<p>Given an <code>m x n</code> <code>board</code>&nbsp;of characters and a list of strings <code>words</code>, return <em>all words on the board</em>.</p>

<p>Each word must be constructed from letters of sequentially adjacent cells, where <strong>adjacent cells</strong> are horizontally or vertically neighboring. The same letter cell may not be used more than once in a word.</p>

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search1.jpg" style="width: 322px; height: 322px;" />
<pre>
<strong>Input:</strong> board = [[&quot;o&quot;,&quot;a&quot;,&quot;a&quot;,&quot;n&quot;],[&quot;e&quot;,&quot;t&quot;,&quot;a&quot;,&quot;e&quot;],[&quot;i&quot;,&quot;h&quot;,&quot;k&quot;,&quot;r&quot;],[&quot;i&quot;,&quot;f&quot;,&quot;l&quot;,&quot;v&quot;]], words = [&quot;oath&quot;,&quot;pea&quot;,&quot;eat&quot;,&quot;rain&quot;]
<strong>Output:</strong> [&quot;eat&quot;,&quot;oath&quot;]
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/07/search2.jpg" style="width: 162px; height: 162px;" />
<pre>
<strong>Input:</strong> board = [[&quot;a&quot;,&quot;b&quot;],[&quot;c&quot;,&quot;d&quot;]], words = [&quot;abcb&quot;]
<strong>Output:</strong> []
</pre>

<p>&nbsp;</p>
<p><strong>Constraints:</strong></p>

<ul>
	<li><code>m == board.length</code></li>
	<li><code>n == board[i].length</code></li>
	<li><code>1 &lt;= m, n &lt;= 12</code></li>
	<li><code>board[i][j]</code> is a lowercase English letter.</li>
	<li><code>1 &lt;= words.length &lt;= 3 * 10<sup>4</sup></code></li>
	<li><code>1 &lt;= words[i].length &lt;= 10</code></li>
	<li><code>words[i]</code> consists of lowercase English letters.</li>
	<li>All the strings of <code>words</code> are unique.</li>
</ul>

class Solution {
public:
    struct Trie{
        struct Node{
            unordered_map<char, int> to;
            int cnt = 0;
            Node(): cnt(0){}
        };
        vector<Node> d;
        Trie(): d(1){}
        void add(int i, const string& s){
            int v = 0;
            for(char c : s){
                if(!d[v].to.count(c)){
                    d[v].to[c] = d.size();
                    d.push_back(Node());
                }
                v = d[v].to[c];
            }
            d[v].cnt = i+1;
        }
        int dx[4] = {0, 0, 1, -1};
        int dy[4] = {1, -1, 0, 0};
        void dfs(int i, int j, vector<vector<bool>>& bm, int l, vector<int>& a, vector<vector<char>>& b){
            if(l == 0){
                if(d[l].to.count(b[i][j])){
                    bm[i][j] = 1;
                    dfs(i, j, bm, d[l].to[b[i][j]], a, b);
                    bm[i][j] = 0;
                }
                return;
            }
            int m = b.size(), n = b[0].size();
            if(d[l].cnt) a.push_back(d[l].cnt-1);
            for(int k = 0; k < 4; k++){
                int xx = i + dx[k], yy = j + dy[k];
                if(xx >= 0 && xx < m && yy >= 0 && yy < n && !bm[xx][yy] && d[l].to.count(b[xx][yy])){
                    bm[xx][yy] = 1;
                    dfs(xx, yy, bm, d[l].to[b[xx][yy]], a, b);
                    bm[xx][yy] = 0;
                }
            }
            return;
        }
    };
 
    vector<string> findWords(vector<vector<char>>& board, vector<string>& words) {
        int m = board.size(), n = board[0].size();
        Trie t;
        vector<int> a;
        for(int i = 0; i < words.size(); i++) t.add(i, words[i]);
        vector bm(m, vector<bool> (n, 0));
        for(int i = 0; i < m; i++) for(int j = 0; j < n; j++) t.dfs(i, j, bm, 0, a, board);
        vector<bool> bb(words.size(), 1);
        vector<string> ans;
        for(auto i : a){
            if(bb[i]){
                ans.push_back(words[i]);
                bb[i] = 0;
            }
        }
        return ans;
    }
};
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