在满足条件的给定矩阵中最大化占用的单元格

给定一个维度为N*M的矩阵，任务是最大化给定矩阵中被占用单元的总数，使得它们遵循给定条件：

• 如果同一行中有两个单元格，则它们之间必须至少有一个空单元格。
• 如果两个单元格被占用在不同的行中，则它们之间必须至少有一个完全空的行
  即，如果第 i行和第 j行中的单元格被占用，使得i < j则必须有第 k行完全空，使得i < k < j 。

例子：

Input: N = 1 ,M = 5
Output: 3
Explanation: There is only one row with five seats. 
Maximum three cells can be occupied. 
See the table below where 1 denotes occupied cell.

1

1

1

Input: N = 3 ,M = 3
Output: 4
Explanation:  There are three rows with three seats each.
Maximum occupied cells can be 4.

1		1
1		1

朴素方法：可以使用贪心方法解决问题。从第一行第一列开始填充，保持一行任意两个占用单元格之间的间隙为1，不同行的两个占用单元格之间保持一行的间隙。

下面是上述方法的实现：

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find
// maximum number of occupied cells
int solve(int N, int M)
{
    int ans = 0;
    while (1) {
        // Count number of occupied cells
        // in one row
        for (int i = 1; i <= M; i += 2) {
            ans++;
        }
        // If row numbers to be filled
        // are greater than or equal to 2
        // decrease by 2 otherwise decrease by 1
        // and if number of rows to be filled
        // are 0 return ans
        if (N >= 2) {
            N--;
            N--;
        }
        else if (N == 1) {
            N--;
        }
        if (N == 0) {
            break;
        }
    }
   
    // Return number of occupied cells
    return ans;
}
 
// Driver code
int main()
{
    int N = 1;
    int M = 5;
    cout << solve(N, M);
    return 0;
}

Java

// Java program for the above approach
import java.util.*;
public class GFG
{
 
  // Function to find
  // maximum number of occupied cells
  static int solve(int N, int M)
  {
    int ans = 0;
    while (true)
    {
 
      // Count number of occupied cells
      // in one row
      for (int i = 1; i <= M; i += 2) {
        ans++;
      }
 
      // If row numbers to be filled
      // are greater than or equal to 2
      // decrease by 2 otherwise decrease by 1
      // and if number of rows to be filled
      // are 0 return ans
      if (N >= 2) {
        N--;
        N--;
      }
      else if (N == 1) {
        N--;
      }
      if (N == 0) {
        break;
      }
    }
 
    // Return number of occupied cells
    return ans;
  }
 
  // Driver code
  public static void main(String[] args)
  {
    int N = 1;
    int M = 5;
    System.out.print(solve(N, M));
  }
}
 
// This code is contributed by Samim Hossain Mondal

Python3

# Python program for the above approach
 
# Function to find
# maximum number of occupied cells
def solve(N, M):
    ans = 0;
    while (1):
     
        # Count number of occupied cells
        # in one row
        for i in range(1, M + 1, 2):
            ans += 1
         
        # If row numbers to be filled
        # are greater than or equal to 2
        # decrease by 2 otherwise decrease by 1
        # and if number of rows to be filled
        # are 0 return ans
        if (N >= 2):
            N -= 1
            N -= 1
        elif (N == 1):
            N -= 1
        if (N == 0):
            break
   
    # Return number of occupied cells
    return ans;
 
# Driver code
N = 1;
M = 5;
print(solve(N, M));
 
# This code is contributed by saurabh_jaiswal.

C#

// C# program for the above approach
using System;
class GFG {
 
  // Function to find
  // maximum number of occupied cells
  static int solve(int N, int M)
  {
    int ans = 0;
    while (true)
    {
 
      // Count number of occupied cells
      // in one row
      for (int i = 1; i <= M; i += 2) {
        ans++;
      }
 
      // If row numbers to be filled
      // are greater than or equal to 2
      // decrease by 2 otherwise decrease by 1
      // and if number of rows to be filled
      // are 0 return ans
      if (N >= 2) {
        N--;
        N--;
      }
      else if (N == 1) {
        N--;
      }
      if (N == 0) {
        break;
      }
    }
 
    // Return number of occupied cells
    return ans;
  }
 
  // Driver code
  public static void Main()
  {
    int N = 1;
    int M = 5;
    Console.Write(solve(N, M));
  }
}
 
// This code is contributed by ukasp.

Javascript

// Javascript program for the above approach
 
// Function to find
// maximum number of occupied cells
function solve(N, M)
{
    let ans = 0;
    while (1)
    {
     
        // Count number of occupied cells
        // in one row
        for (let i = 1; i <= M; i += 2) {
            ans++;
        }
         
        // If row numbers to be filled
        // are greater than or equal to 2
        // decrease by 2 otherwise decrease by 1
        // and if number of rows to be filled
        // are 0 return ans
        if (N >= 2) {
            N--;
            N--;
        }
        else if (N == 1) {
            N--;
        }
        if (N == 0) {
            break;
        }
    }
   
    // Return number of occupied cells
    return ans;
}
 
// Driver code
let N = 1;
let M = 5;
document.write(solve(N, M));
 
// This code is contributed by saurabh_jaiswal.

输出

3

时间复杂度： O(N*M)
辅助空间： O(1)

高效方法：为了最大化占用单元的总数，需要以上述方式占用它们。总数可以从以下观察中获得：

So, the maximum number of cells that can be occupied for each row is ceil(M/2).
And as there is a gap of one row between any two occupied cells of different rows, 
the maximum number of rows that can be occupied is ceil(N/2).
Therefore maximum number of cells that can be occupied is ceil(M/2) * ceil(N/2).

下面是上述方法的实现。

C++

// C++ program for the above approach
#include 
using namespace std;
 
// Function to find
// maximum number of occupied cells
int solve(int N, int M)
{
    int x = (M + 1) / 2;
    int y = (N + 1) / 2;
    int ans = x * y;
 
    // Return number of occupied cells
    return ans;
}
 
// Driver code
int main()
{
    int N = 1;
    int M = 5;
    cout << solve(N, M);
    return 0;
}

Java

// Java program for the above approach
import java.io.*;
class GFG
{
 
  // Function to find
  // maximum number of occupied cells
  public static int solve(int N, int M)
  {
    int x = (M + 1) / 2;
    int y = (N + 1) / 2;
    int ans = x * y;
 
    // Return number of occupied cells
    return ans;
  }
 
  // Driver code
  public static void main (String[] args)
  {
    int N = 1;
    int M = 5;
    System.out.print(solve(N, M));
  }
}
 
// This code is contributed by Shubham Singh

Python3

# Python program for the above approach
 
# Function to find
# maximum number of occupied cells
def solve (N, M):
    x = (M + 1) // 2
    y = (N + 1) // 2
    ans = x * y
     
    # Return number of occupied cells
    return ans
     
# Driver code
N = 1
M = 5
print(solve(N, M));
 
# This code is contributed by Shubham Singh

C＃

// C# program for the above approach
using System;
public class GFG
{
 
  // Function to find
  // maximum number of occupied cells
  public static int solve(int N, int M)
  {
    int x = (M + 1) / 2;
    int y = (N + 1) / 2;
    int ans = x * y;
 
    // Return number of occupied cells
    return ans;
  }
 
  // Driver code
  public static void Main(String[] args)
  {
    int N = 1;
    int M = 5;
    Console.Write(solve(N, M));
  }
}
 
// This code is contributed by shikhasingrajput

Javascript

        // JavaScript program for the above approach
 
        // Function to find
        // maximum number of occupied cells
        const solve = (N, M) => {
            let x = parseInt((M + 1) / 2);
            let y = parseInt((N + 1) / 2);
            let ans = x * y;
 
            // Return number of occupied cells
            return ans;
        }
 
        // Driver code
 
        let N = 1;
        let M = 5;
        document.write(solve(N, M));
 
    // This code is contributed by rakeshsahni
 
    

输出

3

时间复杂度： O(1)
辅助空间： O(1)