标签：lis protoc one 顺序 line sep another fail app
Description
Programs executed concurrently on a uniprocessor system appear to be executed at the same time, but in reality the single CPU alternates between the programs, executing some number of instructions from each program before switching to the next. You are to simulate the concurrent execution of up to ten programs on such a system and determine the output that they will produce.
The program that is currently being executed is said to be running, while all programs awaiting execution are said to be ready. A program consists of a sequence of no more than 200 statements, one per line, followed by an end statement. The statements available are listed below.
| Statement type | Syntax |
|---|---|
| Assignment | ?variable? = ?constant? |
| Output | print ?variable? |
| Begin mutual exclusion | look |
| End mutual exclusion | unlock |
| Stop execution | end |
A ?variable? is any single lowercase alphabetic character and a ?constant? is an unsigned decimal number less than 1000. There are only 26 variables in the computer system, and they are shared among the programs. Thus assignments to a variable in one program affect the value that might be printed by a different program. All variables are initially set to zero.
Each statement requires an integral number of time units to execute. The running program is permitted to continue executing instructions for a period of time called its quantum. When a program’s time quantum expires, another ready program will be selected to run. Any instruction currently being executed when the time quantum expires will be allowed to complete.
Programs are queued first-in-first-out for execution in a ready queue. The initial order of the ready queue corresponds to the original order of the programs in the input file. This order can change, however, as a result of the execution of lock and unlock statements.
The lock and unlock statements are used whenever a program wishes to claim mutually exclusive access to the variables it is manipulating. These 3 statements always occur in pairs, bracketing one ormore other statements. A lock will always precede an unlock, and these statements will never be nested. Once a program successfully executes a lock statement, no other program may successfully execute a lock statement until the locking program runs and executes the corresponding unlock statement. Should a running program attempt to execute a lock while one is already in effect, this program will be placed at the end of the blocked queue. Programs blocked in this fashion lose any of their current time quantum remaining. When an unlock is executed, any program at the head of the blocked queue is moved to the head of the ready queue. The first statement this program will execute when it runs will be the lock statement that previously failed. Note that it is up to the programs involved to enforce themutual exclusion protocol through correct usage of lock and unlock statements. (A renegade program with no lock/unlock pair could alter any variables it wished, despite the proper use of lock/unlock by the other programs.)
Input
The first line of the input file consists of seven integers separated by spaces. These integers specify (in order): the number of programs which follow, the unit execution times for each of the five statements (in the order given above), and the number of time units comprising the time quantum. The remainder of the input consists of the programs, which are correctly formed from statements according to the rules described above.
All program statements begin in the first column of a line. Blanks appearing in a statement should be ignored. Associated with each program is an identification number based upon its location in the input data (the first program has ID=1, the second has ID=2, etc.).
Output
Your output will contain the output generated by the print statements as they occur during the simulation. When a print statement is executed, your program should display the program ID, a colon, a space, and the value of the selected variable. Output from separate print statements should appear on separate lines. A sample input and correct output is shown below.
Sample Input
1 3 1 1 1 1 1 1 a = 4 print a lock b = 9 print b unlock print b end a = 3 print a lock b = 8 print b unlock print b end b = 5 a=17 print a print b lock b = 21 print b unlock print b end
Sample Output
1: 3 2: 3 3: 17 3: 9 1: 9 1: 9 2: 8 2: 8 3: 21 3: 21
本题目的题意是：一共有n个程序，每个程序的运行时间是t，程序的操作共有5种，时间分别是t1,t2,t3,t4,t5,执行完的程序会插入到等待队列中，初始队列按照程序的顺序出现，但是Lock和unlock的出现会打乱这个顺序
lock的作用是对所有变量的独占访问，当一个程序执行完lock后，若其他程序访问，则会被放到阻止队列的队尾，当unlock执行时，阻止队列的一个程序会放在执行队列的一个首部
分析：建立两个双向队列（具体操作参照STL文档），ready和block，如果遇到lock执行block.push_back(i)遇到unlock执行ready.push_front(block.front()),block.pop_front()
双向队列：#include<deque>
deque<type> name;
代码是参照某位大神的代码打的，开始实在没有看懂题目/笑哭
1 #include<iostream>//里面包含string及其函数 2 #include<deque> 3 #include<cstring> 4 using namespace std; 5 6 deque<int> ready,block; 7 int var[26],id[1024],st[5],program,q,t,i;//关于这一段请看下面 8 string code[1024];//存储指令内容 9 bool locked; 10 11 void run(int i){ 12 int time=q; 13 while (time>0){ 14 string now=code[id[i]]; 15 if(now[2] == ‘=‘){ 16 var[now[0] - ‘a‘] = isdigit(now[5]) ? (now[4]-‘0‘)*10+(now[5]-‘0‘) : now[4]-‘0‘; //isdigit检查是否为数字，var依次存储a,b,c等变量当前值 17 time -= st[0]; 18 } 19 else if(now[2] == ‘i‘){ 20 cout<<i+1<<": "<<var[now[6]-‘a‘]<<endl; 21 time -= st[1]; 22 } 23 else if(now[2] == ‘c‘){ 24 if(locked){block.push_back(i);return;} 25 locked = true; 26 time -= st[2]; 27 } 28 else if(now[2] == ‘l‘){ 29 if(!block.empty()) ready.push_front(block.front()),block.pop_front(); 30 time -= st[3]; 31 locked = false; 32 } 33 else if(now[2] == ‘d‘){return ;} 34 id[i]++; 35 } 36 ready.push_back(i); 37 } 38 39 int main(){ 40 cin>>t; 41 while (t--){ 42 cin>>program>>st[0]>>st[1]>>st[2]>>st[3]>>st[4]>>q; 43 memset(var,0,sizeof(var)); 44 ready.clear();block.clear(); 45 int line_num = 0; 46 for (i=0;i<program;i++){ 47 ready.push_back(i); 48 getline(cin,code[line_num++]); 49 id[i]=line_num-1; 50 while (code[line_num-1]!="end"){ 51 getline(cin,code[line_num++]); 52 } 53 } 54 locked=false; 55 while (!ready.empty()){ 56 int now=ready.front();ready.pop_front(); 57 run(now); 58 } 59 if(t) cout<<endl; 60 } 61 return 0; 62 }
标签：lis protoc one 顺序 line sep another fail app
原文地址：http://www.cnblogs.com/muziqiu/p/7231359.html
