码迷,mamicode.com
首页 > 其他好文 > 详细

210 - Concurrency Simulator(WF1991, deque, 模拟)

时间:2014-11-03 22:14:33      阅读:558      评论:0      收藏:0      [点我收藏+]

标签:style   blog   http   io   color   ar   os   for   sp   

题目有点长,理解题花了不少时间

粘下别人的翻译~

你的任务是模拟n个程序(按输入顺序编号为1~n)的并行执行。每个程序包含不超过25条语句,格式一共有5种:

  var=constant(赋值);

  print  var(打印);

  lock;

  unlock;

  end。

变量用单个小写字母表示,初始值为0,为所有程序公有(因此在一个程序里对某个变量赋值可能会影响到另一个程序)。常数是小于100的非负整数。
每个时刻只能有一个程序处于运行态,其他程序均处于等待。上述五种语句分别需要t1、t2、t3、t4、t5单位时间。运行态的程序每次最多运行Q个单位时间(成为配额)。当一个程序的配额用完之后,把当前语句(如果存在)执行完之后该程序会被插入一个等待队列中,然后处理器从队首取出一个程序继续执行。初 始等待队列包含按输入顺序排列的各个程序,但由于lock和unlock语句的出现,这个序列可能会改变。
lock的作用是申请对所有变量的独占访问。lock和unlock总是成对出现,并且不会嵌套。lock总是在unlock的前面。当一个程序成功执行完lock指令后,其他程序一旦试图执行lock指令,就会马上被放到一个所谓的阻止队列的尾部(没有用完的配额就浪费了),当unlock指令执行完毕后,阻止队列的第一个程序进入等待队列的首部。
输入n、t1、t2、t3、t4、t5Q以及n个程序,按照时间顺序输出所有print语句的程序编号和结果。
 
附代码如下:
bubuko.com,布布扣
  1 #include<iostream>
  2 #include<cstdio>
  3 #include<cstdlib>
  4 #include<queue>
  5 #include<cstring>
  6 #include<deque>
  7 using namespace std;
  8 const int maxn = 1000;
  9 int n, t1, t2, t3, t4, t5, Q;
 10 char pro[maxn][10];
 11 int ID[maxn];
 12 int var[30];
 13 bool locked;
 14 deque<int>  ReadyQ;
 15 deque<int>  BlockQ;
 16 
 17 void Run(int id)
 18 {
 19     int q = Q;
 20     while(q > 0)
 21     {
 22         char *ins = pro[ID[id]];
 23         switch(ins[2])
 24         {
 25         case =://var = constant;
 26             {
 27                 /*通过数组var进行各变量值的记录*/
 28                 int buf = 0;
 29                 for(int i = 4; i < strlen(ins)-1; i++)
 30                     buf = buf*10+(ins[i]-0);
 31                 var[ins[0]-a] = buf;
 32                 q -= t1;
 33                 break;
 34             }
 35         case i://print var;
 36             {
 37                 printf("%d: %d\n", id+1, var[ins[6]-a]);
 38                 q -= t2;
 39                 break;
 40             }
 41         case c://lock
 42             {
 43                 /*Once a program successfully executes a lock statement, no other program may successfully execute a lock statement
 44                 until the locking program runs and executes the corresponding unlock statement.
 45                 Should a running program attempt to execute a lock while one is already in effect,
 46                 this program will be placed at the end of the blocked queue.*/
 47                 if(locked)
 48                 {
 49                     BlockQ.push_back(id);
 50                     return;
 51                 }
 52                 locked = true;
 53                 q -= t3;
 54                 break;
 55             }
 56         case l://unlock;
 57             {
 58                 locked = false;
 59                 /*When an unlock is executed, any program at the head of the blocked queue is moved to the head of the ready queue. */
 60                 if(!BlockQ.empty())
 61                 {
 62                     ReadyQ.push_front(BlockQ.front());
 63                     BlockQ.pop_front();
 64                 }
 65                 q -= t4;
 66                 break;
 67             }
 68         case d://end;
 69             {
 70                 q -= t5;
 71                 return;
 72                 break;
 73             }
 74         }//switch;
 75         ID[id]++;
 76     }//while;
 77     /*When a program time quantum expires, another ready program will be selected to run.
 78     Any instruction currently being executed when the time quantum expires will be allowed to complete. */
 79     ReadyQ.push_back(id);
 80 }//Run;
 81 
 82 
 83 int main()
 84 {
 85     int T;
 86     scanf("%d", &T);
 87     for(int cases = 0; cases < T; cases++)
 88     {
 89         memset(var, 0, sizeof(var));
 90         if(cases)   printf("\n");
 91         scanf("%d%d%d%d%d%d%d", &n, &t1, &t2, &t3, &t4, &t5, &Q);
 92         int line = 0;
 93         for(int i = 0; i < n; i++)
 94         {
 95             ///注意记录多行字符串方法
 96             ///================================================
 97             fgets(pro[line++], maxn, stdin);
 98             ID[i] = line-1; ///line可记录某ID开始到最后的操作;
 99                             /*identification number based upon its location in the input data.
100                              (the first program has ID = 1, the second has ID = 2, etc.)*/
101             while(pro[line-1][2] != d)
102                 fgets(pro[line++], maxn, stdin);
103             /*Programs are queued first-in-first-out for execution in a ready queue.
104             The initial order of the ready queue corresponds to the original order of the programs in the input file.*/
105             ///================================================
106             ReadyQ.push_back(i);
107         }
108         locked = false;
109         while(!ReadyQ.empty())
110         {
111             int Pro_id = ReadyQ.front();
112             ReadyQ.pop_front();
113             Run(Pro_id);
114         }
115     }
116     return 0;
117 }
View Code

 

210 - Concurrency Simulator(WF1991, deque, 模拟)

标签:style   blog   http   io   color   ar   os   for   sp   

原文地址:http://www.cnblogs.com/LLGemini/p/4072276.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!