用于计算二进制矩阵中 1 和 0 集合的 C++ 程序

给定一个×m的二进制矩阵，计算一组可以在一行或一列中形成一个或多个相同值的集合的数量。
例子：

Input: 1 0 1
       0 1 0 
Output: 8 
Explanation: There are six one-element sets
(three 1s and three 0s). There are two two-
element sets, the first one consists of the
first and the third cells of the first row.
The second one consists of the first and the 
third cells of the second row. 

Input: 1 0
       1 1 
Output: 6

x 元素的非空子集的数量为 2 ^x – 1。我们遍历每一行并计算 1 和 0 单元格的数量。对于每 u 个零和 v 个 1，总集合为 2 ^u – 1 + 2 ^v – 1。然后我们遍历所有列并计算相同的值并计算总和。我们最终从总和中减去 mxn，因为单个元素被考虑了两次。

CPP

// CPP program to compute number of sets
// in a binary matrix.
#include 
using namespace std;
  
const int m = 3; // no of columns
const int n = 2; // no of rows
  
// function to calculate the number of
// non empty sets of cell
long long countSets(int a[n][m])
{   
    // stores the final answer 
    long long res = 0;
      
    // traverses row-wise 
    for (int i = 0; i < n; i++)
    {
        int u = 0, v = 0; 
        for (int j = 0; j < m; j++) 
            a[i][j] ? u++ : v++;          
        res += pow(2,u)-1 + pow(2,v)-1; 
    }
      
    // traverses column wise 
    for (int i = 0; i < m; i++)
    {
        int u = 0, v = 0; 
        for (int j = 0; j < n; j++) 
             a[j][i] ? u++ : v++;  
        res += pow(2,u)-1 + pow(2,v)-1; 
    }
      
    // at the end subtract n*m as no of
    // single sets have been added twice.
    return res-(n*m);
}
  
// driver program to test the above function.
int main() {
      
    int a[][3] = {(1, 0, 1),
                  (0, 1, 0)};
      
    cout << countSets(a); 
      
    return 0;
}

输出：

8

时间复杂度： O(n * m)
有关详细信息，请参阅有关在二进制矩阵中计数 1 和 0 集的完整文章！