leetcode1277 Count Square Submatrices with All Ones

思路：

使用动态规划，dp[i][j]表示以(i, j)作为右下角的最大的全1矩阵的边长。

实现：

class Solution
{
public:
    int countSquares(vector<vector<int>>& matrix)
    {
        if (matrix.empty()) return ;
        int n = matrix.size(), m = matrix[].size();
        vector<vector<int>> dp(n, vector<int>(m, ));
        for (int i = ; i < n; i++) dp[i][] = (matrix[i][] == );
        for (int i = ; i < m; i++) dp[][i] = (matrix[][i] == );
        for (int i = ; i < n; i++)
        {
            for (int j = ; j < m; j++)
            {
                if (matrix[i][j] == ) dp[i][j] = ;
                else
                {
                    dp[i][j] = min(min(dp[i - ][j - ], dp[i - ][j]), dp[i][j - ]) + ;
                }
            }
        }
        int res = ;
        for (int i = ; i < dp.size(); i++)
            res += accumulate(dp[i].begin(), dp[i].end(), );
        return res;
    }
}