实验要求
要求模拟先来先服务法(First-Come, First-Served,FCFS),最短寻道时间优先法(Shortest Seek Time First, SSTF),电梯法三种磁盘调度算法,输入为一组请求访问磁道序列,输出为每种调度算法的磁头移动轨迹和移动的总磁道数。
代码
#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
int ans, length; // ans是移动的总磁道数,lenght是磁道序列长度
void FCFS(int start, int p[]); // 先来先服务法
void SSTF(int start, int p[]); // 最短寻道时间优先法
void EM(int start, int p[]); // 电梯法
int main() {
int p[] = { 98,183,37,122,14,124,65,67 }; // 磁道序列
int start = 53; // 磁头初始位置
length = sizeof(p) / sizeof(p[0]);
FCFS(start, p);
SSTF(start, p);
EM(start, p);
return 0;
}
// 先来先服务法,时间复杂度O(n)
void FCFS(int start, int p[]) {
cout << "磁头移动轨迹(先来先服务法):";
ans = 0;
int tmp = start;
for (int i = 0; i < length; i++) {
cout << p[i] << " ";
ans += abs(p[i] - tmp);
tmp = p[i];
}
cout << endl << "移动的总磁道数:" << ans << endl;
}
// 最短寻道时间优先法,时间复杂度O(nlogn)
void SSTF(int start, int p[]) {
cout << endl << "磁头移动轨迹(最短寻道时间优先法):";
sort(p, p + length);
ans = 0;
int left, right, tmp = start;
for (int i = 0; i < length; i++) {
if (start < p[i]) {
left = i - 1;
right = i;
break;
}
}
while (!(left < 0 && right > length - 1)) {
if (abs(p[left] - tmp) < abs(p[right] - tmp) && left >= 0) {
cout << p[left] << " ";
ans += abs(p[left] - tmp);
tmp = p[left];
left--;
}
else {
cout << p[right] << " ";
ans += abs(p[right] - tmp);
tmp = p[right];
right++;
}
}
cout << endl << "移动的总磁道数:" << ans << endl;
}
// 电梯法,时间复杂度O(nlogn)
void EM(int start, int p[]) {
cout << endl << "磁头移动轨迹(电梯法):";
sort(p, p + length);
ans = 0;
int loc, tmp = start;
for (int i = 0; i < length; i++) {
if (start < p[i]) {
loc = i;
break;
}
}
if (start < (p[length - 1] - p[0]) / 2) {
for (int i = loc - 1; i >= 0; i--) { // 向左走
cout << p[i] << " ";
ans += abs(p[i] - tmp);
tmp = p[i];
}
for (int i = loc; i < length; i++) { // 向右走
cout << p[i] << " ";
ans += abs(p[i] - tmp);
tmp = p[i];
}
}
else {
for (int i = loc; i < length; i++) { // 向右走
cout << p[i] << " ";
ans += abs(p[i] - tmp);
tmp = p[i];
}
for (int i = loc - 1; i >= 0; i--) { // 向左走
cout << p[i] << " ";
ans += abs(p[i] - tmp);
tmp = p[i];
}
}
cout << endl << "移动的总磁道数:" << ans << endl;
}