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 css注释的影响
ASP.NET与数据库相关技巧
关于HtmlForm控件
三色交替的下拉列表框
精通ASP.NET中弹出窗口技术
ASP.NET Forums与现有系统整合方案示例
ASP.NET操作IIS中的虚拟目录
DataGrid与SQL Server 2000数据绑定
如何让Web应用程序在Client端实现导出报表功能
如何保证web app中的Send Email线程稳定性
关于用ASP.Net识别远程主机服务器种类
ASP.NET中上传下载文件
提高ASP.NET性能的方法
asp.net StreamReader 创建文件
asp.net如何生成图片验证码(简单)
一个.net 压缩位图至JPEG的代码
简单的SQL Server数据库数据读取与数据操作
获取网站的RSS聚合到自己的网页
.Net程序中整站通用的防SQL注入函数
asp.net生成缩略图及给原始图加水印的函数

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: 95 ::
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");: : }