返回一个二维整数数组中最大联通子数组的和

//2016   4  2   王宗泽 程思敏
#include <iostream>
   #include <time.h>
   #define M 3
   #define N 5
  using namespace std;
  
   void main()
   {
       int a[M][N] = {0},b[M][N]={0};            //判断联通性，0为未选中，1为选中,2为连通
      bool flg = 0;                              //判断是否有1存在，存在为O。
      int sum = 0;                               //最后和
  
     srand(unsigned((int)time(0)));
      for (int i = 0;i < M;i++)
      {
          for (int j = 0;j < N;j++)
          {
              a[i][j] = rand()%50 - 20;
             cout << a[i][j] << "	";
             if (a[i][j] >= 0)
             {
                  b[i][j] = 1;
              }
          }
         cout << endl;
     }
     cout << endl;
          
      for (int i = 0;i < M;i++)
      {
          for (int j = 0;j < N;j++)
          {
              if (b[i][j] == 1)
              {
                  if (a[i+1][j] + a[i][j] > 0 && b[i+1][j] == 0)
                  {
                      b[i+1][j] = 2;                            
                  }
                  if (a[i-1][j] + a[i][j] > 0 && b[i-1][j] == 0)
                  {
                      b[i-1][j] = 2;    
                  }
                  if (a[i][j-1] + a[i][j] > 0 && b[i][j-1] == 0)
                  {
                       b[i][j-1] = 2;        
                 }
                 if (a[i][j+1] + a[i][j] > 0 && b[i][j+1] == 0)
                  {
                      b[i][j+1] = 2;    
                  }                  
              }                        
          }
      }
  
      for (int i = 0;i < M;i++)
     {
         for (int j = 0;j < N;j++)
         {                
             flg = 0;
             if (b[i][j] != 0 && a[i][j] < 0)
              {
                 b[i][j] = 0;
                 for (int k = 0;k < M;k++)
                  {
                      for (int l = 0;l < N;l++)
                      {
                          if (b[k][l] != 0)
                          {
                             if ((b[k+1][l] <= 0 || b[k+1][l] > 2)&&
                                 (b[k-1][l] <= 0 || b[k-1][l] > 2)&&
                                 (b[k][l+1] <= 0 || b[k][l+1] > 2)&&
                                  (b[k][l-1] <= 0 || b[k][l-1] > 2))
                              {
                                  flg = 1;
                              }
                          }                                
                     }
                 }
                 if (flg)
                 {
                     b[i][j] = 2;
                 }                        
             }
         }
      }
  
      for (int i = 0;i < M;i++)
      {
         for (int j = 0;j < N;j++)
         {
              if (b[i][j] != 0)
             {
                  cout << a[i][j] << "	";
                 sum += a[i][j];
              }
             else
              {
                  cout << "**" << "	";
              }
         }
         cout << endl;
     }
     
     cout << "sum = " << sum << endl;
 }