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Description: JTable中获取鼠标所在位置的行数
table=new JTable(cells,columnNames) -J mouse to obtain the location of the line a few table = new J (ce X3, columnNames)
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Size: 3233 |
Author: xsw |
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Description: 选择排序是一种比较优化的排序,它定义的k意义在于表示一次循环后找到的最小(大)值的位置,然后让第i个(第几次循环就是第几个)植交换,这样在第一轮循环中就把最小(大)的值换到了最前面,如果只用i,j 的话,就没有办法找出待排序数据中的最值了
比较排序也叫冒泡排序,就是把相邻的数据比较交换,因为其把小的数字从后面往前慢慢移动,感觉象水泡上升一样,所以叫冒泡排序法.
如过你还没理解,留言~-SELECTION SORT is a more optimal sequencing, its definition of k significance lies in a circle said they find the smallest (large) value of the location, i then allow the first (the first cycle was the first several several) plant exchange, this cycle in the first round put the smallest (CUHK) for the value of the front, if only i, j, then there is no way to identify sequencing data to be the most value compared sequencing is also called Bubble Sort. adjacent is to the comparison of data exchange, because of their small numbers from the back slowly moving forward, Soak up feeling like the same, so called Bubble Sort law. No off if you understand that message ~
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Size: 3268 |
Author: 斯多葛 |
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Description: JTable中获取鼠标所在位置的行数
table=new JTable(cells,columnNames) -J mouse to obtain the location of the line a few table = new J (ce X3, columnNames)
Platform: |
Size: 3072 |
Author: xsw |
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Description: 选择排序是一种比较优化的排序,它定义的k意义在于表示一次循环后找到的最小(大)值的位置,然后让第i个(第几次循环就是第几个)植交换,这样在第一轮循环中就把最小(大)的值换到了最前面,如果只用i,j 的话,就没有办法找出待排序数据中的最值了
比较排序也叫冒泡排序,就是把相邻的数据比较交换,因为其把小的数字从后面往前慢慢移动,感觉象水泡上升一样,所以叫冒泡排序法.
如过你还没理解,留言~-SELECTION SORT is a more optimal sequencing, its definition of k significance lies in a circle said they find the smallest (large) value of the location, i then allow the first (the first cycle was the first several several) plant exchange, this cycle in the first round put the smallest (CUHK) for the value of the front, if only i, j, then there is no way to identify sequencing data to be the most value compared sequencing is also called Bubble Sort. adjacent is to the comparison of data exchange, because of their small numbers from the back slowly moving forward, Soak up feeling like the same, so called Bubble Sort law. No off if you understand that message ~
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Size: 3072 |
Author: 斯多葛 |
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Description: J2ME Location API与规范,高性能手机开发本地程序时，必备文档。-J2ME Location API and standardized, high-performance development of the local mobile phone procedures, the necessary documentation.
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Size: 548864 |
Author: 黄狄 |
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Description: Color-based tracking (CAMSHIFT) demo.
Alexandre R.J. Francois
Copyright (C) 2004 University of Southern California
介绍：CamShift是一种应用颜色信息的跟踪算法,在跟踪过程中,CamShift利用目标的颜色直方图模型得到每帧图像的颜色投影图,并根据上一帧跟踪的结果自适应调整搜索窗口的位置和大小,从而得到当前图像中目标的尺寸和中心位置.在该代码的框架基础上可以开发视屏游戏。-Color-based tracking (CAMSHIFT) demo. Alexandre RJ FrancoisCopyright (C) 2004 University of Southern California, introduced: CamShift is an application of color information of the tracking algorithm in the tracking process, CamShift use of the target color histogram model of each frame image color projection, and a track based on the results of adaptive search window location and size, resulting in the current target image size and central location. In the framework of the code can be developed based on the game screen.
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Size: 65536 |
Author: Andy |
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Description: 最大李雅普诺夫指数的计算
该函数用来计算时间序列的最大Lyapunov 指数--Wolf 方法
％ m: 嵌入维数
％ tau:时间延迟
％ data:时间序列
％ N:时间序列长度
％ P:时间序列的平均周期,选择演化相点距当前点的位置差，即若当前相点为I，则演化相点只能在|I－J|>P的相点中搜寻-The largest Lyapunov exponent calculation of the function used to calculate the time series of the largest Lyapunov index Wolf method m: embedding dimension tau: time delay data: time series N: time-series length P: Time-series average cycle , select evolution with distance of the location of the current point difference, that is, if the current points for the I, the evolution of phase points can only be | I-J |
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Size: 2048 |
Author: 刘于江 |
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Description: 生命游戏，为1970年英国数学家J.H.Conway所提出，某一细胞的邻居包括上，下，左，右，左上，左下，右上与右下相邻的细胞，游戏规则如下：
1，孤单死亡：如果细胞的邻居小于一个，则该细胞在下一个状态死亡。
2，拥挤死亡：如果细胞的邻居在四个以上，则该细胞在下一个状态死亡。
3，稳定：如果细胞的邻居为两个或三个，则该细胞在下一个状态稳定。
4，复活：如果某位置原无细胞存活，而该位置的邻居为三个，则该位置将复活一个细胞。-Game of Life for 1970 put forward by the British mathematician JHConway, a neighbor cell, including the upper and lower, left, right, upper left, lower left, upper right and lower right adjacent cells, the rules of the game are as follows: 1, lonely death: If cells in less than a neighbor, then the next state of cell death. 2, overcrowded death: If the cells in four or more neighbors, then the next state of cell death. 3, Stability: If the cells for two or three neighbors, then the cells in the next stable state. 4, Resurrection: If a location of the original cell-free survival, and the location of three neighbors, then the location will be the revival of a cell.
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Size: 1024 |
Author: 白龙 |
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Description: 八皇后游戏的说明：
问题的提出：八皇后是个古老而有趣的游戏，是由高斯于1850年首先提出的。
要求在国际象棋的棋盘上放置八个皇后，使其不能相互攻击，即任意两个皇后不能处于
棋盘的同一行、同一列和同一条对角线上。试问有多少种放法？
基本思想是：先把皇后放在（0，0）位置，然后把1号皇后放在（1，j）位置，
使其满足要求。接着放2号皇后，依此类推。遇到某个皇后如把她无论放在该行的任意
位置均不满足要求，则前一个皇后放置不当，须重新放置前一皇后，如8个皇后均按要
求放置好，这就是一次成功的摆法。 -Eight Queen s Game description:
Of the problem: the Queen s eight months old and are fun games, by Gauss in 1850 first proposed.
Requirements in the international chess board to place eight queen, making it unable to attack each other, that should not in any two of Queen s
Board of the same line, same column and with a diagonal. How many species release law?
The basic idea is: first Queen s on the [0,0] position, then placed on the 1st Queen s [1, j] position,
To meet the requirements. Then put on the 2nd Queen s, and so on. Such as a Queen s encountered her on the line regardless of the arbitrary
Location does not meet the requirements, then placed before a Queen s improper to be re-placed before the First Queen s, 8 s are such as to
Seeking to place good, and this is before a successful method.
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Size: 5120 |
Author: 王清 |
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Description: 添加BDE别名:paikedata
设置path:[存放位置]\paike\data (如:e:\paike\data)
运行:f_ssgl.exe
--★运行说明------------------------------------------------
13
用户名:nwn
密码:j -Add BDE alias: paikedata set path: [location] \ paike \ data (such as: e: \ paike \ data) to run: f_ssgl.exe- ★ run Description------------------------------------------------ 13 user name: nwn password: j
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Size: 2092032 |
Author: maxiangyu |
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Description: 设计一个国际象棋的马踏棋盘的演示程序。以一二维数组Board[8][8]表示国际象棋棋盘，以用户输入任一位置作为马的起始位置，设为(i,j)，其中0≤i，j≤7，马按照走棋的规则在棋盘范围内进行移动。本程序只求出一种遍历路径，用非递归程序实现吗的行走路线。-To design a chess board horse riding demo program. A two-dimensional array Board [8] [8] that the chess board to the user to enter any one location as the initial location of the horse, set (i, j), in which 0 ≤ i, j ≤ 7, in accordance with the following horse the rules of chess board mobile framework. This procedure only a traversal path, the use of non-recursive process routes to achieve it.
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Size: 9216 |
Author: chissie |
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Description: 本程序主要是用栈来实现迷宫问题。所用算法是非递归算法，输入为迷宫的行数和列数，然后在对应位置上输入其值（0表示该位置通，1表示该位置不通）。输出则是以三元组（i,j,d）的形式,其中(i,j)表示迷宫中的一个坐标，d表示走到下一坐标的方向，-This procedure is mainly used to achieve the maze problem stack. Algorithm used in a non-recursive algorithm, enter the number of rows for the maze and the number of rows, and then enter the location in the corresponding value (0 indicates that the location of Qualcomm, said the location of a barrier). Output is based on triples (i, j, d) form, in which (i, j) that one of the coordinates of the maze, d that the next go the direction of coordinates,
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Size: 19456 |
Author: 尹素芳 |
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Description: 八皇后游戏的说明：
问题的提出：八皇后是个古老而有趣的游戏，是由高斯于1850年首先提出的。
要求在国际象棋的棋盘上放置八个皇后，使其不能相互攻击，即任意两个皇后不能处于
棋盘的同一行、同一列和同一条对角线上。试问有多少种放法？
基本思想是：先把皇后放在（0，0）位置，然后把1号皇后放在（1，j）位置，
使其满足要求。接着放2号皇后，依此类推。遇到某个皇后如把她无论放在该行的任意
位置均不满足要求，则前一个皇后放置不当，须重新放置前一皇后，如8个皇后均按要
求放置好，这就是一次成功的摆法。-Queen s description of the eight games:
Of the problem: the eight Queen s is the oldest and interesting games, by Gauss in 1850 first proposed.
Requirements in the international chess board to place eight of the Queen, making it unable to attack each other, that is, any two can not be at Queen s
Board of the same line, same column and on the same diagonal. How many types of release law?
The basic idea is: first Queen s on (0,0) position, then placed on the 1st Queen s (1, j) position,
To meet the requirements. Then put on the 2nd Queen s, and so on. Such as a Queen s encountered her on the line regardless of the arbitrary
Location does not meet the requirements, then placed before a Queen s improper to be re-placed the previous Queen, as Queen s are to be 8
Place for good, that is, before a successful method.
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Size: 6144 |
Author: 颜昌文 |
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Description: 使用字符数组a[N][N]存储布雷的数据，a[i][j]=’+’表示位置（i，j）没有地雷，a[i][j]=’*’表示位置（i，j）有一个地雷，0<=i<m，0<=j<n，m和n是键盘输入的总行数和总列数并且0<m,n<100。设计一个程序，输入m和n，随机初始化布雷数组，并输出布雷数组和每一位置周围的地雷数量，如果该位置是一个地雷，则输出一个’*’。-The use of characters in the array a [N] [N] stored data mine, a [i] [j ]='+' that the location (i, j) is not mine, a [i] [j ]='*' that the location ( i, j) a mine, 0 < = i <m，0<=j<n，m和n是键盘输入的总行数和总列数并且0<m,n<100。设计一个程序，输入m和n，随机初始化布雷数组，并输出布雷数组和每一位置周围的地雷数量，如果该位置是一个地雷，则输出一个’*’。
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Size: 2048 |
Author: 几件 |
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Description: 贪心算法作业的解题报告(含证明)和源代码:磁盘文件最优存储问题，当初证明花了一些时间。问题描述：设磁盘上有n个文件，f1,f2,…,fn，，每个文件占磁盘上1个磁道。这n个文件的检索概率分别是p1,p2,…,pn,
且∑pi=1。磁头从当前磁道移到被检信息磁道所需的时间可用这2 个磁道之间的径向距离来度量。如果文件pi存放在第i道上, ，则检索这n 个文件的期望时间是∑pipjd(i,j)
其中d(i,j) 是第i道与第j 道之间的径向距离|i-j|。磁盘文件的最优存储问题要求确定这n个文件在磁盘上的存储位置，使期望检索时间达到最小。试设计一个解此问题的算法，并分析算法的正确性和计算复杂性。-Greedy algorithm for solving the report of operations (including certificate) and the source code: the optimal disk file storage issues, had spent some time to prove. Problem Description: There are n disk-based files, f1, f2, ..., fn,, accounts for each file on a disk track. This n the probability of a document retrieval are p1, p2, ..., pn,
And Σpi = 1. Head from the current track information to be seized can be used track the time required between these two tracks to measure the radial distance. Pi if the file stored in the first track i,, then retrieve documents that the expectations of n is Σpipjd (i, j)
Which d (i, j) is the first i Road with the first j of the radial distance between the Road | ij |. Optimal storage disk file to seek to establish that n files on the disk storage location, so expect to minimize search time. Try to design a solution algorithm for this problem and to analyze the correctness of algorithms and computational complexity.
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Size: 266240 |
Author: 吴钦阳 |
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Description: 使用说明
使用时打开此例题目录下pic中的图片，然后依次单击按钮“转”、“1”、“2”、“3”、“4”和“5”，就可以实现精确的车牌定位。
具体步骤
1.24位真彩色->256色灰度图。
2.预处理:中值滤波。
3.二值化:用一个初始阈值T对图像A进行二值化得到二值化图像B。
初始阈值T的确定方法是:选择阈值T=Gmax-(Gmax-Gmin)/3,Gmax和Gmin分别是最高、最低灰度值。
该阈值对不同牌照有一定的适应性,能够保证背景基本被置为0,以突出牌照区域。
4.削弱背景干扰。对图像B做简单的相邻像素灰度值相减,得到新的图像G,即Gi,j=|Pi,j-Pi,j-1|i=0,1,…,439 j=0,1,…,639Gi,0=Pi,0,左边缘直接赋值,不会影响整体效果。
5.用自定义模板进行中值滤波
区域灰度基本被赋值为0。考虑到文字是由许多短竖线组成,而背景噪声有一大部分是孤立噪声,用模板(1,1,1,1,1)T对G进行中值滤波,能够得到除掉了大部分干扰的图像C。
6.牌照搜索:利用水平投影法检测车牌水平位置,利用垂直投影法检测车牌垂直位置。
7.区域裁剪,截取车牌图像。-car License plate location
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Size: 728064 |
Author: ziyousecai |
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Description: 画图程序，包含线、点、圆
明位图中的颜色，有若干个表项，每一个表项是一个RGBQUAD类型的结构，定义一种颜色。RGBQUAD结构的定义如下-line, row, round
UCHAR b=buffer[(i*width+j)*3+realPitch]
UCHAR g=buffer[(i*width+j)*3+1+realPitch]
UCHAR r=buffer[(i*width+j)*3+2+realPitch]
pDC->SetPixel(j,height-i,RGB(r,g,b))
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Size: 1938432 |
Author: 陈远锋 |
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Description: MySQL驱动，做jsp课程设计要注意要把它放在正确的位置-MySQL driver, do jsp course design should pay attention to it in the correct location
Platform: |
Size: 465920 |
Author: yeah |
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Description: A simulated annealing heuristic for the capacitated location routing problem
Platform: |
Size: 240640 |
Author: farzad1 |
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Description: 使用说明
使用时打开此例题目录下pic中的图片，然后依次单击按钮“转”、“1”、“2”、“3”、“4”和“5”，就可以实现精确的车牌定位。
具体步骤
1.24位真彩色->256色灰度图。
2.预处理:中值滤波。
3.二值化:用一个初始阈值T对图像A进行二值化得到二值化图像B。
初始阈值T的确定方法是:选择阈值T Gmax-(Gmax-Gmin)/3,Gmax和Gmin分别是最高、最低灰度值。
该阈值对不同牌照有一定的适应性,能够保证背景基本被置为0,以突出牌照区域。
4.削弱背景干扰。对图像B做简单的相邻像素灰度值相减,得到新的图像G,即Gi,j |Pi,j-Pi,j-1|i 0,1,…,439 j 0,1,…,639Gi,0 Pi,0,左边缘直接赋值,不会影响整体效果。
5.用自定义模板进行中值滤波
区域灰度基本被赋值为0。考虑到文字是由许多短竖线组成,而背景噪声有一大部分是孤立噪声,用模板(1,1,1,1,1)T对G进行中值滤波,能够得到除掉了大部分干扰的图像C。
6.牌照搜索:利用水平投影法检测车牌水平位置,利用垂直投影法检测车牌垂直位置。
7.区域裁剪,截取车牌图像。-Instructions for use
When you use this example to open the directory under the picture pic, and then click the button turn , 1 , 2 , 3 , 4 and 5 , you can achieve accurate license plate positioning.
Specific steps
1.24-bit True Color-> 256-color grayscale.
2. Preprocessing: Median filtering.
3. Binarization: Image A is binarized with an initial threshold T to obtain a binarized image B.
The initial threshold T is determined by selecting the threshold T Gmax- (Gmax- Gmin)/3, Gmax and Gmin being the highest and lowest gray values, respectively.
The threshold has a certain adaptability to the different license plates, and can guarantee that the background is basically set to 0 to highlight the license area.
4. To reduce background interference. I 0, 1, ..., 439 j 0, i 0, 1, ..., i 0, i 0, i 0, 1, , 1, ..., 639Gi, 0 Pi, 0, the left edge of the direct assignment, does not affect the overall effect.
5. Use a custom template for
Platform: |
Size: 745472 |
Author: 薛颖杰 |
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