简单多源BFS问题

力扣问题：994. 腐烂的橘子

注意点：多源BFS写法（本质层序遍历）、move数组的使用

class Solution {
public:
    int orangesRotting(vector<vector<int>>& grid) {
        int n = grid.size();
        int m = grid[0].size();
        int fresh = 0;        // 记录未被遍历到的位置个数
        queue<pair<int, int>> q;
        vector<int> move = {-1, 0, 1, 0, -1};
        int ans = 0;

        for (int i = 0; i < n; i++) {
            for (int j = 0; j < m; j++) {
                int type = grid[i][j];
                if (type == 0)
                    continue;
                if (type == 1)
                    fresh++;
                if (type == 2)
                    q.push({i, j});
            }
        }

        while (!q.empty() && fresh) {
            ans++;        // 步数加一
            int count = q.size();
            // 注意循环时机
            for (int i = 1; i <= count; i++) {
                auto current = q.front();
                q.pop();
                for (int j = 0; j < 4; j++) {
                    int x = current.first + move[j];
                    int y = current.second + move[j + 1];
                    if (x >= 0 && x < n && y >= 0 && y < m && grid[x][y] == 1) {
                        grid[x][y] = 2;
                        fresh--;
                        q.push({x, y});
                    }
                }
            }
        }

        return fresh != 0 ? -1 : ans;
    }
};