当前位置: 首页 > 图文教程 > 网络编程 > ASP.NET > 算法讨论:哲学家就餐问题

ASP.NET
asp.net(c#)网页跳转七种方法小结
完美解决在ModalPopupExtender中使用CalendarExtender时被层遮挡的问题
ASP.NET、SharePoint中另存文件的长文件名被截断的原因及解决办法
查看Json输出的*最方便*的方法 (转)
asp.net 代码隐藏的编码模型
ajaxpro.dll 控件实现异步刷新页面
asp.net DbProviderFactory的使用-示例
一个简单的asp.net 单点登录实现
jQuery+Ajax用户登录功能的实现
asp.net 弹出对话框返回多个值
.NET 中英文混合验证码实现代码
一个完整的ASP.NET 2.0 URL重写方案[翻译]
asp.net关于onpropertychange和oninput事件实现代码
asp.net gridview指定某一列滚动
Equals和==的区别 公共变量和属性的区别小结
asp.net 合并GridView中某列相同信息的行(单元格)
ASP.NET(C#) 定时执行一段代码
asp.net 预防SQL注入攻击之我见
asp.net下将Excel转成XML档的实现代码
asp.net url分页类代码

ASP.NET 中的 算法讨论:哲学家就餐问题


出处:互联网   整理: 软晨网(RuanChen.com)   发布: 2009-11-03   浏览: 44 ::
收藏到网摘: n/a

In 1965, Dijkstra posed and solved a synchronization problem he called thedining philosophers problem. ........ The problem can be stated quite simplyas follows. Five philosophers are seated around a circular table. Each philosopher has a plate of spaghetti. The spaghetti is so slippery that a philosopher needs two forks to eat it. Between each pair of plates is one fork. The life of a philosopher consists of alternate periods of eating and thinking. When a philosopher gets hungry, she tries to acquire her left and rightfork, one at a time, in either order. If successful in acquiring two forks,she eats for a while, then puts down the forks and continues to think. Thekey question is: Can you write a program for each philosopher that does what it is supposed to do and never gets stuck? --from written by Andrew S. Tanenbaum typed by foolball :-PProgramme provided by ya: : 法一: 用公共文件,按照严格轮流执行: : #include : : #include : : #include : : #include : : #define N 5: : int i,j,t,status;: : FILE * f;: : char *state[N];: : main(): : {: : f=fopen("/share","w+");: : putc(j,f);: : if (fork()): : { if (fork()): : { if (fork()): : { if (fork()): : { if (fork()): : {waitpid(-1,*status,0);: : fclose(f);}: : else : : philosophy(4);}: : else: : philosophy(3);}: : else: : philosophy(2);}: : else: : philosophy(1);}: : else: : philosophy(0);: : }: : void philosophy(int i): : {: : state[i ="thinking";: : printf("%d%s\n",i,"is thinking");: : for(t=0;t<=rand()+10000;t++);: : state[i = : : printf("%d%s\n",i,"is hungry");: : for(t=0;t<=rand()+10000;t++);: : for(;i!=j;): : {fseek(f,0l,0);: : j=getc(f);} : : state[i ="eating";: : printf("%d%s\n",i,"is eating");: : j=(j+1)% N: : fseek(f,0l,0);: : putc(j,f);: : }: : 法二:通过文件加锁实现: : #include : : #include : : #define N 4: : FILE *f;: : int i, status;: : char *state[N];: : void philosofy(int i);: : void main(): : {: : if ((f=fopen("turn", "w+"))==NULL): : {: : printf("Cann't open this file"); : : exit(0);: : }: : if (fork()): : {: : if(fork()): : {: : if(fork()): : {: : if(fork()): : {: : if(fork()): : {: : waitpid(-1, &status, 0);: : fclose(f);: : }: : else: : philosofy(4);: : }: : else: : philosofy(3);: : }: : else : : philosofy(2);: : }: : else: : philosofy(1);: : }: : else: : philosofy(0);: : }//end of main: : void philosophy(int i): : { int t;: : state[i ="thinking";: : printf("%d%s\n",i,"is thinking");: : for(t=0;t<=rand()+10000;t++);: : state[i ="hungry";: : printf("%d%s\n",i,"is hungry");: : for(t=0;t<=rand()+10000;t++);: : while ((f=fopen("turn.lock","r"))!=NULL);: : link ("turn","turn.lock");: : state[i ="eating"; : : printf("%d%s\n",i," is eating");: : for(t=1; t<=10000+rand();t++) ;: : unlink ("turn");: : }