2d

问题 Where would i go to look for algorithms that take a 2d grid of values that are either 0 or 1 as input and then identifies all possible non-overlapping rectangles in it? In a more practical explanation: I am drawing a grid that is represented by a number of squares, and i wish to find a way to combine as many adjacent squares into rectangles as possible, in order to cut down on the time spent on cycling through each square and drawing it. Maximum efficiency is not needed, speed is more

问题 I have a 2D array of pointer main.c Int32 * pRxSeqAddr[2][2]; func(pRxSeqAddr); / func.c void func( Int32** pRxSeqAddrPtr) { /// } I get this error: argument of type "Int32 *(*)[2]" is incompatible with parameter of type "Int32 ** I know if it was 1D array of pointer then this function call is fine, but 2D is nagging me.. please help 回答1: Change func.c to: void func(Int32 *pRxSeqAddrPtr[][2]) { } In general, to pass 2D arrays to functions you need to specify the array dimensions for all but

问题 I have 3 points in a 3D space of which I know the exact locations. Suppose they are: (x0,y0,z0) , (x1,y1,z1) and (x2,y2,z2) . Also I have a camera that is looking at these 3 points and I know the 2D locations of those three points on camera view plane. So for example (x0,y0,z0) will be (x0',y0') , and (x1,y1,z1) will be (x1',y1') and (x2,y2,z2) will be (x2',y2') from the camera's point of view. What is the easiest way to find the projection matrix that will project those 3D points into 2D

问题 I have 3 points in a 3D space of which I know the exact locations. Suppose they are: (x0,y0,z0) , (x1,y1,z1) and (x2,y2,z2) . Also I have a camera that is looking at these 3 points and I know the 2D locations of those three points on camera view plane. So for example (x0,y0,z0) will be (x0',y0') , and (x1,y1,z1) will be (x1',y1') and (x2,y2,z2) will be (x2',y2') from the camera's point of view. What is the easiest way to find the projection matrix that will project those 3D points into 2D

问题 I have 3 points in a 3D space of which I know the exact locations. Suppose they are: (x0,y0,z0) , (x1,y1,z1) and (x2,y2,z2) . Also I have a camera that is looking at these 3 points and I know the 2D locations of those three points on camera view plane. So for example (x0,y0,z0) will be (x0',y0') , and (x1,y1,z1) will be (x1',y1') and (x2,y2,z2) will be (x2',y2') from the camera's point of view. What is the easiest way to find the projection matrix that will project those 3D points into 2D

问题 So, I have this algorithm to calculate cross-section of 3D shape with plane given with normal vector. However, my current problem is, that the cross-section is set of 3D points (all lying on that given plane) and to display it I need to map this coordinates to XY plane. This works perfect if the plane normal is something like (0,0,c) - I just copy x and y coordinates discarding z. And here is my question: Since I have no idea how to convert any other plain could anybody give me any hint as to

问题 LibGDX has a coordinate system where (0,0) is at the bottom-left. (like this image: http://i.stack.imgur.com/jVrJ0.png) This has me beating my head against a wall, mainly because I'm porting a game I had already made with the usual coordinate system (where 0,0 is in the Top Left Corner). My question: Is there any simple way of changing this coordinate system? 回答1: If you use a Camera (which you should) changing the coordinate system is pretty simple: camera= new OrthographicCamera(Gdx

问题 LibGDX has a coordinate system where (0,0) is at the bottom-left. (like this image: http://i.stack.imgur.com/jVrJ0.png) This has me beating my head against a wall, mainly because I'm porting a game I had already made with the usual coordinate system (where 0,0 is in the Top Left Corner). My question: Is there any simple way of changing this coordinate system? 回答1: If you use a Camera (which you should) changing the coordinate system is pretty simple: camera= new OrthographicCamera(Gdx

问题 I create ImageView with scr run01 (it's first frame of my animation) In drawable I create xml file with this code: <animation-list xmlns:android="http://schemas.android.com/apk/res/android" android:oneshot="true"> <item android:drawable="@drawable/run01" android:duration="200" /> <item android:drawable="@drawable/run02" android:duration="200" /> <item android:drawable="@drawable/run03" android:duration="200" /> <item android:drawable="@drawable/run04" android:duration="200" /> </animation

问题 I need to create a dictionary, with the last and first name of the author as the key, and the quantity on hand, price, and the book's name as the values. [['Shakespeare', 'William', 'Rome And Juliet', '5', '5.99'], ['Shakespeare', 'William', 'Macbeth', '3', '7.99'], ['Dickens', 'Charles', 'Hard Times', '7', '27.00'], ['']] I've compiled this 2-D list, so far and I'm stuck. Any help would be appreciated! 回答1: The following will create a dictionary that maps each author's name to a list of