dfs 就行，列舉每個點作為起點進行搜尋。剩下的就是剪枝的問題了

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word) {
        tag.resize(board.size());
        for(auto &item : tag) item.resize(board[0].size(), false);
        for(int i = 0; i < board.size(); ++ i) {
            
 
for(int j = 0; j < board[0].size(); ++ j) {
                dfs(board, word, 0, i, j);
                if(ans) return true;
            }
        }
        return false;
    }

    void dfs(vector<vector<char>>& board, const string &word, int idx, int x, int y) {
        if(ans) return
 
 ;
        if(idx == word.size()) {
            ans = true;
            return ;
        }
        if(x < 0 || x >= board.size() || y < 0 || y >= board[0].size()) return ;
        if(tag[x][y]) return ;
        if(board[x][y] != word[idx]) return ;
        tag[x][y] = true;
        for(int i = 0
 
; i < 4; ++ i) {
            dfs(board, word, idx + 1, x + to[i][0], y + to[i][1]);
        }
        tag[x][y] = false;
    }

    bool ans = false;
    vector<vector<bool>> tag;
    int to[4][2] = {0, 1, 0, -1, 1, 0, -1, 0};
};