/**
 * 给定一个 m x n 二维字符网格 board 和一个字符串单词 word 。如果 word 存在于网格中，返回 true ；否则，返回 false 。
 *
 * 单词必须按照字母顺序，通过相邻的单元格内的字母构成，其中“相邻”单元格是那些水平相邻或垂直相邻的单元格。同一个单元格内的字母不允许被重复使用。
 *
 *  
 *
 * 示例 1：
 *
 *
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCCED"
 * 输出：true
 * 示例 2：
 *
 *
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "SEE"
 * 输出：true
 * 示例 3：
 *
 *
 * 输入：board = [["A","B","C","E"],["S","F","C","S"],["A","D","E","E"]], word = "ABCB"
 * 输出：false
 *  
 *
 * 提示：
 *
 * m == board.length
 * n = board[i].length
 * 1 <= m, n <= 6
 * 1 <= word.length <= 15
 * board 和 word 仅由大小写英文字母组成
 *  
 *
 * 进阶：你可以使用搜索剪枝的技术来优化解决方案，使其在 board 更大的情况下可以更快解决问题？
 *
 * 来源：力扣（LeetCode）
 * 链接：https://leetcode-cn.com/problems/word-search
 * 著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。
 */

#include <map>
#include <stack>
#include <vector>
#include <iostream>
#include <random>
#include <algorithm>
#include <stdint.h>
using namespace std;

class Solution {
public:
    bool exist(vector<vector<char>> &board, string word)
    {
        this->row = board.size();
        this->col = board[0].size();
        this->word = word;
        for (int i = 0; i < row; i++) {
            for (int j = 0; j < col; j++) {
                visited.resize(row * col, 0);
                if (DFS(board, i, j, 0)) {
                    return true;
                }
            }
        }
        return false;
    }

private:
    int row{};
    int col{};
    string word{};
    vector<uint8_t> visited{};
    bool DFS(const vector<vector<char>> &board, int r, int c, int index)
    {
        // Invalid args;
        if (r < 0 || r >= row || c < 0 || c >= col) {
            return false;
        }

        // Visited already.
        if (visited[r * col + c] != 0) {
            return false;
        }

        // Not same.
        if (board[r][c] != word[index]) {
            return false;
        }

        // Is the same and is the last.
        if (index == word.length() - 1) {
            return true;
        }

        // Is the same and have next.
        visited[r * col + c] = 1;
        static const vector<vector<int>> neighbors{{0, -1}, {0, 1}, {-1, 0}, {1, 0}};
        for (const auto &neighbor : neighbors) {
            if (DFS(board, r + neighbor[0], c + neighbor[1], index + 1)) {
                return true;
            }
        }
        visited[r * col + c] = 0;
        return false;
    }
};