class Solution {
public:int orangesRotting(vector<vector<int>>& grid) {std::queue<std::pair<int, int>> qe; // 保存烂橘子得i,j坐标int fresh = 0; // 记录新鲜水果数量for (int i=0; i<grid.size(); i++) {for (int j=0; j<grid[i].size(); j++) {if (grid[i][j]==2) qe.push({i,j});else if (grid[i][j] == 1) fresh++;}}int count = 0; // 返回次数// 烂橘子移动坐标，上下左右std::vector<std::pair<int, int>> dirts{{-1,0}, {1,0}, {0,1}, {0,-1}};while(!qe.empty()) {int size = qe.size(); // 当前队列大小bool flag = false;    //  标记每层是否出现橘子腐烂，腐烂了count+1while(size>0){auto pair = qe.front();qe.pop();// 对四个方向上得橘子进行判断for (auto direct: dirts) {int i = pair.first + direct.first;int j = pair.second + direct.second;// 校验移动得新坐标i，j，并且移动后当前时新鲜橘子更新为烂橘子if (i>=0 && i<grid.size() && j>=0 && j<grid[0].size() && grid[i][j] == 1) {grid[i][j] = 2; // 更新烂橘子qe.push({i,j}); // 将新烂橘子坐标添加到队列中fresh--;        // 新鲜橘子数量-1flag = true;    // 标记有出现向四周腐烂橘子情况}}size--;}if (flag) count++; // 只有腐烂了才会次数+1,可能出现孤立腐烂橘子}return fresh ? -1 : count;}
};