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 25 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 lock
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 100. There are only 26 variables in the computer system, and thay 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 integeral 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 expired 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 wished to claim mutually exclusive access to the variables it is manipulating. These statements always occur in pairs, bracketing one or more other statements. A lock will always precede an unlock, and these statements will never be nested. One a program successfully execute 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 the mutual exclusion protocol through correct usage of lock and unlock statements. (A renegade program with no lock/unlock pair could alter any variable it wished, despite the proper use of lock/unlock by the other programs)

Input

The input begins with a single positive integer on a line by itself indicating the number of the case following by a blank line, and there is also a blank line between two consecutive inputs.

The first line of the input file consists of seven integers separated by spaces. These integers specify (in order): the number of program which follow, the unit execution time 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 colum of a line.
Blanks appearing in a statement should be ignored.Associated which 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

For each test case, the output must follow the description below. The outputs of two consecutive cases will be seperated by a blank line.

Your output will contain of 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 seperate print statement should appear on seperate lines.

A sample input and correct output are showed below.


Sample Input
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
题意: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。
#include<stdio.h>
#include<algorithm>
#include<iostream>
#include<string>
#include<string.h>
#include<deque>
#include<vector>
using namespace std;
deque<int>qr;
deque<int>qw;
vector<string>v[1005];
int var[26], p[1005], t[1005];
bool lock;
int Q,n,cas;
void run(int i)
{
int up = Q, w;
string cur;
while (up > 0)
{
cur = v[i][p[i]];//读取字符串
if (cur[2] == '=')
{
up -= t[0];
w = cur[4] - '0';
if (cur.size() == 6)
w = w * 10 + cur[5] - '0';
var[cur[0] - 'a'] = w;
}
else if (cur[2] == 'i')
{
up -= t[1];
printf("%d: %d\n", i, var[cur[6] - 'a']);
}
else if (cur[2] == 'c')
{
up -= t[2];
if (lock)
{
qw.push_back(i);
return;
}
lock = true;
}
else if (cur[2] == 'l')
{
lock = false;
up -= t[3];
if (!qw.empty())
{
w = qw.front();
qw.pop_front();
qr.push_front(w);
}
}
else return;
++p[i];
}
qr.push_back(i);//时间配额用完了就放到队尾
}
int main()
{
while (cin >> cas){
while (cas--)
{
cin >> n;
for (int i = 0; i < 5; i++)
cin >> t[i];
cin >> Q;
string s;
for (int i = 1; i <= n; i++)
{
v[i].clear();
while (getline(cin, s))
{
if (s == "")
continue;
v[i].push_back(s);
if (v[i].back() == "end")
break;
}
qr.push_back(i);
}
memset(p, 0, sizeof(p));
memset(var, 0, sizeof(var));
while (!qr.empty())
{
int cur = qr.front();
qr.pop_front();
run(cur);
}
if (cas)printf("\n");
}
}
return 0;
}

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