所有子矩阵的按位或之和

给定一个NxN矩阵，任务是找到其所有矩形子矩阵的按位 OR 的总和。
例子：

Input : arr[][] = {{1, 0, 0},
                   {0, 0, 0},
                   {0, 0, 0}}
Output : 9
Explanation: All the submatrices starting from 
the index (0, 0) will have OR value as 1.
Thus, ans = 9

Input : arr[][] = {{9, 7, 4},
                    {8, 9, 2},
                    {11, 11, 5}}
Output : 398

先决条件：OR值为1的子矩阵数
简单的解决方案：一个简单的解决方案是生成所有子矩阵并为每个子矩阵找到所需的 OR 值。这种方法的时间复杂度为 O(N ⁶ )。
有效的解决方案：为了更好地理解，让我们假设元素的任何位都由变量“i”表示，变量“sum”用于存储最终和。
这里的想法是，我们将尝试找到第 i^个位集的 OR 值(具有按位或(|)的子矩阵)的数量。让我们假设，有s的子矩阵的_i个与^第i位设置。对于^第i 位，sum 可以更新为 sum += (2 ⁱ * S _i )。
对于每个位“i”，如果设置了 arr[R][C] 的^第i 位，则创建一个布尔矩阵set_bit将“1”存储在索引 (R, C) 处。否则，它存储“0”。然后，对于这个布尔数组，我们尝试找到 OR 值为 1(S _i ) 的矩形子矩阵的数量。对于^第i 位，最终和将更新为：

sum += 2i * Si

下面是上述方法的实现：

C++

// C++ program to find sum of Bitwise-OR of
// all submatrices
 
#include 
#include 
 
using namespace std;
 
#define n 3
 
// Function to find prefix-count for each row
// from right to left
void findPrefixCount(int p_arr[][n], bool set_bit[][n])
{
    for (int i = 0; i < n; i++)
    {
        for (int j = n - 1; j >= 0; j--)
        {
            if (set_bit[i][j])
                continue;
 
            if (j != n - 1)
                p_arr[i][j] += p_arr[i][j + 1];
 
            p_arr[i][j] += (int)(!set_bit[i][j]);
        }
    }
}
 
// Function to create a boolean matrix set_bit which
// stores ‘1’ at an index (R, C) if ith bit
// of arr[R][C] is set.
int matrixOrValueOne(bool set_bit[][n])
{
    // array to store prefix count of zeros from
    // right to left for boolean array
    int p_arr[n][n] = { 0 };
 
    findPrefixCount(p_arr, set_bit);
 
    // variable to store the count of
    // submatrices with OR value 0
    int count_zero_submatrices = 0;
 
    // For each index of a column we will try to
    // determine the number of sub-matrices
    // starting from that index
    // and has all 1s
    for (int j = 0; j < n; j++)
    {
        int i = n - 1;
 
        // stack to store elements and the count
        // of the numbers they popped
        // First part of pair will be the
        // value of inserted element.
        // Second part will be the count
        // of the number of elements pushed
        // before with a greater value
        stack > q;
 
        // variable to store the number of submatrices
        // with all 0s
        int to_sum = 0;
        while (i >= 0)
        {
            int c = 0;
 
            while (q.size() != 0 and q.top().first > p_arr[i][j])
            {
                to_sum -= (q.top().second + 1) *
                             (q.top().first - p_arr[i][j]);
                c += q.top().second + 1;
                q.pop();
            }
 
            to_sum += p_arr[i][j];
            count_zero_submatrices += to_sum;
            q.push({ p_arr[i][j], c });
 
            i--;
        }
    }
 
    return (n * (n + 1) * n * (n + 1)) /
                    4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
int sumOrMatrix(int arr[][n])
{
    int sum = 0;
 
    int mul = 1;
 
    for (int i = 0; i < 30; i++)
    {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        bool set_bit[n][n];
 
        for (int R = 0; R < n; R++)
            for (int C = 0; C < n; C++)
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
 
        sum += (mul * matrixOrValueOne(set_bit));
 
        mul *= 2;
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[][n] = { { 9, 7, 4 },
                     { 8, 9, 2 },
                     { 11, 11, 5 } };
 
    cout << sumOrMatrix(arr);
 
    return 0;
}

Java

// Java program to find sum of Bitwise-OR of
// all submatrices
import java.util.*;
 
class GFG
{
 
static int n = 3;
 
// Function to find prefix-count for 
// each row from right to left
static void findPrefixCount(int p_arr[][],
                        boolean set_bit[][])
{
    for (int i = 0; i < n; i++)
    {
        for (int j = n - 1; j >= 0; j--)
        {
            if (set_bit[i][j])
                continue;
 
            if (j != n - 1)
                p_arr[i][j] += p_arr[i][j + 1];
 
            p_arr[i][j] += (!set_bit[i][j]) ? 1 : 0;
        }
    }
}
 
static class pair
{
    int first,second;
    pair(){}
     
    pair(int a,int b)
    {
        first = a;
        second = b;
    }
}
 
// Function to create a booleanean
// matrix set_bit which stores '1'
// at an index (R, C) if ith bit
// of arr[R][C] is set.
static int matrixOrValueOne(boolean set_bit[][])
{
    // array to store prefix count of zeros from
    // right to left for booleanean array
    int p_arr[][] = new int[n][n] ;
     
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
    p_arr[i][j] = 0;
 
    findPrefixCount(p_arr, set_bit);
 
    // variable to store the count of
    // submatrices with OR value 0
    int count_zero_submatrices = 0;
 
    // For each index of a column we will try to
    // determine the number of sub-matrices
    // starting from that index
    // and has all 1s
    for (int j = 0; j < n; j++)
    {
        int i = n - 1;
 
        // stack to store elements and the count
        // of the numbers they popped
        // First part of pair will be the
        // value of inserted element.
        // Second part will be the count
        // of the number of elements pushed
        // before with a greater value
        Stack q = new Stack();
 
        // variable to store the number of submatrices
        // with all 0s
        int to_sum = 0;
        while (i >= 0)
        {
            int c = 0;
 
            while (q.size() != 0 && q.peek().first > p_arr[i][j])
            {
                to_sum -= (q.peek().second + 1) *
                            (q.peek().first - p_arr[i][j]);
                c += q.peek().second + 1;
                q.pop();
            }
 
            to_sum += p_arr[i][j];
            count_zero_submatrices += to_sum;
            q.push(new pair( p_arr[i][j], c ));
 
            i--;
        }
    }
 
    return (n * (n + 1) * n * (n + 1)) /
                4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumOrMatrix(int arr[][])
{
    int sum = 0;
 
    int mul = 1;
 
    for (int i = 0; i < 30; i++)
    {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        boolean set_bit[][] = new boolean[n][n];
 
        for (int R = 0; R < n; R++)
            for (int C = 0; C < n; C++)
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
 
        sum += (mul * matrixOrValueOne(set_bit));
 
        mul *= 2;
    }
 
    return sum;
}
 
// Driver Code
public static void main(String args[])
{
    int arr[][] = { { 9, 7, 4 },
                    { 8, 9, 2 },
                    { 11, 11, 5 } };
 
    System.out.println( sumOrMatrix(arr));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find sum of
# Bitwise-OR of all submatrices
 
# Function to find prefix-count for
# each row from right to left
def findPrefixCount(p_arr, set_bit):
 
    for i in range(0, n):
        for j in range(n - 1, -1, -1):
 
            if set_bit[i][j]:
                continue
            if j != n - 1:
                p_arr[i][j] += p_arr[i][j + 1]
 
            p_arr[i][j] += int(not set_bit[i][j])
 
# Function to create a boolean matrix
# set_bit which stores ‘1’ at an index
# (R, C) if ith bit of arr[R][C] is set.
def matrixOrValueOne(arr):
 
    # Array to store prefix count of zeros
    # from right to left for boolean array
    p_arr = [[0 for i in range(n)]
                for j in range(n)]
 
    findPrefixCount(p_arr, arr)
 
    # Variable to store the count of
    # submatrices with OR value 0
    count_zero_submatrices = 0
 
    # Loop to evaluate each column of
    # the prefix matrix uniquely.
    # For each index of a column we will try
    # to determine the number of sub-matrices
    # starting from that index and has all 1s
    for j in range(0, n):
 
        i = n - 1
         
        # stack to store elements and the
        # count of the numbers they popped
 
        # First part of pair will be the
        # value of inserted element.
        # Second part will be the count
        # of the number of elements pushed
        # before with a greater value
        q = []
 
        # Variable to store the number
        # of submatrices with all 0s
        to_sum = 0
         
        while i >= 0:
 
            c = 0
            while (len(q) != 0 and
                   q[-1][0] > p_arr[i][j]):
 
                to_sum -= ((q[-1][1] + 1) *
                           (q[-1][0] - p_arr[i][j]))
 
                c += q.pop()[1] + 1
 
            to_sum += p_arr[i][j]
            count_zero_submatrices += to_sum
 
            q.append((p_arr[i][j], c))
            i -= 1
 
    # Return the final answer
    return ((n * (n + 1) * n * (n + 1)) //
             4 - count_zero_submatrices)
 
# Function to find sum of
# Bitwise-OR of all submatrices
def sumOrMatrix(arr):
 
    Sum, mul = 0, 1
    for i in range(0, 30):
     
        # matrix to store the status
        # of ith bit of each element
        # of matrix arr
        set_bit = [[False for i in range(n)]
                          for j in range(n)]
 
        for R in range(0, n):
            for C in range(0, n):
                set_bit[R][C] = ((arr[R][C] &
                                 (1 << i)) != 0)
 
        Sum += (mul * matrixOrValueOne(set_bit))
        mul *= 2
 
    return Sum
 
# Driver Code
if __name__ == "__main__":
     
    n = 3
    arr = [[9, 7, 4],
        [8, 9, 2],
        [11, 11, 5]]
 
    print(sumOrMatrix(arr))
 
# This code is contributed by Rituraj Jain

C#

// C# program to find sum of Bitwise-OR of
// all submatrices
using System;
using System.Collections.Generic;
 
class GFG
{
static int n = 3;
 
// Function to find prefix-count for
// each row from right to left
static void findPrefixCount(int [,]p_arr,
                            Boolean [,]set_bit)
{
    for (int i = 0; i < n; i++)
    {
        for (int j = n - 1; j >= 0; j--)
        {
            if (set_bit[i, j])
                continue;
 
            if (j != n - 1)
                p_arr[i, j] += p_arr[i, j + 1];
 
            p_arr[i, j] += (!set_bit[i, j]) ? 1 : 0;
        }
    }
}
 
public class pair
{
    public int first,second;
    public pair(){}
     
    public pair(int a, int b)
    {
        first = a;
        second = b;
    }
}
 
// Function to create a booleanean
// matrix set_bit which stores '1'
// at an index (R, C) if ith bit
// of arr[R,C] is set.
static int matrixOrValueOne(Boolean [,]set_bit)
{
    // array to store prefix count of zeros from
    // right to left for booleanean array
    int [,]p_arr = new int[n, n];
     
    for(int i = 0; i < n; i++)
    for(int j = 0; j < n; j++)
    p_arr[i, j] = 0;
 
    findPrefixCount(p_arr, set_bit);
 
    // variable to store the count of
    // submatrices with OR value 0
    int count_zero_submatrices = 0;
 
    // For each index of a column we will try to
    // determine the number of sub-matrices
    // starting from that index
    // and has all 1s
    for (int j = 0; j < n; j++)
    {
        int i = n - 1;
 
        // stack to store elements and the count
        // of the numbers they popped
        // First part of pair will be the
        // value of inserted element.
        // Second part will be the count
        // of the number of elements pushed
        // before with a greater value
        Stack q = new Stack();
 
        // variable to store the number of
        // submatrices with all 0s
        int to_sum = 0;
        while (i >= 0)
        {
            int c = 0;
 
            while (q.Count != 0 &&
                   q.Peek().first > p_arr[i, j])
            {
                to_sum -= (q.Peek().second + 1) *
                          (q.Peek().first - p_arr[i, j]);
                c += q.Peek().second + 1;
                q.Pop();
            }
 
            to_sum += p_arr[i, j];
            count_zero_submatrices += to_sum;
            q.Push(new pair(p_arr[i, j], c));
 
            i--;
        }
    }
 
    return (n * (n + 1) * n * (n + 1)) /
            4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumOrMatrix(int [,]arr)
{
    int sum = 0;
 
    int mul = 1;
 
    for (int i = 0; i < 30; i++)
    {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        Boolean [,]set_bit = new Boolean[n, n];
 
        for (int R = 0; R < n; R++)
            for (int C = 0; C < n; C++)
                set_bit[R, C] = ((arr[R, C] &
                                 (1 << i)) != 0);
 
        sum += (mul * matrixOrValueOne(set_bit));
 
        mul *= 2;
    }
 
    return sum;
}
 
// Driver Code
public static void Main(String []args)
{
    int [,]arr = {{ 9, 7, 4 },
                  { 8, 9, 2 },
                  { 11, 11, 5 }};
 
    Console.WriteLine( sumOrMatrix(arr));
}
}
 
// This code is contributed by Rajput-Ji

Javascript

// Javascript program to find sum of Bitwise-OR of
// all submatrices
 
let n = 3;
// Function to find prefix-count for 
// each row from right to left
function findPrefixCount(p_arr,set_bit)
{
    for (let i = 0; i < n; i++)
    {
        for (let j = n - 1; j >= 0; j--)
        {
            if (set_bit[i][j])
                continue;
   
            if (j != n - 1)
                p_arr[i][j] += p_arr[i][j + 1];
   
            p_arr[i][j] += (!set_bit[i][j]) ? 1 : 0;
        }
    }
}
 
class pair
{
    constructor(a,b)
    {
        this.first=a;
        this.second=b;
    }
}
 
// Function to create a booleanean
// matrix set_bit which stores '1'
// at an index (R, C) if ith bit
// of arr[R][C] is set.
function matrixOrValueOne(set_bit)
{
    // array to store prefix count of zeros from
    // right to left for booleanean array
    let p_arr = new Array(n);
     
       
    for(let i = 0; i < n; i++)
    {
        p_arr[i]=new Array(n);
        for(let j = 0; j < n; j++)
            p_arr[i][j] = 0;
      }
    findPrefixCount(p_arr, set_bit);
   
    // variable to store the count of
    // submatrices with OR value 0
    let count_zero_submatrices = 0;
   
    // For each index of a column we will try to
    // determine the number of sub-matrices
    // starting from that index
    // and has all 1s
    for (let j = 0; j < n; j++)
    {
        let i = n - 1;
   
        // stack to store elements and the count
        // of the numbers they popped
        // First part of pair will be the
        // value of inserted element.
        // Second part will be the count
        // of the number of elements pushed
        // before with a greater value
        let q = [];
   
        // variable to store the number of submatrices
        // with all 0s
        let to_sum = 0;
        while (i >= 0)
        {
            let c = 0;
   
            while (q.length != 0 && q[q.length-1].first > p_arr[i][j])
            {
                to_sum -= (q[q.length-1].second + 1) *
                            (q[q.length-1].first - p_arr[i][j]);
                c += q[q.length-1].second + 1;
                q.pop();
            }
   
            to_sum += p_arr[i][j];
            count_zero_submatrices += to_sum;
            q.push(new pair( p_arr[i][j], c ));
   
            i--;
        }
    }
   
    return (n * (n + 1) * n * (n + 1)) /
                4 - count_zero_submatrices;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
function sumOrMatrix(arr)
{
    let sum = 0;
   
    let mul = 1;
   
    for (let i = 0; i < 30; i++)
    {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        let set_bit = new Array(n);
        for(let i=0;i<n;i++)
            set_bit[i]=new Array(n);
   
        for (let R = 0; R < n; R++)
            for (let C = 0; C < n; C++)
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
   
        sum += (mul * matrixOrValueOne(set_bit));
   
        mul *= 2;
    }
   
    return sum;
}
 
// Driver Code
let arr=[[ 9, 7, 4 ],
                    [ 8, 9, 2 ],
                    [ 11, 11, 5 ]];
document.write( sumOrMatrix(arr));
 
// This code is contributed by avanitrachhadiya2155

输出：

398

时间复杂度：O(N ² )。

如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。