 1:

Input: [[1,0,0,0],[0,0,0,0],[0,0,2,-1]]
Output: 2
Explanation: We have the following two paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2)
2. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2)

Example 2:

Input: [[1,0,0,0],[0,0,0,0],[0,0,0,2]]
Output: 4
Explanation: We have the following four paths: 
1. (0,0),(0,1),(0,2),(0,3),(1,3),(1,2),(1,1),(1,0),(2,0),(2,1),(2,2),(2,3)
2. (0,0),(0,1),(1,1),(1,0),(2,0),(2,1),(2,2),(1,2),(0,2),(0,3),(1,3),(2,3)
3. (0,0),(1,0),(2,0),(2,1),(2,2),(1,2),(1,1),(0,1),(0,2),(0,3),(1,3),(2,3)
4. (0,0),(1,0),(2,0),(2,1),(1,1),(0,1),(0,2),(0,3),(1,3),(1,2),(2,2),(2,3)

Example 3:

Input: [[0,1],[2,0]]
Output: 0
Explanation: 
There is no path that walks over every empty square exactly once.
Note that the starting and ending square can be anywhere in the grid.

Note:

1 <= grid.length * grid[0].length <= 20

Solution

Time complexity : O(4^mn)
Space complexity : O(mn)

class Solution {
public:
    int uniquePathsIII(vector<vector<int>>& grid) {
        int start_x = -1;
        int start_y = -1;
        int n = 1;
        for (int y=0; y<grid.size(); ++y) {
            for (int x=0; x<grid[0].size(); ++x) {
                if (grid[y][x] == 0) ++n;
                else if (grid[y][x] == 1) {
                    start_x = x;
                    start_y = y;
                }
            }
        }
        return dfs(grid, start_x, start_y, n);
    }
private:
    int dfs(vector<vector<int>>& grid, int x, int y, int n) {
        if (x<0 || x==grid[0].size() ||
            y<0 || y==grid.size() ||
            grid[y][x] == -1) return 0;
        else if (grid[y][x] == 2) return n==0;
        grid[y][x] = -1;
        int paths = dfs(grid, x+1, y, n-1) +
                    dfs(grid, x-1, y, n-1) +
                    dfs(grid, x, y+1, n-1) +
                    dfs(grid, x, y-1, n-1);
        grid[y][x] = 0;
        return paths;
    }
};

一開始找到起始位置、和空位數n，然後從起始位置開始dfs & Backtracking。

若超出範圍、或是該值為-1(走過)，則回傳0。
若為目的地、且剩餘空位n==0，代表走好走滿走到目的地，回傳1，反之回傳0。

都不是的話，繼續往下走。
將目前格設為已走過(-1)，然後四個方位都走一遍。
走完後回復該格(0)，回傳這個位置的走法結果。