c - C中的非抢占式优先级调度
问题描述
算法解释：
非抢占式优先级调度 每个进程都有（到达时间、优先级和突发（执行）时间），首先到达时间（到达时间较少的进程）的进程会先被执行，如果两个进程到达时间相同，则比较优先级（最高进程优先）。此外，如果两个进程具有相同的优先级，则与进程号进行比较（首先是较少的进程号）。在执行所有过程时重复此过程。
我使用了下面的代码，但没有得到正确的答案。我已经尝试解决了 2 周，但不幸的是我不知道错误在哪里（这是一个逻辑错误，但我无法识别它）。我试图调试它很多次，但我仍然找不到导致它的原因。
谢谢。
#include <stdio.h>
void main()
{
int pn = 0; //Processes Number
int CPU = 0; //CPU Current time
int allTime = 0; // Time neded to finish all processes
printf("Enrer Processes Count: ");
scanf("%d",&pn);
int AT[pn];
int ATt[pn];
int NoP = pn;
int PT[pn]; //Processes Time
int PP[pn]; //Processes piriorty
int waittingTime[pn];
int turnaroundTime[pn];
//Scanning Time and Piriorty
for(int i=0 ;i<pn ;i++){
printf("\nProcessing time for P%d: ",i+1);
scanf("%d",&PT[i]);
printf("Piriorty for P%d: ",i+1);
scanf("%d",&PP[i]);
printf("Arrival Time for P%d: ",i+1);
scanf("%d",&AT[i]);
ATt[i] = AT[i];
}
int LAT = 0; //LastArrivalTime
for(int i = 0; i < pn; i++)
if(AT[i] > LAT)
LAT = AT[i];
int ATv = AT[0]; //Pointing to Arrival Time Value
int ATi = 0; //Pointing to Arrival Time indix
int P1 = PP[0]; //Pointing to 1st piriorty Value
int P2 = PP[0]; //Pointing to 2nd piriorty Value
//findding the First Arrival Time and Highst piriorty Process
while(NoP > 0 && CPU <= 1000){
for(int i = 0; i < pn; i++){
if(ATt[i] < ATv){
ATi = i;
ATv = ATt[i];
P1 = PP[i];
P2 = PP[i];
}
else if(ATt[i] == ATv || ATt[i] <= CPU){
if(PP[i] != (pn+1))
P2 = PP[i];
if(P2 < P1){
ATi = i;
ATv = ATt[i];
P1 = PP[i];
P2 = PP[i];
}
}
}
if(CPU < ATv){
CPU = CPU+1;
continue;
}else{
waittingTime[ATi] = CPU - ATt[ATi];
CPU = CPU + PT[ATi];
turnaroundTime[ATi] = CPU - ATt[ATi];
ATt[ATi] = LAT +10;
ATv = LAT +10; //Pointing to Arrival Time Value
ATi = 0; //Pointing to Arrival Time indix
PP[ATi] = pn + 1;
P1 = PP[0]; //Pointing to 1st piriorty Value
P2 = PP[0]; //Pointing to 2nd piriorty Value
printf("Iam in");
NoP = NoP - 1;
}
}
printf("\nPN\tPT\tPP\tWT\tTT\n\n");
for(int i = 0; i < pn; i++){
printf("P%d\t%d\t%d\t%d\t%d\n",i+1,PT[i],PP[i],waittingTime[i],turnaroundTime[i]);
}
int AvgWT = 0;
int AVGTaT = 0;
for(int i = 0; i < pn; i++){
AvgWT = waittingTime[i] + AvgWT;
AVGTaT = turnaroundTime[i] + AVGTaT;
}
printf("AvgWaittingTime = %d\nAvgTurnaroundTime = %d\n",AvgWT/pn,AVGTaT/pn);
}
/*
Test Cases:
PT: Processing Time
PP: Process priority
WT Waitting Time
TaT: Turnaround Time
Arrival time for 1st 2 cases is 0
PN PT PP WT TaT
P1 10 3 6 16
P2 1 1 0 1
P3 2 4 16 18
P4 1 5 18 19
P5 5 2 1 6
PN PT PP WT TaT
P1 1 1 0 1
P2 2 2 1 3
P3 3 3 3 6
P4 4 4 6 10
P5 5 5 10 15
PN PP AT PT WT TaT
1 2 0 3 0 3
2 6 2 5 11 16
3 3 1 4 2 6
4 5 4 2 7 9
5 7 6 9 12 21
6 4 5 4 2 6
7 10 7 10 18 30
*/```
解决方案
正确的代码是：
#include <stdlib.h>
#include <stdio.h>
void main()
{
int pn = 0; //Processes Number
int CPU = 0; //CPU Current time
int allTime = 0; // Time neded to finish all processes
printf("Enrer Processes Count: ");
scanf("%d",&pn);
int AT[pn];
int ATt[pn];
int NoP = pn;
int PT[pn]; //Processes Time
int PP[pn]; //Processes piriorty
int PPt[pn];
int waittingTime[pn];
int turnaroundTime[pn];
int i=0;
//Scanning Time and Piriorty
for(i=0 ;i<pn ;i++){
printf("\nProcessing time for P%d: ",i+1);
scanf("%d",&PT[i]);
printf("Piriorty for P%d: ",i+1);
scanf("%d",&PP[i]);
PPt[i] = PP[i];
printf("Arrival Time for P%d: ",i+1);
scanf("%d",&AT[i]);
ATt[i] = AT[i];
}
int LAT = 0; //LastArrivalTime
for(i = 0; i < pn; i++)
if(AT[i] > LAT)
LAT = AT[i];
int MAX_P = 0; //Max Piriorty
for(i = 0; i < pn; i++)
if(PPt[i] > MAX_P)
MAX_P = PPt[i];
int ATi = 0; //Pointing to Arrival Time indix
int P1 = PPt[0]; //Pointing to 1st piriorty Value
int P2 = PPt[0]; //Pointing to 2nd piriorty Value
//findding the First Arrival Time and Highst piriorty Process
int j = -1;
while(NoP > 0 && CPU <= 1000){
for(i = 0; i < pn; i++){
if((ATt[i] <= CPU) && (ATt[i] != (LAT+10))){
if(PPt[i] != (MAX_P+1)){
P2 = PPt[i];
j= 1;
if(P2 < P1){
j= 1;
ATi = i;
P1 = PPt[i];
P2 = PPt[i];
}
}
}
}
if(j == -1){
CPU = CPU+1;
continue;
}else{
waittingTime[ATi] = CPU - ATt[ATi];
CPU = CPU + PT[ATi];
turnaroundTime[ATi] = CPU - ATt[ATi];
ATt[ATi] = LAT +10;
j = -1;
PPt[ATi] = MAX_P + 1;
ATi = 0; //Pointing to Arrival Time indix
P1 = MAX_P+1; //Pointing to 1st piriorty Value
P2 = MAX_P+1; //Pointing to 2nd piriorty Value
NoP = NoP - 1;
}
}
printf("\nPN\tPT\tPP\tAT\tWT\tTT\n\n");
for(i = 0; i < pn; i++){
printf("P%d\t%d\t%d\t%d\t%d\t%d\n",i+1,PT[i],PP[i],AT[i],waittingTime[i],turnaroundTime[i]);
}
int AvgWT = 0;
int AVGTaT = 0;
for(i = 0; i < pn; i++){
AvgWT = waittingTime[i] + AvgWT;
AVGTaT = turnaroundTime[i] + AVGTaT;
}
printf("AvgWaittingTime = %d\nAvgTurnaroundTime = %d\n",AvgWT/pn,AVGTaT/pn);
}
/*
Test Cases:
PT: Processing Time
PP: Process priority
WT Waitting Time
TaT: Turnaround Time
Arrival time for 1st 2 cases is 0
PN PT PP WT TaT
P1 10 3 6 16
P2 1 1 0 1
P3 2 4 16 18
P4 1 5 18 19
P5 5 2 1 6
PN PT PP WT TaT
P1 1 1 0 1
P2 2 2 1 3
P3 3 3 3 6
P4 4 4 6 10
P5 5 5 10 15
PN PT PP AT WT TT
P1 3 2 0 0 3
P2 5 6 2 11 16
P3 4 3 1 2 6
P4 2 5 4 7 9
P5 9 7 6 12 21
P6 4 4 5 2 6
P7 10 10 7 20 30
PN PT PP AT WT TT
P1 4 2 0 0 4
P2 3 3 1 3 6
P3 1 4 2 5 6
P4 5 5 3 5 10
P5 2 5 4 9 11
*/```
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