<p>Given an <code>m x n</code> grid of characters <code>board</code> and a string <code>word</code>, return <code>true</code> <em>if</em> <code>word</code> <em>exists in the grid</em>.</p>

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

<p>&nbsp;</p>
<p><strong class="example">Example 1:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word2.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCCED&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 2:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/11/04/word-1.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;SEE&quot;
<strong>Output:</strong> true
</pre>

<p><strong class="example">Example 3:</strong></p>
<img alt="" src="https://assets.leetcode.com/uploads/2020/10/15/word3.jpg" style="width: 322px; height: 242px;" />
<pre>
<strong>Input:</strong> board = [[&quot;A&quot;,&quot;B&quot;,&quot;C&quot;,&quot;E&quot;],[&quot;S&quot;,&quot;F&quot;,&quot;C&quot;,&quot;S&quot;],[&quot;A&quot;,&quot;D&quot;,&quot;E&quot;,&quot;E&quot;]], word = &quot;ABCB&quot;
<strong>Output:</strong> false
</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;= 6</code></li>
	<li><code>1 &lt;= word.length &lt;= 15</code></li>
	<li><code>board</code> and <code>word</code> consists of only lowercase and uppercase English letters.</li>
</ul>

<p>&nbsp;</p>
<p><strong>Follow up:</strong> Could you use search pruning to make your solution faster with a larger <code>board</code>?</p>

class Solution {
public:
    bool exist(vector<vector<char>>& board, string word)
    {
        std::vector<std::vector<int>> table(board.size(), std::vector<int>(board[0].size(), 1));
        std::stack<std::tuple<int, int, bool>> sta;
        int idx = 0;
        for (int y = 0; y<board.size(); ++y)
        {
            for (int x = 0; x<board[0].size(); ++x)
            {
                if (board[y][x]==word[idx])
                {
                    sta.emplace(x, y, 0);
                    sta.emplace(x, y, 1);
                }
            }
        }

        std::vector<int> dirs{0, 1, 0, -1, 0};
        while (!sta.empty())
        {
            auto t = sta.top();
            sta.pop();
            if (idx==word.size()) return true;

            int x = std::get<0>(t);
            int y = std::get<1>(t);
            int b = std::get<2>(t);

            table[y][x] = false;

            if (b==0)
            {
                table[y][x] = true;
                --idx;
                continue;
            }

            ++idx;
            for (int d = 0; d<4; ++d)
            {
                if (x+dirs[d]>=0 && y+dirs[d+1]>=0 &&
                    x+dirs[d]<board[0].size() && y+dirs[d+1]<board.size() &&
                    table[y+dirs[d+1]][x+dirs[d]] &&
                    board[y+dirs[d+1]][x+dirs[d]]==word[idx])
                {
                    sta.emplace(x+dirs[d], y+dirs[d+1], 0);
                    sta.emplace(x+dirs[d], y+dirs[d+1], 1);
                }
            }
        }

        return false;
    }
};
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