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Maze. With an m* n matrix, said the long labyrinth of 0 and 1, respectively, the...
迷宫问题。以一个 m*n的长方阵表示迷宫,0和1分别表示迷宫中的通路和障碍。设计一个程序,对任意设定的迷宫,求出一条从入口到出口的通路,或得出没有通路的结论
起点就是方阵中的(1,1)终点就是(m-1,n-1) -Maze. With an m* n matrix, said the long labyrinth of 0 and 1, respectively, the maze of paths and obstacles. Designing a program, for any set of the maze, find a path from the entrance to the export, or come to the conclusion there is no access point is a square matrix of the (1,1) The end is (m-1, n-1)
- 2022-04-15 13:14:53下载
- 积分:1
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One uses tcp communication. 2 Des realize freedom to choose the main library. ba...
1采用了tcp通信。
2实现自由选择主辅库。
backup:数据库监视备份软件(backup.exe,backup.ini,backup.log,sendclient.log)-One uses tcp communication. 2 Des realize freedom to choose the main library. backup: backup software, database monitoring (backup.exe, backup.ini, backup.log, sendclient.log)
- 2022-03-25 02:10:16下载
- 积分:1
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vfp enterprise produces
vfp企业生产-vfp enterprise produces
- 2023-04-10 19:55:03下载
- 积分:1
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方括号做的程序,用户可以输入括号,以及是否可以使用。
用C做的括号匹配程序,可以检测用户输入的括号是否匹配并给出提示-C matching brackets do procedures, the user can input the brackets and whether the match is suggested
- 2023-01-21 12:30:03下载
- 积分:1
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本人上传的是用贪心算法实现找零钱问题。贪婪算法的根本思想是逐步获得最优解。...
本人上传的是用贪心算法实现找零钱问题。贪婪算法的根本思想是逐步获得最优解。-I uploaded is to use greedy algorithm to find small change problem. Greedy algorithm is the fundamental idea is to gradually get the optimal solution.
- 2022-03-02 15:39:29下载
- 积分:1
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这是一个产品管理系统的程序。用java编写的希望的热情…
这是一个产品管理系统程序。用java 编写
希望各位踊跃分享-This is a product management system procedures. Prepared using java hope enthusiastically sharing
- 2022-01-28 04:18:45下载
- 积分:1
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A*算法的演示程序,可以自由设置障碍和地形,单步演示A*算法的寻路过程。包含源码,是学习A*算法的最佳工具。...
A*算法的演示程序,可以自由设置障碍和地形,单步演示A*算法的寻路过程。包含源码,是学习A*算法的最佳工具。-A* algorithm demo program, they can set up obstacles and terrain, single-step demonstration of the A* search algorithm road course. Includes source is studying A* algorithm for the best tools.
- 2023-04-24 02:20:03下载
- 积分:1
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求解约瑟夫问题。设有n个人围成一个圆圈坐下,对所有围从的人从某个位置开始编号为1,2,3,……,n,从编号为1的人开始报数1,报数依交进行,报数n的人即出列,下...
求解约瑟夫问题。设有n个人围成一个圆圈坐下,对所有围从的人从某个位置开始编号为1,2,3,……,n,从编号为1的人开始报数1,报数依交进行,报数n的人即出列,下一个人从1开始报数,再报数m的人便是第二个出列的人如此重复下去,直到最后一个人出列为止,于是便得到一个出列的顺序,这称之为约瑟夫(Josephu)问题。-solving problems. N individuals have formed a circle to sit down, all right Wai from the start from a position of No. 1, 2, 3, ..., n, for a number of people reported a few, according to several newspaper pay, the newspaper n ie the out, the next person was from the beginning a few , and reported m is the second out of this series continue to repeat until the final out in a series, then get a series out of order, which called Joseph (Josephu) problem.
- 2022-05-21 23:13:13下载
- 积分:1
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数据结构中的求强连通分量的源程序,用vc6.0来编写
数据结构中的求强连通分量的源程序,用vc6.0来编写-data structure for the strongly connected component of the source used to prepare vc6.0
- 2023-05-10 23:05:03下载
- 积分:1
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十天学会ASP,很好~
十天学会ASP,很好~- Ten days learn ASP, very good ~
- 2022-10-04 08:30:03下载
- 积分:1