C++ C中更简短的条件

所以，我的问题是: 我有一个平方矩阵，代表一个有疫情的社区，矩阵中的每个细胞代表一所房子.房子有三种不同的情况，S生病了，H痊愈了，C传染了.我有几天的时间，然后我改变了社区的初始情况，每天我都会按照即将到来的规则改变社区:

• 如果房子是‘H’，它的上/下/左/右邻居是‘C’，它就变成了‘C’房子(不是对角线).
• 如果房子是‘C’，并且至少有两个‘C’邻居变成了‘S’房子(在这种情况下，邻居是上/下/左/右/对角线，这里我们有8个不同的 Select ).
• 如果房子是‘S’，那么它就变成‘H’房子.

这不是一个棘手的问题，但我的问题是，如果我能为边缘情况写一个更简单的条件.

这是一个函数，用于判断房屋为‘C’时的边缘情况:

void limit_C(char **neigh, int i, int j, int d, char **help)
{
    int counter = 0;

    if ((i == 0) && (j == 0))
   {
       if(help[i + 1][j] == CONTAGIOUS)
       {counter++;}
           
          if(help[i][j + 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j + 1] == CONTAGIOUS)
          {counter++;}
       
           if(counter >= 2)
           {neigh[i][j] = SICK;}
       
   }
   else if ((i == d - 1) && (j == d - 1))
   {
         if(help[i - 1][j] == CONTAGIOUS)
        {counter++;}
       
         if(help[i][j - 1] == CONTAGIOUS)
         {counter++;}
       
         if(help[i - 1][j - 1] == CONTAGIOUS)
        {counter++;}
       
       if(counter >= 2)
            {neigh[i][j] = SICK;}
   }
       
   else if ((i == d - 1) && (j == 0))
   {
          if(help[i - 1][j] == CONTAGIOUS)
          {counter++;}
       
          if(help[i][j + 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i - 1][j + 1] == CONTAGIOUS)
          {counter++;}
       
       if(counter >= 2)
            {neigh[i][j] = SICK;}
   }
   else if ((i == 0) && (j == d - 1))
   {
          if(help[i + 1][j] == CONTAGIOUS)
          {counter++;}
       
          if(help[i][j - 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j - 1] == CONTAGIOUS)
          {counter++;}
       
       if(counter >= 2)
            {neigh[i][j] = SICK;}
       
   }
   
  else if((i == 0) && (j != 0))
   {
          if(help[i][j + 1] == CONTAGIOUS)
          {counter++;}
           
          if(help[i][j - 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j + 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j - 1] == CONTAGIOUS)
          {counter++;}
       
       if(counter >= 2)
       {neigh[i][j] = SICK;}
       
   }
   
   else if ((i != 0) && (j == 0))
   {
          if(help[i][j + 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i - 1][j] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j] == CONTAGIOUS)
          {counter++;}
       
          if(help[i - 1][j + 1] == CONTAGIOUS)
          {counter++;}
       
          if(help[i + 1][j + 1] == CONTAGIOUS)
          {counter++;}
      
       if(counter >= 2)
       {neigh[i][j] = SICK;}
       
   }
    else if((i == d - 1) && (j != d - 1))
    {
        if(help[i - 1][j] == CONTAGIOUS)
        {counter++;}
        
        if(help[i][j + 1] == CONTAGIOUS)
        {counter++;}
        
        if(help[i][j - 1] == CONTAGIOUS)
        {counter++;}
        
        if(help[i - 1][j + 1] == CONTAGIOUS)
        {counter++;}
        
        if(help[i - 1][j - 1] == CONTAGIOUS)
        {counter++;}
        
        if(counter >= 2)
        {neigh[i][j] = SICK;}
        
        
    }
    
    else if((i != d - 1) && (j == d - 1))
    {
        if(help[i - 1][j] == CONTAGIOUS)
        {counter++;}
        
        if(help[i + 1][j] == CONTAGIOUS)
        {counter++;}
        
        if(help[i][j - 1] == CONTAGIOUS)
        {counter++;}
        
        if(help[i - 1][j - 1] == CONTAGIOUS)
        {counter++;}
        
        if(help[i + 1][j - 1] == CONTAGIOUS)
        {counter++;}
        
        if(counter >= 2)
        {neigh[i][j] = SICK;}
        
    }
}

那么，有没有更简单的方法来编写相同的代码呢？