力扣79. 单词搜索（网格DFS + 回溯）

 比较常规的多起点搜索问题

class Solution {
public:
    typedef pair<int, int> pii;
    int m, n;
    vector<pii> start;
    vector<vector<char>> board;
    vector<int> next = {1, 0, -1, 0, 0, 1, 0, -1};
    bool visited[10][10] = {false};
    bool result = false;
    string word;
    void dfs_back_travel(pii p, int key_index) { //（当前点位，当前需要匹配字符的下标）
        if (key_index == word.size()) {
            result = true;
            return;
        }
        visited[p.first][p.second] = true;
        char key = word[key_index];
        int x = p.first, y = p.second;
        for (int i = 0; i < 4; ++i) {
            if (result) return;
            int next_x = x + next[i * 2], next_y = y + next[i * 2 + 1];
            if (next_x >= 0 && next_x < m && next_y >= 0 && next_y < n) {
                if (!visited[next_x][next_y] && board[next_x][next_y] == key) {
                    dfs_back_travel(make_pair(next_x, next_y), key_index + 1);
                    visited[next_x][next_y] = false;
                }
            }
        }
    }
    bool exist(vector<vector<char>>& board, string word) {
        this -> board = board;
        this -> word = word;
        m = board.size();
        n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == word[0] && !result) {
                    memset(visited, false, sizeof(visited));
                    dfs_back_travel(make_pair(i, j), 1); //下标从1开始，因为第0个字符已经匹配
                }
            }
        }
        return result;
    }
};