LeetCode "Maximal Square"

Intuition: 2D DP. Basic idea: compose square at dp[i][j] from dp[i-1][j-1]. You need 2 facility 2D matrix: accumulated horizontalvertical number of 1s.

class Solution
{
public:
    int maximalSquare(vector<vector<char>>& m)
    {
        size_t row = m.size();
        if (row == 0) return 0;
        size_t col = m[0].size();

        vector<vector<int>> hori(row, vector<int>(col, 0));
        vector<vector<int>> vert(row, vector<int>(col, 0));
        vector<vector<int>> dp(row, vector<int>(col, 0));

        for (int i = 0; i < row; i++)
        for (int j = 0; j < col; j++)
        {
            int v = m[i][j] - '0';
            if (j == 0)            hori[i][j] = v;
            else if (v == 0)    hori[i][j] = 0;
            else                hori[i][j] = hori[i][j - 1] + 1;
        }

        for (int j = 0; j < col; j++)
        for (int i = 0; i < row; i++)
        {
            int v = m[i][j] - '0';
            if (i == 0)            vert[i][j] = v;
            else if (v == 0)    vert[i][j] = 0;
            else                vert[i][j] = vert[i - 1][j] + 1;
        }

        //
        int ret = 0;
        for (int i = 0; i < row; i++)
        {
            dp[i][0] = m[i][0] - '0';
            if (dp[i][0]) ret = 1;
        }
        for (int j = 0; j < col; j++)
        {
            dp[0][j] = m[0][j] - '0';
            if (dp[0][j]) ret = 1;
        }

        for (int i = 1; i < row; i++)
        for (int j = 1; j < col; j++)
        {
            int v = m[i][j] - '0';
            if (v)
            {
                int a = dp[i - 1][j - 1] + 1;
                int b = std::min(hori[i][j], vert[i][j]);
                dp[i][j] = std::min(a, b);

                ret = std::max(ret, dp[i][j]);
            }
            else
            {
                dp[i][j] = m[i][j] - '0';
            }
        }
        return ret * ret;
    }
};