Instead of representing states explicitly, we may represent a world using variables: A state is described by a set of variables (i.e., features, attributes). We use capital letters, e.g., X, to represent variables. Each variable X can take a value from its domain D. A factored representation consists of a set of variables X1, X2, . . . , Xk where each variable Xi is associated with a domain Di A candidate solution (or assignment) of a set of variables is a function that maps each variable in the set with a value in its domain. Let Ω be the set of all candidate solutions.理解分析这些内容

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On the set of equations representing a Bézier curve and its applications

这段话讨论了一种描述状态的方式,即使用变量来表示一个世界。每个状态由一组变量描述,每个变量都可以从其域中取值。这个表示方法称为分解表示,由一组变量X1,X2,...,Xk组成,其中每个变量Xi都与域Di相关联。候选解决方案是将变量集中每个变量映射到其域中的值的函数。所有候选解决方案的集合称为Ω。这种表示方法可以用于描述各种问题,如搜索、规划等。
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Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO def check_adjacent_for_one(flat_image: list[int], flat_pixel: int) -> bool: """ Checks if a pixel has an adjacent pixel with the value of 1. :param flat_image: 1D list of integers representing a flattened image . :param flat_pixel: Integer representing the index of the pixel in question. :return: Boolean. """ #TODO

Create a function pixel_flip(lst, orig_lst, budget, results, i=0) that uses recursion to generate all possible new unique images from the input orig_lst, following these rules: • The input lst is the current list being processed. Initially, this will be the same as orig_lst which is the original flattened image. • The input budget represents the number of pixels that can still be flipped. When the budget reaches 0, no more pixels can be flipped. • The input results is a list of resulting flattened images with flipped pixels. Initially, this will be an empty list. • The input i represents the index of the pixel being processed, by default set to 0, which is used to drive the recursive function towards its base case (i.e., initially starting from i=0). At termination of the function, the argument results should contain all possibilities of the input orig_lst by only flipping pixels from 0 to 1 under both the budget and the adjacency constraints. fill code at #TODO def pixel_flip(lst: list[int], orig_lst: list[int], budget: int, results: list, i: int = 0) -> None: """ Uses recursion to generate all possibilities of flipped arrays where a pixel was a 0 and there was an adjacent pixel with the value of 1. :param lst: 1D list of integers representing a flattened image . :param orig_lst: 1D list of integers representing the original flattened image. :param budget: Integer representing the number of pixels that can be flipped . :param results: List of 1D lists of integers representing all possibilities of flipped arrays, initially empty. :param i: Integer representing the index of the pixel in question. :return: None. """ #TODO

The human visual cortex is biased towards shape components while CNNs produce texture biased features. This fact may explain why the performance of CNN significantly degrades with low-labeled input data scenarios. In this paper, we propose a frequency re-calibration U-Net (FRCU-Net) for medical image segmentation. Representing an object in terms of frequency may reduce the effect of texture bias, resulting in better generalization for a low data regime. To do so, we apply the Laplacian pyramid in the bottleneck layer of the U-shaped structure. The Laplacian pyramid represents the object proposal in different frequency domains, where the high frequencies are responsible for the texture information and lower frequencies might be related to the shape. Adaptively re-calibrating these frequency representations can produce a more discriminative representation for describing the object of interest. To this end, we first propose to use a channel-wise attention mechanism to capture the relationship between the channels of a set of feature maps in one layer of the frequency pyramid. Second, the extracted features of each level of the pyramid are then combined through a non-linear function based on their impact on the final segmentation output. The proposed FRCU-Net is evaluated on five datasets ISIC 2017, ISIC 2018, the PH2, lung segmentation, and SegPC 2021 challenge datasets and compared to existing alternatives, achieving state-of-the-art results.请详细介绍这段话中的技术点和实现方式

代码We now want to always redraw all the points that have ever been drawn in the panel, not just the last point. To do this, we must save the coordinates of all these points so that we can redraw them all one by one in the paintComponent method every time this method is called. To save the coordinates of the various mouse positions we click, replace the x and y instance variables of the MyPanel class with a single private instance variable called points of type ArrayList. The Point class is provided to you by Swing. In the constructor of MyPanel, initialize the points instance variable with a new arraylist object of the same type. In the mouseClicked method of the mouse listener, use the getPoint method of the mouse event object to get a Point object representing the position of the mouse click (that Point object internally stores both the x and y coordinates of the mouse click event). Then add this Point object to the arraylist using the arraylist’s add method. Then, in the paintComponent method, add a loop to draw in the panel all the points of the arraylist. You can get the number of elements in the arraylist by using the size method of the arraylist; you can access a specific element of the arraylist at index i by using the get(i) method of the arraylist (element indexes start at zero in an arraylist). The Point class has getX and getY methods to get the coordinates of the point (these two methods return values of type double so you need to cast the returned values into the int type before you can use them to draw a point).

用C++编写程序,实现以下问题2、题目ID Codes(POJ1146) Time Limit: 1000MS Memory Limit: 10000K 描述: It is 2084 and the year of Big Brother has finally arrived, albeit a century late. In order to exercise greater control over its citizens and thereby to counter a chronic breakdown in law and order, the Government decides on a radical measure--all citizens are to have a tiny microcomputer surgically implanted in their left wrists. This computer will contains all sorts of personal information as well as a transmitter which will allow people's movements to be logged and monitored by a central computer. (A desirable side effect of this process is that it will shorten the dole queue for plastic surgeons.) An essential component of each computer will be a unique identification code, consisting of up to 50 characters drawn from the 26 lower case letters. The set of characters for any given code is chosen somewhat haphazardly. The complicated way in which the code is imprinted into the chip makes it much easier for the manufacturer to produce codes which are rearrangements of other codes than to produce new codes with a different selection of letters. Thus, once a set of letters has been chosen all possible codes derivable from it are used before changing the set. For example, suppose it is decided that a code will contain exactly 3 occurrences of a', 2 of b' and 1 of c', then three of the allowable 60 codes under these conditions are: abaabc abaacb ababac These three codes are listed from top to bottom in alphabetic order. Among all codes generated with this set of characters, these codes appear consecutively in this order. Write a program to assist in the issuing of these identification codes. Your program will accept a sequence of no more than 50 lower case letters (which may contain repeated characters) and print the successor code if one exists or the message No Successor' if the given code is the last in the sequence for that set of characters. 输入: Input will consist of a series of lines each containing a string representing a code. The entire file will be terminated by a line consisting of a single #. 输出: Output will consist of one line for each code read containing the successor code or the words 'No Successor'. 样例输入 abaacb cbbaa # 样例输出 ababac No Successor

Q21: Which of the following is a valid user-defined output stream manipulator header? a. ostream& tab( ostream& output ) b. ostream tab( ostream output ) c. istream& tab( istream output ) d. void tab( ostream& output ) Q22: What will be output by the following statement? cout << showpoint << setprecision(4) << 11.0 << endl; a. 11 b. 11.0 c. 11.00 d. 11.000 Q23: Which of the following stream manipulators causes an outputted number’s sign to be left justified, its magnitude to be right justified and the center space to be filled with fill characters? a. left b. right c. internal d. showpos Q24: Which of the following statements restores the default fill character? a. cout.defaultFill(); b. cout.fill(); c. cout.fill( 0 ); d. cout.fill( ' ' ); Q25: When the showbase flag is set: a. The base of a number precedes it in brackets. b. Decimal numbers are not output any differently. c. "oct" or "hex" will be displayed in the output stream. d. Octal numbers can appear in one of two ways. Q26: What will be output by the following statements? double x = .0012345; cout << fixed << x << endl; cout << scientific << x << endl; a. 1.234500e-003 0.001235 b. 1.23450e-003 0.00123450 c. .001235 1.234500e-003 d. 0.00123450 1.23450e-003 Q27: Which of the following outputs does not guarantee that the uppercase flag has been set? a. All hexadecimal numbers appear in the form 0X87. b. All numbers written in scientific notation appear the form 6.45E+010. c. All text outputs appear in the form SAMPLE OUTPUT. d. All hexadecimal numbers appear in the form AF6. Q28: Which of the following is not true about bool values and how they're output with the output stream? a. The old style of representing true/false values used -1 to indicate false and 1 to indicate true. b. A bool value outputs as 0 or 1 by default. c. Stream manipulator boolalpha sets the output stream to display bool values as the strings "true" and "false". d. Both boolalpha and noboolalpha are “sticky” settings.

Another example is the SRIOV_NET_VF resource class, which is provided by SRIOV-enabled network interface cards. In the case of multiple SRIOV-enabled NICs on a compute host, different qualitative traits may be tagged to each NIC. For example, the NIC called enp2s0 might have a trait “CUSTOM_PHYSNET_PUBLIC” indicating that the NIC is attached to a physical network called “public”. The NIC enp2s1 might have a trait “CUSTOM_PHYSNET_INTRANET” that indicates the NIC is attached to the physical network called “Intranet”. We need a way of representing that these NICs each provide SRIOV_NET_VF resources but those virtual functions are associated with different physical networks. In the resource providers data modeling, the entity which is associated with qualitative traits is the resource provider object. Therefore, we require a way of representing that the SRIOV-enabled NICs are themselves resource providers with inventories of SRIOV_NET_VF resources. Those resource providers are contained on a compute host which is a resource provider that has inventory records for other types of resources such as VCPU, MEMORY_MB or DISK_GB. This spec proposes that nested resource providers be created to allow for distinguishing details of complex components of some resource providers. During review the question came up about “rolling up” amounts of these nested providers to the root level. Imagine this scenario: I have a NIC with two PFs, each of which has only 1 VF available, and I get a request for 2 VFs without any traits to distinguish them. Since there is no single resource provider that can satisfy this request, it will not select this root provider, even though the root provider “owns” 2 VFs. This spec does not propose any sort of “rolling up” of inventory, but this may be something to consider in the future. If it is an idea that has support, another BP/spec can be created then to add this behavior.

Solve the problem with c++ code, and give your code: Ack Country has N cities connected by M one-way channels. The cities occupied by the rebels are numbered 1, while the capital of Ack country is numbered N. In order to reduce the loss of effective force, you are permitted to use self-propelled bombers for this task. Any bomber enters the capital, your job is done. This seems simple enough, but the only difficulty is that many cities in Ack Country are covered by shields. If a city is protected by a shield, all shield generators that maintain the shield need to be destroyed before the bomber can enter or pass through the city. Fortunately, we know the cities where all the shield generators are located, and which cities' shields are being charged. If the bomber enters a city, all of its shield generators can be destroyed instantly. You can release any number of Bombermen and execute any command at the same time, but it takes time for bombermen to pass through the roads between cities. Please figure out how soon you can blow up Ack Nation's capital. The clock is ticking. Input: Two positive integers N,M in the first row. The next M lines, each with three positive integers, indicate that there is a road leading from the city to the city. It takes w time for the bomber to cross this road. Then N lines, each describing a city's shield. The first is a positive integer n, representing the number of shield generators that maintain shields in the city. Then n_i city numbers between 1 and N, indicating the location of each shield generator. In other words, if your bomber needs to enter the city, the bomber needs to enter all the entered cities in advance. If n_i=0, the city has no shields. Guarantee n_i=0.Output: a positive integer, the minimum time to blow up the capital. e.g., Input: 6 6 1 2 1 1 4 3 2 3 3 2 5 2 4 6 2 5 3 2 0 0 0 1 3 0 2 3 5, Output: 6.

Also create a ControllerCreate class that extends Controller.The create method takes as arguments the name of a new library user, a number of books (as a string), and an integer representing the role of user to create (where the integer 0 means a lender and the integer 1 means a borrower). The create method of the controller then transforms the book number from a string to an integer (using the Integer.parseInt static method), creates an object from the correct class (based on the role specified by the user input: lender or borrower) and calls the addUser method of the library to add the new user object to the library. • If no exception occurs then the create method of the controller returns the empty string. • If the constructor of the Borrower class throws a NotALenderException then the create method of the controller must catch this exception and return as result the error message from the exception object. • If the parseInt method of the Integer class throws a NumberFormatException (because the user typed something which is not an integer) then the create method of the controller must catch this exception and return as result the error message from the exception object. Modify the run method of the GUI class to add a ViewCreate view that uses a ControllerCreate controller and the same model as before (not a new model!) Do not delete the previous views. Note: if at the end of Question 7 you had manually added to your library (model object) some users for testing, then you must now remove those users from the run method of the anonymous class inside the GUI class. You do not need these test users anymore because you have now a graphical user interface to create new users! Run your GUI and check that you can correctly use the new view to create different users for your library, with different types of roles. • Check that, when you create a new user, the simple view is automatically correctly updated to show the new total number of books borrowed by all users. • Also use the “get book” view to check that the users are correctly created with the correct names and correct number of books. • Also check that trying to create a borrower with a negative number of books correctly shows an error message. Also check that trying to create a user with a number of books which is not an integer correctly shows an error message (do not worry about the content of the error message). After you created a new user, you can also check whether it is a lender or a borrower using the “more book” view to increase the number of books of the user by a big negative number: • if the new user you created is a lender, then increasing the number of books by a big negative value will work and the number of books borrowed by the user will just become a larger value (you can then check that using the “get book” view); • if the new user you created is a borrower, then increasing the number of books by a big negative value will fail with an error message and the number of books borrowed by the user will not change (you can then check that using the “get book” view). 完成符合以上要求的java代码

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