查找具有给定和的子矩阵

给定一个N x N矩阵和两个整数S和K ，任务是找出是否存在总和等于S的K x K子矩阵。

例子：

Input: K = 2, S = 14, mat[][] = { 
{ 1, 2, 3, 4 }, 
{ 5, 6, 7, 8 }, 
{ 9, 10, 11, 12 }, 
{ 13, 14, 15, 16 }} 
Output: Yes 
1 2 
5 6 
is the required 2 x 2 sub-matrix with element sum = 14

Input: K = 1, S = 5, mat[][] = {{1, 2}, {7, 8}} 
Output: No 

方法：动态规划可以用来解决这个问题，

• 创建一个数组dp[N + 1][N + 1] ，其中dp[i][j]存储行在1到i之间、列在1到j之间的所有元素的总和。
• 一旦生成了二维矩阵，现在假设我们希望找到从(i, j)到(i + x, j + x)的平方和。所需的总和将为dp[i + x][j + x] – dp[i][j + x] – dp[i + x][j] + dp[i][j]其中，
  1. 第一项表示1到i + x之间的行和1到j + x之间的列中存在的所有元素的总和。这个区域有我们需要的正方形。
  2. 后两项是去除我们所需区域之外但在第一步计算的区域内的区域。
  3. 1到i之间的行和1到j之间的列的元素和在第二步中减去两次，因此添加一次。

下面是上述方法的实现：

C++

// C++ implementation of the approach
#include 
using namespace std;
#define ll long long int
#define N 4
 
// Function to return the sum of the sub-matrix
int getSum(int r1, int r2, int c1, int c2,
           int dp[N + 1][N + 1])
{
    return dp[r2][c2] - dp[r2][c1] - dp[r1][c2]
           + dp[r1][c1];
}
 
// Function that returns true if it is possible
// to find the sub-matrix with required sum
bool sumFound(int K, int S, int grid[N][N])
{
 
    // 2-D array to store the sum of
    // all the sub-matrices
    int dp[N + 1][N + 1];
 
    // Filling of dp[][] array
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++)
            dp[i + 1][j + 1] = dp[i + 1][j] + dp[i][j + 1]
                               - dp[i][j] + grid[i][j];
 
    // Checking for each possible sub-matrix of size k X k
    for (int i = 0; i < N; i++)
        for (int j = 0; j < N; j++) {
            int sum = getSum(i, i + K, j, j + K, dp);
 
            if (sum == S)
                return true;
        }
 
    // Sub-matrix with the given sum not found
    return false;
}
 
// Driver code
int main()
{
    int grid[N][N] = { { 1, 2, 3, 4 },
                       { 5, 6, 7, 8 },
                       { 9, 10, 11, 12 },
                       { 13, 14, 15, 16 } };
    int K = 2;
    int S = 14;
 
    // Function call
    if (sumFound(K, S, grid))
        cout << "Yes" << endl;
    else
        cout << "No" << endl;
}
// Modified by Kartik Verma

Java

// Java implementation of the approach
class GfG {
    static int N = 4;
 
    // Function to return the sum of the sub-matrix
    static int getSum(int r1, int r2, int c1, int c2,
                      int dp[][])
    {
        return dp[r2][c2] - dp[r2][c1] - dp[r1][c2]
            + dp[r1][c1];
    }
 
    // Function that returns true if it is possible
    // to find the sub-matrix with required sum
    static boolean sumFound(int K, int S, int grid[][])
    {
 
        // 2-D array to store the sum of
        // all the sub-matrices
        int dp[][] = new int[N + 1][N + 1];
 
        // Filling of dp[][] array
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                dp[i + 1][j + 1] = dp[i + 1][j]
                                   + dp[i][j + 1] - dp[i][j]
                                   + grid[i][j];
            }
        }
 
        // Checking for each possible sub-matrix of size k X
        // k
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                int sum = getSum(i, i + K, j, j + K, dp);
 
                if (sum == S) {
                    return true;
                }
            }
        }
 
        // Sub-matrix with the given sum not found
        return false;
    }
 
    // Driver code
    public static void main(String[] args)
    {
        int grid[][] = { { 1, 2, 3, 4 },
                         { 5, 6, 7, 8 },
                         { 9, 10, 11, 12 },
                         { 13, 14, 15, 16 } };
        int K = 2;
        int S = 14;
 
        // Function call
        if (sumFound(K, S, grid)) {
            System.out.println("Yes");
        }
        else {
            System.out.println("No");
        }
    }
}
 
// This code contributed by Rajput-Ji
// Modified by Kartik Verma

Python3

# Python implementation of the approach
N = 4
 
# Function to return the sum of the sub-matrix
 
 
def getSum(r1, r2, c1, c2, dp):
    return dp[r2][c2] - dp[r2][c1] - dp[r1][c2] + dp[r1][c1]
 
 
# Function that returns true if it is possible
# to find the sub-matrix with required sum
def sumFound(K, S, grid):
 
    # 2-D array to store the sum of
    # all the sub-matrices
    dp = [[0 for i in range(N+1)] for j in range(N+1)]
 
    # Filling of dp[][] array
    for i in range(N):
        for j in range(N):
            dp[i + 1][j + 1] = dp[i + 1][j] + \
                dp[i][j + 1] - dp[i][j] + grid[i][j]
 
    # Checking for each possible sub-matrix of size k X k
    for i in range(0, N):
        for j in range(0, N):
            sum = getSum(i, i + K, j, j + K, dp)
            if (sum == S):
                return True
 
    # Sub-matrix with the given sum not found
    return False
 
 
# Driver code
grid = [[1, 2, 3, 4],
        [5, 6, 7, 8],
        [9, 10, 11, 12],
        [13, 14, 15, 16]]
K = 2
S = 14
 
# Function call
if (sumFound(K, S, grid)):
    print("Yes")
else:
    print("No")
 
# This code is contributed by ankush_953
# Modified by Kartik Verma

C#

// C# implementation of the approach
using System;
 
class GfG {
    static int N = 4;
 
    // Function to return the sum of the sub-matrix
    static int getSum(int r1, int r2, int c1, int c2,
                      int[, ] dp)
    {
        return dp[r2, c2] - dp[r2, c1] - dp[r1, c2]
            + dp[r1, c1];
    }
 
    // Function that returns true if it is possible
    // to find the sub-matrix with required sum
    static bool sumFound(int K, int S, int[, ] grid)
    {
 
        // 2-D array to store the sum of
        // all the sub-matrices
        int[, ] dp = new int[N + 1, N + 1];
 
        // Filling of dp[,] array
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                dp[i + 1, j + 1] = dp[i + 1, j]
                                   + dp[i, j + 1] - dp[i, j]
                                   + grid[i, j];
            }
        }
 
        // Checking for each possible sub-matrix of size k X
        // k
        for (int i = 0; i < N; i++) {
            for (int j = 0; j < N; j++) {
                int sum = getSum(i, i + K, j, j + K, dp);
 
                if (sum == S) {
                    return true;
                }
            }
        }
 
        // Sub-matrix with the given sum not found
        return false;
    }
 
    // Driver code
    public static void Main(String[] args)
    {
        int[, ] grid = { { 1, 2, 3, 4 },
                         { 5, 6, 7, 8 },
                         { 9, 10, 11, 12 },
                         { 13, 14, 15, 16 } };
        int K = 2;
        int S = 14;
 
        // Funciton call
        if (sumFound(K, S, grid)) {
            Console.WriteLine("Yes");
        }
        else {
            Console.WriteLine("No");
        }
    }
}
 
// This code has been contributed by 29AjayKumar
// Modified by Kartik Verma

Javascript

 
// Javascript implementation of the approach
 
var N = 4
 
// Function to return the sum of the sub-matrix
function getSum(r1, r2, c1, c2, dp)
{
    return dp[r2][c2] - dp[r2][c1] - dp[r1][c2]
           + dp[r1][c1];
}
 
// Function that returns true if it is possible
// to find the sub-matrix with required sum
function sumFound(K, S, grid)
{
 
    // 2-D array to store the sum of
    // all the sub-matrices
    var dp = Array.from(Array(N+1), ()=> Array(N+1).fill(0));
 
    // Filling of dp[][] array
    for (var i = 0; i < N; i++)
        for (var j = 0; j < N; j++)
            dp[i + 1][j + 1] = dp[i + 1][j] + dp[i][j + 1]
                               - dp[i][j] + grid[i][j];
 
    // Checking for each possible sub-matrix of size k X k
    for (var i = 0; i < N; i++)
        for (var j = 0; j < N; j++) {
            var sum = getSum(i, i + K, j, j + K, dp);
 
            if (sum == S)
                return true;
        }
 
    // Sub-matrix with the given sum not found
    return false;
}
 
// Driver code
var grid = [ [ 1, 2, 3, 4 ],
                   [ 5, 6, 7, 8 ],
                   [ 9, 10, 11, 12 ],
                   [ 13, 14, 15, 16 ] ];
var K = 2;
var S = 14;
// Function call
if (sumFound(K, S, grid))
    document.write( "Yes");
else
    document.write( "No" );
 

输出：

Yes

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