給定一個二維網格和一個單詞，找出該單詞是否存在於網格中。

單詞必須按照字母順序，通過相鄰的單元格內的字母構成，其中“相鄰”單元格是那些水平相鄰或垂直相鄰的單元格。同一個單元格內的字母不允許被重複使用。

示例:

board =
[
  ['A','B','C','E'],
  ['S','F','C','S'],
  ['A','D','E','E']
]

給定 word = "ABCCED", 返回 true
給定 word = "SEE", 返回 true
給定 word = "ABCB", 返回 false

來源：力扣（LeetCode）
連結：

應該dfs寫了個BFS，毛病。deepcopy佔用很多時間。
trick

ans = []
        for b in board:
            ans += b
        need = Counter(word)
        ans = Counter(ans)
        for k, v in need.items():
            if v > ans[k]:
                return False
 

import copy
from collections import defaultdict
class Solution:
    def exist(self, board, word: str) -> bool:
        m = len(board)
        n = len(board[0])
        stack = []
        flags = set([])
       
        for i in range(m):
            for j in range(n):
                if board[i][j] == word[0]:
                    t = copy.deepcopy(flags)
                    t.add((i,j))
                    stack.append((t,(i,j),1))
                    
        while len(stack)>0:
            #print(stack)
            top = stack[-1]
            stack = stack[:-1]
            t, (i,j), k = top
            if k>=len(word):
                return True
            for p, q in [(0,1),(0,-1),(1,0),(-1,0)]:
                if 0<=i+p<m and 0<=j+q<n and (i+p,j+q) not in t and word[k]==board[i+p][j+q]:
                    new_t = copy.deepcopy(t)
                    new_t.add((i+p,j+q))
                    stack.append((new_t,(i+p,j+q),k+1))

        return False