给定的二进制矩阵V [] []尺寸的N * M个，其中每个单元为空或者已被一个0和1分别标，任务是计数的方式的数量以从同一行选择K个连续的空单元或列。
例子：
Input: V[][] = {{1, 1, 0}, {0, 0, 0}}, K = 2
Output: 3
Explanation:
Considering 1-based indexing, 2 consecutive empty cells can be selected in the following ways:
[(1, 3), (2, 3)], [(2, 1), (2, 2)], [(2, 2), (2, 3)]
Input: V[][] = {{1, 1, 0}, {0, 0, 0}, {0, 0, 0}}, K = 1
Output: 9
Explanation:
It is possible to select all the cells since every cell is empty.
方法：
想法是逐行遍历矩阵，并为每一行计数空的连续单元格的总数。
请按照以下步骤解决问题：
- 逐行遍历矩阵并计算连续单元的数量。如果计数等于或超过K，则将计数增加1。
- 每次遇到阻塞的单元格时，请将连续的空单元格计数重置为0。
- 如果K≥，则在遍历矩阵时重复上述步骤。 1。
- 返回获得的细胞的最终计数。
下面是上述方法的实现：
C++
// C++ program to find no of ways
// to select K consecutive empty
// cells from a row or column
#include
using namespace std;
// Function to Traverse
// the matrix row wise
int rowWise(char* v, int n,
int m, int k)
{
// Initialize ans
int ans = 0;
// Traverse row wise
for (int i = 0; i < n; i++) {
// Initialize no of
// consecutive empty
// cells
int countcons = 0;
for (int j = 0; j < m; j++) {
// Check if blocked cell is
// encountered then reset
// countcons to 0
if (*(v + i * m + j) == '1') {
countcons = 0;
}
// Check if empty cell is
// encountered, then
// increment countcons
else {
countcons++;
}
// Check if number of empty
// consecutive cells
// is greater or equal
// to K, increment the ans
if (countcons >= k) {
ans++;
}
}
}
// Return the count
return ans;
}
// Function to Traverse the
// matrix column wise
int colWise(char* v, int n,
int m, int k)
{
// Initialize ans
int ans = 0;
// Traverse column wise
for (int i = 0; i < m; i++) {
// Initialize no of
// consecutive empty cells
int countcons = 0;
for (int j = 0; j < n; j++) {
// Check if blocked cell is
// encountered then reset
// countcons to 0
if (*(v + j * n + i) == '1') {
countcons = 0;
}
// Check if empty cell is
// encountered, increment
// countcons
else {
countcons++;
}
// Check if number of empty
// consecutive cells
// is greater than or equal
// to K, increment the ans
if (countcons >= k) {
ans++;
}
}
}
// Return the count
return ans;
}
// Driver Code
int main()
{
int n = 3, m = 3, k = 1;
char v[n][m] = { '0', '0', '0',
'0', '0', '0',
'0', '0', '0' };
// If k = 1 only traverse row wise
if (k == 1) {
cout << rowWise(v[0], n, m, k);
}
// Traverse both row and column wise
else {
cout << colWise(v[0], n, m, k)
+ rowWise(v[0], n,
m, k);
}
return 0;
} Java
// Java program to find no of ways
// to select K consecutive empty
// cells from a row or column
import java.util.*;
class GFG{
// Function to Traverse
// the matrix row wise
static int rowWise(char [][]v, int n,
int m, int k)
{
// Initialize ans
int ans = 0;
// Traverse row wise
for(int i = 0; i < n; i++)
{
// Initialize no of
// consecutive empty
// cells
int countcons = 0;
for(int j = 0; j < m; j++)
{
// Check if blocked cell is
// encountered then reset
// countcons to 0
if (v[i][j] == '1')
{
countcons = 0;
}
// Check if empty cell is
// encountered, then
// increment countcons
else
{
countcons++;
}
// Check if number of empty
// consecutive cells
// is greater or equal
// to K, increment the ans
if (countcons >= k)
{
ans++;
}
}
}
// Return the count
return ans;
}
// Function to Traverse the
// matrix column wise
static int colWise(char [][]v, int n,
int m, int k)
{
// Initialize ans
int ans = 0;
// Traverse column wise
for(int i = 0; i < m; i++)
{
// Initialize no of
// consecutive empty cells
int countcons = 0;
for(int j = 0; j < n; j++)
{
// Check if blocked cell is
// encountered then reset
// countcons to 0
if (v[j][i] == '1')
{
countcons = 0;
}
// Check if empty cell is
// encountered, increment
// countcons
else
{
countcons++;
}
// Check if number of empty
// consecutive cells
// is greater than or equal
// to K, increment the ans
if (countcons >= k)
{
ans++;
}
}
}
// Return the count
return ans;
}
// Driver Code
public static void main(String[] args)
{
int n = 3, m = 3, k = 1;
char v[][] = { { '0', '0', '0' },
{ '0', '0', '0' },
{ '0', '0', '0' } };
// If k = 1 only traverse row wise
if (k == 1)
{
System.out.print(rowWise(v, n, m, k));
}
// Traverse both row and column wise
else
{
System.out.print(colWise(v, n, m, k) +
rowWise(v, n, m, k));
}
}
}
// This code is contributed by amal kumar choubeyPython3
# Python 3 program to find no of ways
# to select K consecutive empty
# cells from a row or column
# Function to Traverse
# the matrix row wise
def rowWise(v, n, m, k):
# Initialize ans
ans = 0
# Traverse row wise
for i in range (n):
# Initialize no of
# consecutive empty
# cells
countcons = 0
for j in range (m):
# Check if blocked cell is
# encountered then reset
# countcons to 0
if (v[i][j] == '1'):
countcons = 0
# Check if empty cell is
# encountered, then
# increment countcons
else:
countcons += 1
# Check if number of empty
# consecutive cells
# is greater or equal
# to K, increment the ans
if (countcons >= k):
ans += 1
# Return the count
return ans
# Function to Traverse the
# matrix column wise
def colWise(v, n, m, k):
# Initialize ans
ans = 0
# Traverse column wise
for i in range (m):
# Initialize no of
# consecutive empty cells
countcons = 0
for j in range (n):
# Check if blocked cell is
# encountered then reset
# countcons to 0
if (v[j][i] == '1'):
countcons = 0
# Check if empty cell is
# encountered, increment
# countcons
else:
countcons += 1
# Check if number of empty
# consecutive cells
# is greater than or equal
# to K, increment the ans
if (countcons >= k):
ans += 1
# Return the count
return ans
# Driver Code
if __name__ == "__main__":
n = 3
m = 3
k = 1
v = [['0', '0', '0'],
['0', '0', '0'],
['0', '0', '0']]
# If k = 1 only
# traverse row wise
if (k == 1):
print (rowWise(v, n, m, k))
# Traverse both row
# and column wise
else:
print (colWise(v, n, m, k) +
rowWise(v, n, m, k))
# This code is contributed by ChitranayalC#
// C# program to find no of ways
// to select K consecutive empty
// cells from a row or column
using System;
class GFG{
// Function to Traverse
// the matrix row wise
static int rowWise(char [,]v, int n,
int m, int k)
{
// Initialize ans
int ans = 0;
// Traverse row wise
for(int i = 0; i < n; i++)
{
// Initialize no of
// consecutive empty
// cells
int countcons = 0;
for(int j = 0; j < m; j++)
{
// Check if blocked cell is
// encountered then reset
// countcons to 0
if (v[i, j] == '1')
{
countcons = 0;
}
// Check if empty cell is
// encountered, then
// increment countcons
else
{
countcons++;
}
// Check if number of empty
// consecutive cells
// is greater or equal
// to K, increment the ans
if (countcons >= k)
{
ans++;
}
}
}
// Return the count
return ans;
}
// Function to Traverse the
// matrix column wise
static int colWise(char [,]v, int n,
int m, int k)
{
// Initialize ans
int ans = 0;
// Traverse column wise
for(int i = 0; i < m; i++)
{
// Initialize no of
// consecutive empty cells
int countcons = 0;
for(int j = 0; j < n; j++)
{
// Check if blocked cell is
// encountered then reset
// countcons to 0
if (v[j, i] == '1')
{
countcons = 0;
}
// Check if empty cell is
// encountered, increment
// countcons
else
{
countcons++;
}
// Check if number of empty
// consecutive cells
// is greater than or equal
// to K, increment the ans
if (countcons >= k)
{
ans++;
}
}
}
// Return the count
return ans;
}
// Driver Code
public static void Main(String[] args)
{
int n = 3, m = 3, k = 1;
char [,]v = { { '0', '0', '0' },
{ '0', '0', '0' },
{ '0', '0', '0' } };
// If k = 1 only traverse row wise
if (k == 1)
{
Console.Write(rowWise(v, n, m, k));
}
// Traverse both row and column wise
else
{
Console.Write(colWise(v, n, m, k) +
rowWise(v, n, m, k));
}
}
}
// This code is contributed by amal kumar choubey输出：
9
时间复杂度： O(N * M)
空间复杂度： O(N * M)