求银行家算法判断是否存在安全序列程序

[堕落的天使]

#include<stdio.h>
#define M 3
#define N 5
void main()
{
int max[N][M],allocate[N][M],need[N][M],available[M];
int i,j,flag=1;
printf("\nThere are %d process in system!\n",N);
printf("\nPlease input the maxresource!\n");
for(i=0;i<N;i++)
{ printf("\nInput the maxresource of p%d:",i);
for(j=0;j<M;j++)
scanf("%d",&max[i][j]);
}
printf("\nPlease input the allocation!\n");
for(i=0;i<N;i++)
{ printf("\nInput the allocation of p%d:",i);
for(j=0;j<M;j++)
scanf("%d",&allocate[i][j]);
}
printf("\nplease Input the available of resources:");
for(i=0;i<M;i++)
scanf("%d",&available[i]);
for(i=0;i<N;i++)
{for(j=0;j<M;j++)
need[i][j]=max[i][j]-allocate[i][j];
}
printf("---------------------------------------------------------------\n");
printf("|ProcessId| Max | Allocation | Need | Available|\n");
printf("---------------------------------------------------------------\n");
for(i=0;i<N;i++)
{ printf("| P%d | %2d%2d%2d | %2d%2d%2d | %2d%2d%2d | %2d%2d%2d |\n",
i,max[i][0],max[i][1],max[i][2],
allocate[i][0],allocate[i][1],allocate[i][2],
need[i][0],need[i][1],need[i][2],
available[0],available[1],available[2]);
printf("---------------------------------------------------------------\n");
}
printf("\nThe order of request successfully process is:");













}
上面是一个银行家算法程序，在printf("\nThe order of request successfully process is:");语句后面缺少一个判断再输入数据后是否存在一个安全序列的程序 请高手指点！！！！！！谢谢！！！！！

[太乙]

呵呵~~~~up一下~~~~

[DouBreak]

UP 下

[luffyke]

给你我写的程序，正确的话楼主要给分哦！在VC++6.0上编译运行就行了
#include <stdio.h>
#include <stdlib.h>
#define N 5

struct bank
{char num[6];
int max[3];
int allocation[3];
int need[3];
int available[3];
int work[3];
int wa[3];
char finish;
}bank[N];


main()
{int i,j,k,number,p,t;
int request[3];
char h;
char a='T';
char b='F';
int q=0;
char finish1[5];
int m;
int record[5];
int work1[3];
char f[5];

printf(" ┏━━━━━━━━━━━━━━━━━━━━━━━┓"); printf("\n");
printf(" ┃　　　　　　 　银 行 家 算 法　 　　　　　　　┃"); printf("\n");
printf(" ┠───────────────────────┨"); printf("\n");
printf(" ┃ 操作系统作业 ┃ "); printf("\n");
printf(" ┃　　　　　　 　　 　　　　　　 ┃ "); printf("\n");
printf(" ┃　　　　　　　 luffyke 　　　　 ┃ "); printf("\n");
printf(" ┗━━━━━━━━━━━━━━━━━━━━━━━┛ "); printf("\n");


for(i=0;i<N;i++)
finish1[i]=b;

for(i=0;i<N;i++)
{printf("\nInput data of bank %d:\n",i);
printf("NUMBER.:");
scanf("%s",&bank[i].num);

for(j=0;j<3;j++)
{printf("max %d:\n",j+1);
scanf("%d",&bank[i].max[j]);
}

for(j=0;j<3;j++)
{printf("allocation %d:\n",j+1);
scanf("%d",&bank[i].allocation[j]);
}

printf("\n");
}

for(i=0;i<N;i++)
{for(j=0;j<3;j++)
bank[i].need[j]=bank[i].max[j]-bank[i].allocation[j];
}

for(j=0;j<3;j++)
{printf("enter data of available:\n");
scanf("%d",&bank[0].available[j]);
}

for(i=0;i<N;i++)
{f[i]=finish1[i];
}

printf(" max allocation need available\n");
printf(" NO. A B C A B C A B C A B C\n");

for(i=0;i<N;i++)
{printf("%5s",bank[i].num);
for(j=0;j<3;j++)
printf("%4d",bank[i].max[j]);
for(j=0;j<3;j++)
printf("%4d",bank[i].allocation[j]);
for(j=0;j<3;j++)
printf("%4d",bank[i].need[j]);
if(i==0)
{for(j=0;j<3;j++)
printf("%4d",bank[i].available[j]);
printf("\n");
}
else
{printf("\n");
}
}

loop: {printf("\n which number of request?\n");
scanf("%d",&number);
printf("\n");
}

for(j=0;j<3;j++)
{printf("enter data of request %d:\n",j+1);
scanf("%d",&request[j]);
}

q=0;
for(j=0;j<3;j++)
{if ((request[j]<=bank[number].need[j])&&(request[j]<=bank[0].available[j]))
q++;
else
q+=0;
}

{if (q==3)
{for(j=0;j<3;j++)
{bank[number].need[j]=bank[number].need[j]-request[j];
bank[0].available[j]=bank[0].available[j]-request[j];
bank[number].allocation[j]=bank[number].allocation[j]+request[j];
}

printf(" max allocation need available\n");
printf(" NO. A B C A B C A B C A B C\n");

for(i=0;i<N;i++)
{printf("%5s",bank[i].num);
for(j=0;j<3;j++)
printf("%4d",bank[i].max[j]);
for(j=0;j<3;j++)
printf("%4d",bank[i].allocation[j]);
for(j=0;j<3;j++)
printf("%4d",bank[i].need[j]);
if (i==0)
{for(j=0;j<3;j++)
printf("%4d",bank[i].available[j]);
printf("\n");
}
else
printf("\n");
}


for(j=0;j<3;j++)
{work1[j]=bank[0].available[j];
}

m=0;
t=0;
while (t<5)
{for(i=0;i<5;i++)
{if ((work1[0]>=bank[i].need[0])&&(work1[1]>=bank[i].need[1])&&(work1[2]>=bank[i].need[2])&&(finish1

[i]==b))
{record[m]=i;
bank[i].work[0]=work1[0];
bank[i].work[1]=work1[1];
bank[i].work[2]=work1[2];

work1[0]=work1[0]+bank[i].allocation[0];
work1[1]=work1[1]+bank[i].allocation[1];
work1[2]=work1[2]+bank[i].allocation[2];

finish1[i]=a;

bank[i].wa[0]=work1[0];
bank[i].wa[1]=work1[1];
bank[i].wa[2]=work1[2];

bank[i].finish=finish1[i];

m++;
}
else
continue;
}
t++;
}


printf("\n");
{if (m==5)
{printf(" work need allocation work+alloct finish\n");
printf(" NO. A B C A B C A B C A B C \n");

for(k=0;k<N;k++)
{p=record[k];
printf("%5s",bank[p].num);
for(j=0;j<3;j++)
printf("%4d",bank[p].work[j]);
for(j=0;j<3;j++)
printf("%4d",bank[p].need[j]);
for(j=0;j<3;j++)
printf("%4d",bank[p].allocation[j]);
for(j=0;j<3;j++)
printf("%4d",bank[p].wa[j]);
printf("%4c",bank[p].finish);
printf("\n");
}
printf("The safe path is :");
printf("\n");
for(k=0;k<N;k++)
printf("%4d",record[k]);

printf("\n Contiune or not(Y/N)?");
getchar();
h=getchar();
if (h=='N'||h=='n')
{getchar();
exit(0);
}
else
for(i=0;i<N;i++)
{finish1[i]=f[i];
}
goto loop;

}
else
printf("it is not safe to assign");
printf("\n Contiune or not(Y/N)?");
getchar();
h=getchar();
if (h=='N'||h=='n')
{getchar();
exit(0);}
else
{printf("\n Enter the request\n");

for(j=0;j<3;j++)
{bank[number].need[j]=bank[number].need[j]+request[j];
bank[0].available[j]=bank[0].available[j]+request[j];
bank[number].allocation[j]=bank[number].allocation[j]-request[j];
}

for(i=0;i<N;i++)
{finish1[i]=f[i];
}

goto loop;
}
}
}
else
{printf("\n Waring: Out the extent!");
printf("\n Contiune or not(Y/N)?");
getchar();
h=getchar();
if (h=='N'||h=='n')
{getchar();
exit(0);
}
else
{printf("\n Rewrite the request\n");
goto loop;
}
}
}}

[堕落的天使]

谢谢你！呵呵！就是程序太长了啊！不过还是要谢谢你啊！

[青葱岁月不可追]

楼主要的是最最关键的思路……

这个得先自己想想吧？？？？

先自己写，然后再说别的，o(∩_∩)o...

[堕落的天使]

谢谢啊！我一直在想啊！可是我对c才刚刚起步啊！所以有很多东西是无法用程序写出来啊！所以才出来求救的啊！

[liufabin]

操作系统经常考试找个题目