所有子矩阵的按位与之和

给定一个NxN矩阵，任务是找到所有矩形子矩阵的按位与之和。

例子：

Input : arr[][] = {{1, 1, 1},
                   {1, 1, 1},
                   {1, 1, 1}}
Output : 36
Explanation: All the possible submatrices will have AND value 1.
Since, there are 36 submatrices in total, ans = 36

Input : arr[][] = {{9, 7, 4},
                   {8, 9, 2},
                   {11, 11, 5}}
Output : 135 

先决条件：二进制全为1的矩形子矩阵的数目。
天真的解决方案：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AND。该方法的时间复杂度将为O(N ⁶ )。

高效的方法：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，并且变量“ sum”用于存储最终和。
这里的想法是，我们将尝试找到第i^个位设置为AND的数量(按位与(＆)的子矩阵)。让我们假设，存在^第i^个位设置为’S _i ‘的子矩阵。对于，^第i位，总和可以更新为总和+ =(2 ⁱ * S _i ) 。
对于每个位“ i”，如果设置了arr [R] [C]的^第i^个位，则创建一个布尔矩阵set_bit ，将其存储在索引(R，C)处为“ 1”。否则，它存储为“ 0”。然后，对于该布尔数组，我们尝试找到全为1s(S _i )的矩形子矩阵的数量。因为，^第i位，最后一笔将被更新为：

sum += 2i * Si

下面是上述方法的实现：

C++

// C++ program to find sum of Bit-wise AND
// 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 find the number of submatrices
// with all 1s
int matrixAllOne(bool set_bit[][n])
{
    // Array to store required prefix count of 1s from
    // right to left for boolean array
    int p_arr[n][n] = { 0 };
 
    findPrefixCount(p_arr, set_bit);
 
    // Variable to store the final answer
    int ans = 0;
 
    // For each index of a column, 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 is value of inserted element
        // Second part is count of the number of elements
        // pushed before with a greater value
        stack > q;
 
        // variable to store the number of submatrices
        // with all 1s
        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];
            ans += to_sum;
 
            q.push({ p_arr[i][j], c });
            i--;
        }
    }
 
    return ans;
}
 
// Function to find the sum of Bitwise-AND
// of all submatrices
int sumAndMatrix(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 * matrixAllOne(set_bit));
 
        mul *= 2;
    }
 
    return sum;
}
 
// Driver Code
int main()
{
    int arr[][n] = { { 9, 7, 4 },
                     { 8, 9, 2 },
                     { 11, 11, 5 } };
 
    cout << sumAndMatrix(arr);
 
    return 0;
}

Java

// Java program to find sum of Bit-wise AND
// 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 find the number of
// submatrices with all 1s
static int matrixAllOne(boolean set_bit[][])
{
    // Array to store required prefix count of 1s from
    // right to left for boolean 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 final answer
    int ans = 0;
 
    // For each index of a column, 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 is value of inserted element
        // Second part is 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 1s
        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];
            ans += to_sum;
 
            q.push(new pair( p_arr[i][j], c ));
            i--;
        }
    }
    return ans;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumAndMatrix(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 * matrixAllOne(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( sumAndMatrix(arr));
}
}
 
// This code is contributed by Arnab Kundu

Python3

# Python3 program to find sum of
# Bitwise-AND 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 not 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.
def matrixAllOne(set_bit):
 
    # 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, set_bit)
 
    # Variable to store the final answer
    ans = 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 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]
            ans += to_sum
 
            q.append((p_arr[i][j], c))
            i -= 1
 
    # Return the final answer
    return ans
 
# Function to find sum of
# Bitwise-AND of all submatrices
def sumAndMatrix(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 * matrixAllOne(set_bit))
        mul *= 2
 
    return Sum
 
# Driver Code
if __name__ == "__main__":
     
    n = 3
    arr = [[9, 7, 4],
        [8, 9, 2],
        [11, 11, 5]]
 
    print(sumAndMatrix(arr))
 
# This code is contributed by Rituraj Jain

C#

// C# program to find sum of Bit-wise AND
// 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,
                            bool [,]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 find the number of
// submatrices with all 1s
static int matrixAllOne(bool [,]set_bit)
{
    // Array to store required prefix count of 1s from
    // right to left for boolean 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 final answer
    int ans = 0;
 
    // For each index of a column, 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 is value of inserted element
        // Second part is 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 1s
        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];
            ans += to_sum;
 
            q.Push(new pair( p_arr[i,j], c ));
            i--;
        }
    }
    return ans;
}
 
// Function to find sum of Bitwise-OR of
// all submatrices
static int sumAndMatrix(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
        bool [,]set_bit = new bool[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 * matrixAllOne(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(sumAndMatrix(arr));
}
}
 
// This code contributed by Rajput-Ji

Javascript

 
// Javascript program to find sum of Bit-wise AND
// of all submatrices
 
var n = 3;
 
// Function to find prefix-count for each row
// from right to left
function findPrefixCount(p_arr, set_bit)
{
    for (var i = 0; i < n; i++) {
        for (var 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];
        }
    }
}
 
// Function to find the number of submatrices
// with all 1s
function matrixAllOne(set_bit)
{
    // Array to store required prefix count of 1s from
    // right to left for boolean array
    var p_arr = Array.from(Array(n), ()=> Array(n).fill(0));
 
    findPrefixCount(p_arr, set_bit);
 
    // Variable to store the final answer
    var ans = 0;
 
    // For each index of a column, determine the number
    // of sub-matrices starting from that index
    // and has all 1s
    for (var j = 0; j < n; j++) {
        var i = n - 1;
 
        // Stack to store elements and the count
        // of the numbers they popped
        // First part of pair is value of inserted element
        // Second part is count of the number of elements
        // pushed before with a greater value
        var q = [];
 
        // variable to store the number of submatrices
        // with all 1s
        var to_sum = 0;
 
        while (i >= 0) {
            var c = 0;
            while (q.length != 0 && q[q.length - 1][0] > p_arr[i][j]) {
                to_sum -= (q[q.length - 1][1] + 1) * (q[q.length - 1][0] - p_arr[i][j]);
                c += q[q.length - 1][1] + 1;
                q.pop();
            }
 
            to_sum += p_arr[i][j];
            ans += to_sum;
 
            q.push([p_arr[i][j], c ]);
            i--;
        }
    }
 
    return ans;
}
 
// Function to find the sum of Bitwise-AND
// of all submatrices
function sumAndMatrix(arr)
{
    var sum = 0;
 
    var mul = 1;
 
    for (var i = 0; i < 30; i++) {
        // matrix to store the status
        // of ith bit of each element
        // of matrix arr
        var set_bit = Array.from(Array(n), ()=> Array(n));
 
        for (var R = 0; R < n; R++)
            for (var C = 0; C < n; C++)
                set_bit[R][C] = ((arr[R][C] & (1 << i)) != 0);
 
        sum += (mul * matrixAllOne(set_bit));
 
        mul *= 2;
    }
 
    return sum;
}
 
// Driver Code
var arr = [ [ 9, 7, 4 ],
                 [ 8, 9, 2 ],
                 [ 11, 11, 5 ] ];
document.write( sumAndMatrix(arr));
 

输出：

135

时间复杂度：O(N ² )。

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