leetcode 980. Unique Paths III

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

思路：深搜， 终止条件到达目标位置，以及可到达的位置全部走了一遍，算一条路径。

 class Solution {
     int dx[] = {, -, , };
     int dy[] = {, , -, };
 public:
     int uniquePathsIII(vector<vector<int>>& grid) {
         int m = grid.size();
         if (m == )
             return ;
         int n = grid[].size();
         int todo = ;
         int start_x, start_y, end_x, end_y;
         for (int i = ; i < m; i++) {
             for (int j = ; j < n; j++) {
                 if (grid[i][j] != -) { //记录要走的总的位置数
                     todo++;
                     if (grid[i][j] == ) { //记录起始位置
                         start_x = i;
                         start_y = j;
                     } else if (grid[i][j] == ) { //记录终点
                         end_x = i;
                         end_y = j;
                     }
                 }
             }
         }
         int ans = ;
         dfs(grid, start_x, start_y, end_x, end_y, todo, ans, m, n);
         return ans;
     }
     void dfs(vector<vector<int> > &grid, int sx, int sy, const int ex, const int ey, int todo, int &ans, int row, int col) {
         todo--;
         if (todo < )
             return ;
         if (sx == ex && sy == ey) {
             if (todo == ) ans++;
             return;
         }
         //上下左右四个方向
         for (int k = ; k < ; k++) {
             int new_x = sx + dx[k];
             int new_y = sy + dy[k];
             if (new_x >=  && new_x < row && new_y >=  && new_y < col) {
                 if (grid[new_x][new_y] ==  || grid[new_x][new_y] == ) {
                     grid[new_x][new_y] = -;
                     dfs(grid, new_x, new_y, ex, ey, todo, ans, row, col);
                     grid[new_x][new_y] = ;
                 }
             }
         }
     }
 };