求矩阵中 M 次运算后 (Xi, Yi) 处的权重

给定笛卡尔平面上的 nxn个点，任务是在m 次运算后找到 (x _i , y _j ) 处的权重。 x _i , y _j , w表示在 x = x _i和 y = y _j线上的所有点上添加权重w的操作。

例子：

Input: n = 3, m = 2
0 0 1
1 1 2
x = 1, y = 0
Output: 3
Explanation: Initially, weights are
0 0 0
0 0 0
0 0 0
After 1st operation
2 1 1
1 0 0
1 0 0
After 2nd operation
2 3 1
3 4 2
1 0 0
Clearly, weight at (x₁, y₀) is 3

Input: n = 2, m = 2
0 1 1
1 0 2
x = 1, y = 1
Output: 3

天真的方法：
考虑一个二维数组 arr[n][n] = {0} 并执行给定的操作，然后检索(x _i , y _j )处的权重。这种方法将花费O(n*m)时间。

有效的方法：

• 考虑数组 arrX[n] = arrY[n] = {0}。
• 将操作 x _i , y _j , w 重新定义为

  arrX[i] += w 
  and
  arrY[j] += w
  

• 使用 (x _i , y _{j ) 求重量}

  w = arrX[i] + arrY[j]

  下面是上述方法的实现：
  

  C++

  // C++ program to find the
  // weight at xi and yi
    
  #include 
  using namespace std;
    
  // Function to calculate weight at (xFind, yFind)
  int findWeight(vector >& operations,
                 int n, int m,
                 int xFind, int yFind)
  {
      int row[n] = { 0 };
      int col[n] = { 0 };
    
      // Loop to perform operations
      for (int i = 0; i < m; i++) {
    
          // Updating row
          row[operations[i][0]]
              += operations[i][2];
    
          // Updating column
          col[operations[i][0]]
              += operations[i][2];
      }
    
      // Find weight at (xi, yj) using
      // w = arrX[i] + arrY[j]
      int result = row[xFind] + col[yFind];
    
      return result;
  }
    
  // Driver code
  int main()
  {
      vector > operations
          = {
              { 0, 0, 1 },
              { 1, 1, 2 }
            };
      int n = 3,
          m = operations.size(),
          xFind = 1,
          yFind = 0;
      cout << findWeight(operations,
                         n, m, xFind,
                         yFind);
      return 0;
  }

  
  

  Python3

  # Python3 program to find the
  # weight at xi and yi
    
  # Function to calculate weight at (xFind, yFind)
  def findWeight(operations, n, m, xFind, yFind) :
    
      row = [ 0 ] * n
      col = [ 0 ] * n
     
      # Loop to perform operations
      for i in range(m) :
     
          # Updating row
          row[operations[i][0]]+= operations[i][2]
     
          # Updating column
          col[operations[i][0]]+= operations[i][2]
     
      # Find weight at (xi, yj) using
      # w = arrX[i] + arrY[j]
      result = row[xFind] + col[yFind]
     
      return result
     
  # Driver code
  operations = [[ 0, 0, 1 ],[ 1, 1, 2 ]]
  n = 2
  m = len(operations)
  xFind = 1
  yFind = 0
  print(findWeight(operations,n, m, xFind, yFind))
    
  # This code is contributed by divyamohan123

  
  
  
  输出：

  3

  时间复杂度： $O(m)$其中 m 是操作次数

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