最短作業(yè)優(yōu)先調(diào)度算法(SJF算法)的C++實(shí)現(xiàn)

1、在作業(yè)調(diào)度中，該算法每次從后備作業(yè)隊(duì)列中挑選估計(jì)服務(wù)時(shí)間最短的一個(gè)或幾個(gè)作業(yè)，將他們調(diào)入內(nèi)存，分配必要的資源，創(chuàng)建進(jìn)程并放入就緒隊(duì)列。與在進(jìn)程調(diào)度中的原理類(lèi)似。假設(shè)有n項(xiàng)作業(yè)位于就緒隊(duì)列中，這些作業(yè)的請(qǐng)求時(shí)間用數(shù)組requestTimes按照提交時(shí)間的先后順序存儲(chǔ)，對(duì)應(yīng)的作業(yè)服務(wù)時(shí)間（也稱持續(xù)時(shí)間）用數(shù)組durations存儲(chǔ)。當(dāng)每個(gè)作業(yè)被完成后，接下來(lái)優(yōu)先選擇服務(wù)時(shí)間（持續(xù)時(shí)間）短的，如果多個(gè)服務(wù)時(shí)間一樣，則優(yōu)先選擇請(qǐng)求時(shí)間最先的。采用SJF算法，計(jì)算n項(xiàng)作業(yè)的平均等待時(shí)間。所有要執(zhí)行的任務(wù)的請(qǐng)求時(shí)間必須在上一個(gè)任務(wù)完成的時(shí)間內(nèi)。假設(shè)0= n = 100。測(cè)試用例：requestTimes

2、= 0, 2, 4, 5;durations = 7, 4, 1, 4;可以理解為一定首先執(zhí)行第一個(gè)任務(wù)，請(qǐng)求時(shí)間為0，執(zhí)行了7個(gè)時(shí)間單位；接下來(lái)在服務(wù)時(shí)間最短的中選擇，顯然是任務(wù)3，請(qǐng)求時(shí)間為4，執(zhí)行1個(gè)時(shí)間單位；再接下來(lái)發(fā)現(xiàn)2和4這兩個(gè)任務(wù)服務(wù)時(shí)間一樣短，則優(yōu)先選擇請(qǐng)求時(shí)間最先的，顯然是選擇任務(wù)2，執(zhí)行4個(gè)時(shí)間單位最后是任務(wù)4（并且只有任務(wù)4了），請(qǐng)求時(shí)間為5，執(zhí)行4個(gè)時(shí)間單位等待時(shí)間：任務(wù)1是0，任務(wù)3是3，任務(wù)2是5，任務(wù)4是7，所以平均等待時(shí)間是4。int findmin_duration(vector &a,int *b,int end_time, int n)int min =

3、INT_MAX;int pos = 0;for (int i = 1; i n; i+)if (ai min & bi = end_time)min = ai;pos = i;apos = INT_MAX;return pos;float average_wait_time(int *requestTimes, int *durations, int n)vector durate(n,0);vector startTime(n, 0);vector endTime(n, 0);vector waitTime(n, 0);int i;for (i = 0; i n; i+)duratei =

4、durationsi;startTime0 = requestTimes0;endTime0 = durations0 + startTime0;waitTime0 = startTime0 - requestTimes0;int minIndex = 0;int lastIndex = minIndex;for (i = 1; i n; i+)minIndex = findmin_duration(durate, requestTimes, endTimelastIndex, n);startTimeminIndex = endTimelastIndex;endTimeminIndex =

5、durationsminIndex + startTimeminIndex;waitTimeminIndex = startTimeminIndex - requestTimesminIndex;lastIndex = minIndex;int s = 0;for (i = 0; i n; i+)s += waitTimei;float avageTime = (float)s / (float)n;return avageTime;int main(int argc, char* argv)int requestTimes4 = 0, 1, 3, 9 ;int durations4 = 2, 1, 7, 5 ;/int requestTimes = 0, 2, 4, 5 ;/int durations = 7, 4,