UVA 210 并行程序模拟

这道题模拟程序执行过程，就绪队列使用STL的双端队列deque，阻止队列使用普通队列。在遇到unlock时，把阻止队列的头节点放入就绪队列的队首。

代码如下：

#include<cstdio>#include<iostream>#include<algorithm>#include<cstring>#include<string>#include<deque>#include<queue>using namespace std;typedef struct{    int ID;    char cmd[30][100];    int next;}Process;Process ps[1005];int var[26];int n;//n programint timeuse[5];// 5 kind statement timeuse costint period;bool locked;queue<int> blockq;deque<int> readyq;void run(int p){    int tleft = period;    while (tleft>0){        int n = ps[p].next;        char* str = ps[p].cmd[n];        if (str[2] == '='){//var = constant            int s = 0;            for (int i = 4; str[i] != '\0'; i++){                s = s * 10 + (str[i] - '0');            }            var[str[0] - 'a'] = s;            tleft -= timeuse[0];        }        else if (str[2] == 'i'){//print            int d = str[6] - 'a';            cout << p << ": " << var[d] << endl;            tleft -= timeuse[1];        }        else if (str[2] == 'c'){//lock            if (locked){                blockq.push(p);                return;            }            else{                locked = true;                tleft -= timeuse[2];            }        }        else if (str[2] == 'l'){//unlock            locked = false;            if (!blockq.empty()){                int p = blockq.front();                blockq.pop();                readyq.push_front(p);            }            tleft -= timeuse[3];        }        else//end            return;        ps[p].next++;    }    readyq.push_back(p);    return;}int main(){    int t;    cin >> t;    while (t--){        cin >> n;        for (int i = 0; i < 5; i++)            cin >> timeuse[i];        cin >> period;        getchar();        locked = false;        memset(var, 0, sizeof(var));        readyq.clear();        for (int i = 1; i <= n; i++){            int j = 0;            while (true){                cin.getline(ps[i].cmd[j], 100);                if (!strcmp(ps[i].cmd[j], "end"))                    break;                j++;            }            ps[i].ID = i;            ps[i].next = 0;            readyq.push_back(i);        }        while (!readyq.empty()){            int cur = readyq.front();            readyq.pop_front();            run(cur);        }        if (t)            cout << endl;    }    return 0;}