求一个很短的先来先服务或者最短作业优先算法的代码。。还有解释。急求！！！

是C++的……能运行的……越短越好。同时求大体的对代码的解释

#include <stdio.h>
struct fcfs
{
char name[10];
float arrivetime;
float servicetime;
float starttime;
float finishtime;
float zztime;
float dqzztime;
};
//读取数字
fcfs a[100];
void input(fcfs *p,int N)
{
int i;
printf("intput the process's name & arrivetime & servicetime:\nfor exmple: a 0 100\n");
for(i=0;i<=N-1;i++)
{
printf("input the %dth process's information:\n",i+1);
scanf("%s%f%f",&p[i].name,&p[i].arrivetime,&p[i].servicetime);
}
}
//打印输出
void Print(fcfs *p,float arrivetime,float servicetime,float starttime,float finishtime,float zztime,float dqzztime,int N)
{
int k;

printf("run order:");
printf("%s",p[0].name);
for(k=1;k<N;k++)
{
printf("-->%s",p[k].name);
}
printf("\nthe process's information:\n");
printf("\nname\tarrive\tservice\tstart\tfinish\tzz\tdqzz\n");

for(k=0;k<=N-1;k++)
{
printf("%s\t%-.2f\t%-.2f\t%-.2f\t%-.2f\t%-.2f\t%-.2f\t\n",p[k].name,p[k].arrivetime,p[k].servicetime,p[k].starttime,p[k].finishtime,p[k].zztime,p[k].dqzztime);
}
}
//按时间排序
void sort(fcfs *p,int N)
{
for(int i=0;i<=N-1;i++)
{
for(int j=0;j<=i;j++)
{
if(p[i].arrivetime < p[j].arrivetime)
{
fcfs temp;
temp=p[i];
p[i]=p[j];
p[j]=temp;
}
}
}
}

void deal(fcfs *p, float arrivetime,float servicetime,float starttime,float finishtime,float &zztime,float &dqzztime,int N)
{
int k;
for(k=0;k<=N-1;k++)
{
if(k==0)
{
p[k].starttime=p[k].arrivetime;
p[k].finishtime=p[k].arrivetime+p[k].servicetime;
}
else
{
p[k].starttime=p[k-1].finishtime;
p[k].finishtime=p[k-1].finishtime+p[k].servicetime;
}
}

for(k=0;k<=N-1;k++)
{
p[k].zztime=p[k].finishtime-p[k].arrivetime;
p[k].dqzztime=p[k].zztime/p[k].servicetime;
}
}
void FCFS(fcfs *p,int N)
{
float arrivetime=0,servicetime=0,starttime=0,finishtime=0,zztime=0,dqzztime=0;
sort(p,N);
deal(p,arrivetime,servicetime,starttime,finishtime,zztime,dqzztime,N);
Print(p,arrivetime,servicetime,starttime,finishtime,zztime,dqzztime,N);
}
void main()
{
int N;
printf("------先来先服务调度算法------\n");
printf("input the process's number:\n");
scanf("%d",&N);
input(a,N);
FCFS(a,N);
}