广度优先搜索; 把访问过的位置标记为1; def bfs(A, i, j): m,n = len(A), len(A[0]) if i == m-1 and j == n-1: return 0, [(i,j)] if i < 0 or i >= m or j < 0 or j >= n or A[i][j] == '1': return m*n, None A[i][j] = '1' d_min = m*n p_min = [] for (ai,aj) in [(i-1,j),(i+1,j),(i,j-1),(i,j+1)]: d_pq, p_pq = bfs(A,...