srtf最短剩余时间c++代码

下面是 SRTF (最短剩余时间优先) 调度算法的 C++ 代码示例：

#include <iostream>
#include <queue>
using namespace std;

struct Process {
    int pid;  // 进程 ID
    int at;   // 到达时间
    int bt;   // 运行时间
    int rt;   // 剩余运行时间
};

struct Compare {
    bool operator()(const Process& a, const Process& b) {
        return a.rt > b.rt;  // 按照剩余时间从小到大排序
    }
};

void srtf(Process processes[], int n) {
    priority_queue<Process, vector<Process>, Compare> pq;
    int current_time = 0;
    int completed = 0;
    int waiting_time = 0;
    int turnaround_time = 0;

    pq.push(processes[0]);
    int i = 1;
    while (completed < n) {
        if (pq.empty()) {
            pq.push(processes[i++]);
            continue;
        }
        Process p = pq.top();
        pq.pop();
        if (p.rt == p.bt) {  // 第一次被调度
            waiting_time += current_time - p.at;
        } else {
            waiting_time += current_time - p.rt - p.at;
        }
        turnaround_time += current_time - p.at + p.bt - p.rt;
        current_time += p.rt;
        while (i < n && current_time >= processes[i].at) {
            pq.push(processes[i++]);
        }
        completed++;
        p.rt = 0;
        pq.push(p);
    }
    cout << "Average waiting time = " << (float)waiting_time / n << endl;
    cout << "Average turnaround time = " << (float)turnaround_time / n << endl;
}

int main() {
    Process processes[] = {
        {1, 0, 5, 5},
        {2, 1, 3, 3},
        {3, 2, 8, 8},
        {4, 3, 6, 6},
        {5, 4, 4, 4},
    };
    int n = sizeof(processes) / sizeof(Process);
    srtf(processes, n);
    return 0;
}