題目：EPI

int max_fishing(vector<vector<int>> A)
{
	if (A.empty() || A[0].empty())
		throw new exception;
	for (int i = 0; i < A.size(); i++)
	{
		for (int j = 0; j < A[i].size(); j++)
		{
			int count1 = i - 1 >= 0 ? A[i - 1][j] : 0;
			int count2 = j - 1 >= 0 ? A[i][j - 1] : 0;
			A[i][j] += max(count1, count2);//應該從數字最大的那個格子到達A[i][j]
		}
	}
	return A.back().back();
}