给定大小为M * N的二元矩阵mat [] [] ，并以{pi，pj，qi，qj}的形式进行Q个查询，每个查询的任务是计算单元格子矩阵中0 s的个数( pi，pj)至(qi，qj) 。
例子：
Input: mat[][] = {{0, 1, 0, 1, 1, 1, 0}, {1, 0, 1, 1, 1, 0, 1}, {1, 1, 0, 0, 1, 1, 0}, {1, 1, 1, 1, 1, 0, 1}, {0, 0, 1, 0, 1, 1, 1}, {1, 1, 0, 1, 1, 0, 1 }}, Q[] = {{0, 1, 3, 2}, {2, 2, 4, 5}, {4, 3, 5, 6}}
Output: 3 4 2
Explanation:
The submatrix from indices (0, 1) to (3, 2) contains 3 0s.
The submatrix from indices (2, 2) to (4, 5) contains 4 0s.
The submatrix from indices (4, 3) to (5, 6) contains 2 0s.
Input: mat[][] = {{0, 1, 0}, {1, 0, 1}}, Q[] = {{0, 0, 2, 2}}
Output: 3
天真的方法：解决问题的最简单方法是通过迭代子矩阵并为每个查询打印计数来计算每个查询的0秒数。
时间复杂度： O(M * N * Q)
辅助空间： O(1)
高效方法：为了优化上述方法，其思想是对给定的矩阵mat [] []进行预处理。创建一个计数矩阵，例如prefixCnt [M] [N] ，以便prefix_cnt [i] [j]将0 s的计数存储在索引(0，0)至(i，j)内的子矩阵中。请按照以下步骤计算prefixCnt [] [] ：
- 如果mat [i] [j]为0，则将prefixCnt [i] [j]初始化为1 。否则，用0初始化。
- 找到每一行和每一列的前缀总和，并相应地更新矩阵。
一旦计算出矩阵prefixCnt [] [] ，就可以在O(1)时间中评估任何子矩阵中的0计数。下面是对从(pi，pj)到(qi，qj)的每个子矩阵中的0进行计数的表达式可以通过以下公式计算：
cnt = prefixCnt[qi][qj] – prefixCnt[pi-1][qj] – prefixCnt[qi][pj – 1] + prefixCnt[pi – 1][pj – 1]
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
#define M 6
#define N 7
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
void preCompute(int mat[M][N],
int prefixCnt[M][N])
{
for (int i = 0; i < M; i++) {
for (int j = 0; j < N; j++) {
// Initialize prefixCnt[i][j]
// with 1 if mat[i][j] is 0
if (mat[i][j] == 0) {
prefixCnt[i][j] = 1;
}
// Otherwise, assign with 0
else {
prefixCnt[i][j] = 0;
}
}
}
// Calculate prefix sum for each row
for (int i = 0; i < M; i++)
for (int j = 1; j < N; j++)
prefixCnt[i][j] += prefixCnt[i][j - 1];
// Calculate prefix sum for each column
for (int i = 1; i < M; i++)
for (int j = 0; j < N; j++)
prefixCnt[i][j] += prefixCnt[i - 1][j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
int countQuery(int prefixCnt[M][N],
int pi, int pj,
int qi, int qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
int cnt = prefixCnt[qi][qj];
// Substract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1][qj];
// Substract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi][pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1][pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
void count0s(int mat[M][N], int Q[][4],
int sizeQ)
{
// Stores the prefix sum of each
// row and column
int prefixCnt[M][N];
// Compute matrix prefixCnt[][]
preCompute(mat, prefixCnt);
for (int i = 0; i < sizeQ; i++) {
// Function Call for each query
cout << countQuery(prefixCnt, Q[i][0],
Q[i][1], Q[i][2],
Q[i][3])
<< ' ';
}
}
// Driver Code
int main()
{
// Given matrix
int mat[M][N] = {
{ 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1 },
{ 1, 1, 0, 0, 1, 1, 0 },
{ 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1 }
};
int Q[][4] = { { 0, 1, 3, 2 },
{ 2, 2, 4, 5 },
{ 4, 3, 5, 6 } };
int sizeQ = sizeof(Q) / sizeof(Q[0]);
// Function Call
count0s(mat, Q, sizeQ);
return 0;
} Java
// Java program for the above approach
class GFG{
final static int M = 6;
final static int N = 7;
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
static void preCompute(int mat[][], int prefixCnt[][])
{
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// Initialize prefixCnt[i][j]
// with 1 if mat[i][j] is 0
if (mat[i][j] == 0)
{
prefixCnt[i][j] = 1;
}
// Otherwise, assign with 0
else
{
prefixCnt[i][j] = 0;
}
}
}
// Calculate prefix sum for each row
for(int i = 0; i < M; i++)
for(int j = 1; j < N; j++)
prefixCnt[i][j] += prefixCnt[i][j - 1];
// Calculate prefix sum for each column
for(int i = 1; i < M; i++)
for(int j = 0; j < N; j++)
prefixCnt[i][j] += prefixCnt[i - 1][j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
static int countQuery(int prefixCnt[][],
int pi, int pj,
int qi, int qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
int cnt = prefixCnt[qi][qj];
// Substract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1][qj];
// Substract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi][pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added;
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1][pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
static void count0s(int mat[][], int Q[][],
int sizeQ)
{
// Stores the prefix sum of each
// row and column
int prefixCnt[][] = new int[M][N];
// Compute matrix prefixCnt[][]
preCompute(mat, prefixCnt);
for(int i = 0; i < sizeQ; i++)
{
// Function Call for each query
System.out.print(countQuery(prefixCnt, Q[i][0],
Q[i][1],
Q[i][2],
Q[i][3]) + " ");
}
}
// Driver Code
public static void main (String[] args)
{
// Given matrix
int mat[][] = { { 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1 },
{ 1, 1, 0, 0, 1, 1, 0 },
{ 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1 } };
int Q[][] = { { 0, 1, 3, 2 },
{ 2, 2, 4, 5 },
{ 4, 3, 5, 6 } };
int sizeQ = Q.length;
// Function Call
count0s(mat, Q, sizeQ);
}
}
// This code is contributed by AnkThonPython3
# Python3 program for the above approach
M = 6
N = 7
# Function to compute the matrix
# prefixCnt[M][N] from mat[M][N] such
# that prefixCnt[i][j] stores the
# count of 0's from (0, 0) to (i, j)
def preCompute(mat, prefixCnt):
for i in range(M):
for j in range(N):
# Initialize prefixCnt[i][j]
# with 1 if mat[i][j] is 0
if (mat[i][j] == 0):
prefixCnt[i][j] = 1
# Otherwise, assign with 0
else:
prefixCnt[i][j] = 0
# Calculate prefix sum for each row
for i in range(M):
for j in range(1, N):
prefixCnt[i][j] += prefixCnt[i][j - 1]
# Calculate prefix sum for each column
for i in range(1, M):
for j in range(N):
prefixCnt[i][j] += prefixCnt[i - 1][j]
return prefixCnt
# Function to compute count of 0's
# in submatrix from (pi, pj) to
# (qi, qj) from prefixCnt[M][N]
def countQuery(prefixCnt, pi, pj, qi, qj):
# Initialize that count of 0's
# in the sub-matrix within
# indices (0, 0) to (qi, qj)
cnt = prefixCnt[qi][qj]
# Substract count of 0's within
# indices (0, 0) and (pi-1, qj)
if (pi > 0):
cnt -= prefixCnt[pi - 1][qj]
# Substract count of 0's within
# indices (0, 0) and (qi, pj-1)
if (pj > 0):
cnt -= prefixCnt[qi][pj - 1]
# Add prefixCnt[pi - 1][pj - 1]
# because its value has been added
# once but subtracted twice
if (pi > 0 and pj > 0):
cnt += prefixCnt[pi - 1][pj - 1]
return cnt
# Function to count the 0s in the
# each given submatrix
def count0s(mat, Q, sizeQ):
# Stores the prefix sum of each
# row and column
prefixCnt = [[ 0 for i in range(N)] for i in range(M)]
# Compute matrix prefixCnt[][]
prefixCnt = preCompute(mat, prefixCnt)
for i in range(sizeQ):
# Function Call for each query
print(countQuery(prefixCnt, Q[i][0], Q[i][1], Q[i][2], Q[i][3]), end=" ")
# Driver Code
if __name__ == '__main__':
# Given matrix
mat = [[ 0, 1, 0, 1, 1, 1, 0 ],
[ 1, 0, 1, 1, 1, 0, 1 ],
[ 1, 1, 0, 0, 1, 1, 0 ],
[ 1, 1, 1, 1, 1, 0, 1 ],
[ 0, 0, 1, 0, 1, 1, 1 ],
[ 1, 1, 0, 1, 1, 0, 1 ] ]
Q= [ [ 0, 1, 3, 2 ],
[ 2, 2, 4, 5 ],
[ 4, 3, 5, 6 ] ]
sizeQ = len(Q)
# Function Call
count0s(mat, Q, sizeQ)
# This code is contributed by mohit kumar 29C#
// C# program for the above approach
using System;
class GFG
{
static int M = 6;
static int N = 7;
// Function to compute the matrix
// prefixCnt[M][N] from mat[M][N] such
// that prefixCnt[i][j] stores the
// count of 0's from (0, 0) to (i, j)
static void preCompute(int [,]mat, int [,]prefixCnt)
{
for(int i = 0; i < M; i++)
{
for(int j = 0; j < N; j++)
{
// Initialize prefixCnt[i][j]
// with 1 if mat[i,j] is 0
if (mat[i, j] == 0)
{
prefixCnt[i, j] = 1;
}
// Otherwise, assign with 0
else
{
prefixCnt[i, j] = 0;
}
}
}
// Calculate prefix sum for each row
for(int i = 0; i < M; i++)
for(int j = 1; j < N; j++)
prefixCnt[i, j] += prefixCnt[i, j - 1];
// Calculate prefix sum for each column
for(int i = 1; i < M; i++)
for(int j = 0; j < N; j++)
prefixCnt[i, j] += prefixCnt[i - 1, j];
}
// Function to compute count of 0's
// in submatrix from (pi, pj) to
// (qi, qj) from prefixCnt[M][N]
static int countQuery(int [,]prefixCnt,
int pi, int pj,
int qi, int qj)
{
// Initialize that count of 0's
// in the sub-matrix within
// indices (0, 0) to (qi, qj)
int cnt = prefixCnt[qi, qj];
// Substract count of 0's within
// indices (0, 0) and (pi-1, qj)
if (pi > 0)
cnt -= prefixCnt[pi - 1, qj];
// Substract count of 0's within
// indices (0, 0) and (qi, pj-1)
if (pj > 0)
cnt -= prefixCnt[qi, pj - 1];
// Add prefixCnt[pi - 1][pj - 1]
// because its value has been added;
// once but subtracted twice
if (pi > 0 && pj > 0)
cnt += prefixCnt[pi - 1, pj - 1];
return cnt;
}
// Function to count the 0s in the
// each given submatrix
static void count0s(int [,]mat, int [,]Q,
int sizeQ)
{
// Stores the prefix sum of each
// row and column
int [,]prefixCnt = new int[M, N];
// Compute matrix prefixCnt[,]
preCompute(mat, prefixCnt);
for(int i = 0; i < sizeQ; i++)
{
// Function Call for each query
Console.Write(countQuery(prefixCnt, Q[i, 0],
Q[i, 1],
Q[i, 2],
Q[i, 3]) + " ");
}
}
// Driver Code
public static void Main(string[] args)
{
// Given matrix
int [,]mat = { { 0, 1, 0, 1, 1, 1, 0 },
{ 1, 0, 1, 1, 1, 0, 1 },
{ 1, 1, 0, 0, 1, 1, 0 },
{ 1, 1, 1, 1, 1, 0, 1 },
{ 0, 0, 1, 0, 1, 1, 1 },
{ 1, 1, 0, 1, 1, 0, 1 } };
int [,]Q = { { 0, 1, 3, 2 },
{ 2, 2, 4, 5 },
{ 4, 3, 5, 6 } };
int sizeQ = Q.GetLength(0);
// Function Call
count0s(mat, Q, sizeQ);
}
}
// This code is contributed by AnkThon输出：
3 4 2时间复杂度： O(M * N + Q)
辅助空间： O(M * N)