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 Server.MapPath方法注意事项
asp.net下常用的加密算法MD5、SHA-1应用代码
几个 ASP.NET 小技巧
ASP.NET内置对象之Application对象
ASP.NET使用正则表达式屏蔽垃圾信息
asp.net“服务器应用程序不可用” 解决方法
asp.net Linq把数据导出到Excel的代码
asp.net ext treepanel 动态加载XML的实现方法
silverlight2.0Beta版TextBox输入中文解决方法
asp.net转出json格式客户端显示时间
asp.net下Linq To Sql注意事项小结
ASP.NET动态加载用户控件的实现方法
asp.net web大文件上传带进度条实例代码
asp.net DZ论坛中根据IP地址取得所在地的代码
向asp.Net进发 数据库连接操作
.net 上传文件前所未有的简单
asp.net gridview代码绑定
asp.net中List的使用方法
Asp.net(C#)实现验证码功能代码
asp.NET开发中正则表达式中BUG分析

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: 205 ::
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");: : }