并行程序模拟（Concurrency Simulator, UVA 210）

2019-04-14 19:20发布

站内文章 / 模拟电子
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题目链接：https://vjudge.net/problem/UVA-210    这一题给我的感觉是，难度并不是很大，但难点在于理解题目，读懂题目之后将题目里描述的规则用代码写出来就行了。    核心有两点：一是维护ready queue和block queue；二是根据题目对模拟规则的描述进行各个statement执行的实现。 在ac之后，我看了一个lrj书中提供的代码，整体方法是一致的，但是他的代码非常短，只有77行，而我的代码达到了100多行，这说明我的钥匙思维还不够凝练。仔细分析一下代码的差异，主要在输入数据的存储上和对时间片部分的维护上，我的重复代码相当的多。lrj的解法提供了解决这种有时间限制问题的一种解法，以后还可以借鉴。 附上我的代码。 #include #include #include #include #include #include #include using namespace std; const int max_statements=25; const int max_programs = 50; int program_num = 0; map quantum; class Program { public: vector statements; int id; int execute_line; }; Program read_Q[max_programs]; int Qfront = 0; int Qend = 0; int sizeQ = 0; void push_front(Program P) { Qfront = (Qfront - 1 + max_programs) % max_programs; read_Q[Qfront] = P; sizeQ++; } void push_end(Program P) { read_Q[Qend] = P; Qend = (Qend + 1) % max_programs; sizeQ++; } Program pop() { Program p = read_Q[Qfront]; Qfront = (Qfront + 1) % max_programs; sizeQ--; return p; } queue block_Q; void clear() { Qfront = 0; Qend = 0; quantum.clear(); } int main() { //freopen("input.txt","r",stdin); //freopen("output.txt", "w", stdout); int case_num = 0; cin >> case_num; int pt = case_num; while (case_num--) { getchar(); clear(); int table[26] = {0}; cin >> program_num; int qt = 0; int quantumT = 0; cin >> qt; quantum['='] = qt; cin >> qt; quantum['i'] = qt; cin >> qt; quantum['c'] = qt; cin >> qt; quantum['l'] = qt; cin >> qt; quantum['d'] = qt; cin >> quantumT; for (int i = 0; i < program_num; i++) { Program program; program.id = i+1; program.execute_line = 0; string s; while (getline(cin, s)) { if (s == "")continue; if (s != "end") { program.statements.push_back(s); } else { program.statements.push_back(s); break; } } push_end(program); } if (case_num < (pt - 1))cout << endl; //simulate the program execute int lock = 0; while (sizeQ > 0) { Program p = pop(); int time = 0; for (int i = p.execute_line; i < p.statements.size(); i++) { //judge the current statement is lock or unlock if (p.statements[i] == "lock") { //if current statement is lock , it need to judge whether the state is lock. if (lock == 0) { lock = p.id; time += quantum[p.statements[i][2]]; p.execute_line++; if (time >= quantumT) { push_end(p); break; } } else { block_Q.push(p);//the quantum remained is given up. p.execute_line++; break; } } else if (p.statements[i] == "unlock") { lock = 0; if (block_Q.size() > 0) { push_front(block_Q.front()); block_Q.pop(); } time += quantum[p.statements[i][2]]; p.execute_line++; if (time >= quantumT) { push_end(p); break; } } else if (p.statements[i][2] == '=') { stringstream stream(p.statements[i].substr(4)); int num; stream >> num; table[p.statements[i][0] - 'a'] = num; time += quantum[p.statements[i][2]]; p.execute_line++; if (time >= quantumT) { push_end(p); break; } } else if (p.statements[i][2] == 'i') { cout << p.id << ": " << table[p.statements[i][6] - 'a'] << endl;; time += quantum[p.statements[i][2]]; p.execute_line++; if (time >= quantumT) { push_end(p); break; } } else if(p.statements[i][2] == 'd') { break; } } } } return 0; }  

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