[LeetCode][沙雕氵]如何用LeetCode130写一个故事？

Code算是程序员的基本功，但是有些时候，code还是有那么一点无聊的，因而总是希望给自己找些乐子。今天在写LeetCode130的时候就突然来了一点点有趣的想法。
首先先来看一下题面：

这是一个非常简单的题，我们可以用deque+set轻松的解决他，但是~
各位看官，不觉得我们可以用这个题写一个故事吗
在末日，我们爆发了尸潮，凶残的僵尸包围住了民众，此时英勇的救援队出现了：

deque<location> rescueTeam；// using loaction = pair

我们需要为它募集成员，谁是一开始的成员呢？当然是在外围，而且没有被感染的人们：

for(int i = 0; i < rows; ++i){
     
	for(int j = 0; j < cols; ++j){
     
		if((!i || !j || i == rows - 1 || j == cols - 1) && board[i][j] == 'O')
			rescueTeam.push_back({
     i, j});

然后嘛，我们需要有一个丧尸题材里面常见的设备——安全屋

set<location> safeHouse;

接下来，救援队需要去救援，很明显大家都不是什么一拳超人，所以，在救援了自己身边的人之后，马上撤回安全屋，由被救出去的民众继续救人，把勇气传递下去：

while(!rescueTeam.empty()){
     
    location rescuer = rescueTeam.front(); rescueTeam.pop_front();
    if(safeHouse.find({
     rescuer.first - 1, rescuer.second}) == safeHouse.end())
        if(rescuer.first - 1 >= 0 && board[rescuer.first - 1][rescuer.second] == 'O')
            rescueTeam.push_back({
     rescuer.first - 1, rescuer.second});
    if(safeHouse.find({
     rescuer.first + 1, rescuer.second}) == safeHouse.end())
        if(rescuer.first + 1 < rows && board[rescuer.first + 1][rescuer.second] == 'O')
            rescueTeam.push_back({
     rescuer.first + 1, rescuer.second});
    if(safeHouse.find({
     rescuer.first, rescuer.second - 1}) == safeHouse.end())
        if(rescuer.second - 1 >= 0 && board[rescuer.first][rescuer.second - 1] == 'O')
            rescueTeam.push_back({
     rescuer.first, rescuer.second - 1});
    if(safeHouse.find({
     rescuer.first, rescuer.second + 1}) == safeHouse.end())
        if(rescuer.second + 1 < cols && board[rescuer.first][rescuer.second + 1] == 'O')
            rescueTeam.push_back({
     rescuer.first, rescuer.second + 1});
    safeHouse.insert(rescuer);
}

当然，并不是所有的事情都会有一个浪漫的结局，没能被救援的人还是无法逃脱变为丧尸的命运，rip

for(int i = 0; i < rows; ++i)
   for(int j = 0; j < cols; ++j)
        if(safeHouse.find({
     i, j}) == safeHouse.end())
            board[i][j] = 'X';

最后是完整的故事：

class Solution {
     
	using location = pair<int, int>;
public:
	void solve(vector<vector<char>>& board) {
     
		deque<location> rescueTeam;
		set<location> safeHouse;
		int rows = board.size();
		if (!rows) return;
		int cols = board[0].size();
		for (int i = 0; i < rows; ++i)
			for (int j = 0; j < cols; ++j)
				if ((!i || !j || i == rows - 1 || j == cols - 1) && board[i][j] == 'O')
					rescueTeam.push_back({
      i, j });
		while (!rescueTeam.empty()) {
     
			location rescuer = rescueTeam.front(); rescueTeam.pop_front();
			safeHouse.insert(rescuer);
			if (safeHouse.find({
      rescuer.first - 1, rescuer.second }) == safeHouse.end())
				if (rescuer.first - 1 >= 0 && board[rescuer.first - 1][rescuer.second] == 'O')
					rescueTeam.push_back({
      rescuer.first - 1, rescuer.second });
			if (safeHouse.find({
      rescuer.first + 1, rescuer.second }) == safeHouse.end())
				if (rescuer.first + 1 < rows && board[rescuer.first + 1][rescuer.second] == 'O')
					rescueTeam.push_back({
      rescuer.first + 1, rescuer.second });
			if (safeHouse.find({
      rescuer.first, rescuer.second - 1 }) == safeHouse.end())
				if (rescuer.second - 1 >= 0 && board[rescuer.first][rescuer.second - 1] == 'O')
					rescueTeam.push_back({
      rescuer.first, rescuer.second - 1 });
			if (safeHouse.find({
      rescuer.first, rescuer.second + 1 }) == safeHouse.end())
				if (rescuer.second + 1 < cols && board[rescuer.first][rescuer.second + 1] == 'O')
					rescueTeam.push_back({
      rescuer.first, rescuer.second + 1 });
		}
		for (int i = 0; i < rows; ++i)
			for (int j = 0; j < cols; ++j)
				if (safeHouse.find({
      i, j }) == safeHouse.end())
					board[i][j] = 'X';
		return;
	}
};

时间beat60，空间beat100，算是一个可取的方法吧XD