一个简单的JAVA小旅馆的程序，功能简单，课程设计的作业

单循环赛中选手胜负序列求解问题 名称：单循环赛中选手胜负序列求解问题 内容：有n个选手 P 1 ,P 2 ,P 3 ,… ,P n 参加了的单循环赛，每对选手之间非胜即负。现要求求出一个选手序列 P 1 ,P 2 ,P 3 ,… ,P n , 使其满足 P i 胜 P i+ 1 (i=1,… ,n-1) 。 -Single round robin tournament players winning sequence to solve the problem name: the outcome of single round-robin tournament players to solve the problem sequence elements: There are n-player P 1, P 2, P 3, ..., P n participated in a single round-robin between each pair of players Non-win or negative. Players are now required to find a sequence of P 1 " , P 2" , P 3 " , ..., P n" , to meet the P i " W P i+ 1" (i = 1, ..., n-1).

此程序是485通信控制程序，在实物中以经调试通过．各项性能指标均合格．-This program is 485 communication control procedures, in-kind in order to debug passed by. The performance indicators are satisfactory.

将用户指定的符合iso-9660标准的文件的信息dump出来。-Will be user-specified in line with the iso-9660 standard document dump out the information.

VC++关于MFC-VC on the MFC

在非汉字系统下显示汉字-system in Chinese characters displaying Chinese characters

将格式为SEG-Y的地震数据的某道信息截出来,程序为c++程序,经过编译了-format will SEG-Y seismic data information of a certain cut, and procedures for c procedures are followed, the compiler

this program is for controling MAC_messages in the simulator, through we can control MAC msges

基于vc下的中国象棋代码，我是学人工只能的-vc on the Chinese chess code, I study the manual only

[输入] 图的顶点个数N，图中顶点之间的关系及起点A和终点B [输出] 若A到B无路径，则输出“There is no path” 否则输出A到B路径上个顶点 [存储结构] 图采用邻接矩阵的方式存储。 [算法的基本思想] 采用广度优先搜索的方法，从顶点A开始，依次访问与A邻接的顶点VA1,VA2,...,VAK, 访问遍之后,若没有访问B，则继续访问与VA1邻接的顶点VA11,VA12,...,VA1M，再访问与VA2邻接顶点...，如此下去，直至找到B，最先到达B点的路径，一定是边数最少的路径。实现时采用队列记录被访问过的顶点。每次访问与队头顶点相邻接的顶点，然后将队头顶点从队列中删去。若队空，则说明到不存在通路。在访问顶点过程中，每次把当前顶点的序号作为与其邻接的未访问的顶点的前驱顶点记录下来，以便输出时回溯。 #include int number //队列类型 typedef struct{ int q[20] -[imported] map of the number of vertices N, Vertex map of the relationship between the starting point and end point A and B [output] if A to B without path then exporting "There is no path" Otherwise output A to B on the path vertices [storage structure] plan adopted adjacency matrix of storage. [Algorithm for the basic idea] BFS way from the vertex A, A visit followed with adjacent vertices VA1, VA2 ,..., VAK, visiting times, if not visit B, continue to visit with the VA1 adjacent vertices VA11, VA12 ,..., VA1M. VA2 and then

一个应邀而写的玩的游戏 ，一个摸奖游戏，其中的对一个timer等控件有一个好的应用，应用的算法值得参考。-an invitation to write the play of the game, a game Our troops, the right controls such as a timer is a good application, the application of algorithms is worth noting.