思路是找没有被1围起来的棋子，遍历四条边，如果是0，则与它相邻的元素0肯定没有被围起来（继续深度优先搜索到所有相邻的0，且改变其状态，免得下一个位置遍历的时候重复搜索）。然后求剩下的面积即可
class Solution:
    def weiqi(self, grid: List[List[int]]) -> int:
        m = len(grid)
        n = len(grid[0])
        def dfs(x, y):
            if x<0 or x>=m or y<0 or y>=n or grid[x][y] == 1:
                return 0
            grid[x][y] = 1
            return 1 + dfs(x-1,y) + dfs(x+1,y) + dfs(x, y-1) + dfs(x, y+1)
        
        num_free = 0
        for r in range(m):
            if grid[r][0] == 0:
                num_free += dfs(r, 0)
            if grid[r][n-1] == 0:
                num_free += dfs(r, n-1)
        for c in range(n):
            if grid[0][c] == 0:
                num_free += dfs(0, c)
            if grid[m-1][c] == 0:
                num_free += dfs(m-1, c)
        
        return m*n - num_free

输入
[[0,1,1,0,1], [1,0,0,1,1],[0,1,1,1,0],[0,0,1,0,0]]
输出
12