Question

二进制索引树也可以在二维中实现。但与一维实现不同，它需要辅助阵列。这个算法中这个辅助阵列的目的是什么

中描述了该实现

using namespace std;

#define N 4 // N-->max_x and max_y

struct Query
{
   int x1, y1; // x and y co-ordinates of bottom left
   int x2, y2; // x and y co-ordinates of top right
};


void updateBIT(int BIT[][N+1], int x, int y, int val)
{
   for (; x <= N; x += (x & -x))
   {
      for (; y <= N; y += (y & -y))
         BIT[x][y] += val;
   }
   return;
}

// A function to get sum from (0, 0) to (x, y)
int getSum(int BIT[][N+1], int x, int y)
{
   int sum = 0;

   for(; x > 0; x -= x&-x)
   {
      // This loop sum through all the 1D BIT
      // inside the array of 1D BIT = BIT[x]
      for(; y > 0; y -= y&-y)
      {
         sum += BIT[x][y];
      }
   }
   return sum;
}

void constructAux(int mat[][N], int aux[][N+1])
{
   // Initialise Auxiliary array to 0
   for (int i=0; i<=N; i++)
   for (int j=0; j<=N; j++)
   aux[i][j] = 0;

   // Construct the Auxiliary Matrix
   for (int j=1; j<=N; j++)
   for (int i=1; i<=N; i++)
   aux[i][j] = mat[N-j][i-1];

   return;
}

// A function to construct a 2D BIT
void construct2DBIT(int mat[][N], int BIT[][N+1])
{
   // Create an auxiliary matrix
   int aux[N+1][N+1];
   constructAux(mat, aux);

   // Initialise the BIT to 0
   for (int i=1; i<=N; i++)
   for (int j=1; j<=N; j++)
   BIT[i][j] = 0;

   for (int j=1; j<=N; j++)
   {
      for (int i=1; i<=N; i++)
      {
         // Creating a 2D-BIT using update function
         // everytime we/ encounter a value in the
         // input 2D-array
         int v1 = getSum(BIT, i, j);
         int v2 = getSum(BIT, i, j-1);
         int v3 = getSum(BIT, i-1, j-1);
         int v4 = getSum(BIT, i-1, j);

         // Assigning a value to a particular element
         // of 2D BIT
         updateBIT(BIT, i, j, aux[i][j]-(v1-v2-v4+v3));
      }
   }

   return;
}

// A function to answer the queries
void answerQueries(Query q[], int m, int BIT[][N+1])
{
   for (int i=0; i<m; i++)
   {
      int x1 = q[i].x1 + 1;
      int y1 = q[i].y1 + 1;
      int x2 = q[i].x2 + 1;
      int y2 = q[i].y2 + 1;

      int ans = getSum(BIT, x2, y2)-getSum(BIT, x2, y1-1)-
      getSum(BIT, x1-1, y2)+getSum(BIT, x1-1, y1-1);

      printf ("Query(%d, %d, %d, %d) = %d\n",
      q[i].x1, q[i].y1, q[i].x2, q[i].y2, ans);
   }
   return;
}

// Driver program
int main()
{
   int mat[N][N] = {{1, 2, 3, 4},
                    {5, 3, 8, 1},
                    {4, 6, 7, 5},
                    {2, 4, 8, 9}};

   // Create a 2D Binary Indexed Tree
   int BIT[N+1][N+1];
   construct2DBIT(mat, BIT);

   Query q[] = {{1, 1, 3, 2}, {2, 3, 3, 3}, {1, 1, 1, 1}};
   int m = sizeof(q)/sizeof(q[0]);

   answerQueries(q, m, BIT);

   return(0);
}

Answer 1

此算法中此辅助阵列的用途是什么？

此算法需要辅助阵列，因为它们使用原点作为矩阵的左侧底部，如果它们使用左侧顶部作为原点，则不需要它。

使用辅助数组的二维二进制索引树实现

1 个答案: