363 Max Sum of Rectangle No Larger Than K 最大矩阵和不超过K

Given a non-empty 2D matrix matrix and an integer k, find the max sum of a rectangle in the matrix such that its sum is no larger than k.
Example:
Given matrix = [
  [1,  0, 1],
  [0, -2, 3]
]
k = 2
The answer is 2. Because the sum of rectangle [[0, 1], [-2, 3]] is 2 and 2 is the max number no larger than k (k = 2).
Note:
    The rectangle inside the matrix must have an area > 0.
    What if the number of rows is much larger than the number of columns?
详见：https://leetcode.com/problems/max-sum-of-rectangle-no-larger-than-k/description/

C++：

class Solution {
public:
    int maxSumSubmatrix(vector<vector<int>>& matrix, int k)
    {
        if (matrix.empty() || matrix[0].empty())
        {
            return 0;
        }
        int m = matrix.size(), n = matrix[0].size(), res = INT_MIN;
        int sum[m][n];
        for (int i = 0; i < m; ++i) 
        {
            for (int j = 0; j < n; ++j) 
            {
                int t = matrix[i][j];
                if (i > 0)
                {
                    t += sum[i - 1][j];
                }
                if (j > 0)
                {
                    t += sum[i][j - 1];
                }
                if (i > 0 && j > 0) 
                {
                    t -= sum[i - 1][j - 1];
                }
                sum[i][j] = t;
                for (int r = 0; r <= i; ++r) 
                {
                    for (int c = 0; c <= j; ++c) 
                    {
                        int d = sum[i][j];
                        if (r > 0) 
                        {
                            d -= sum[r - 1][j];
                        }
                        if (c > 0)
                        {
                            d -= sum[i][c - 1];
                        }
                        if (r > 0 && c > 0) 
                        {
                            d += sum[r - 1][c - 1];
                        }
                        if (d <= k)
                        {
                            res = max(res, d);
                        }
                    }
                }
            }
        }
        return res;
    }
};

 参考：https://www.cnblogs.com/grandyang/p/5617660.html