织金以那穿青人庆坛:怎样用C语言实现银行家算法?

#include<stdio.h>
struct claim
{
int user;
int num[3];
}claims;
int input()
{
printf("please input your request:user(0~4):\n");
scanf("%d",&claims.user);
printf("input the number of resource a:\n");
scanf("%d",&claims.num[0]);
printf("input the number of resource b:\n");
scanf("%d",&claims.num[1]);
printf("input the number of resource c:\n");
scanf("%d",&claims.num[2]);
return 1;
}
int safety_chk(int alloc[][3],int need[][3],int avail[3])
{
int work[3],finish[5];
for(int p=0;p<5;p++)//i大于2后对WORK是无意义的
{
work[p]=avail[p];
finish[p]=0;
}
for(int i=0;i<5;i++)
{
if(finish[i]==0&&
need[i][0]<=work[0]&&
need[i][1]<=work[1]&&
need[i][2]<=work[2] )
{
for(int j=0;j<3;j++)
work[j]=alloc[i][j]+work[j];
finish[i]=1;
i=-1;//重头再来
}
}
for(i=0;i<5;i++)
{
if(finish[i]==0)
return 0;
}
return 1;
}

int process(int alloc[][3],int need[][3],int avail[3])
{
int ret;
input();
for(int i=0;i<3;i++) //out of resource number
{
if(claims.num[i]>need[claims.user][i]||claims.num[i]>avail[i])
return 0;
}
for(i=0;i<3;i++)//trying
{
avail[i]=avail[i]-claims.num[i];
alloc[claims.user][i]=alloc[claims.user][i]+claims.num[i];
need[claims.user][i]=need[claims.user][i]-claims.num[i];
}
if((ret=safety_chk(alloc,need,avail)==0))
{

printf("safety_chk's result %d \n",0);
for(i=0;i<3;i++)
{
avail[i]=avail[i]+claims.num[i];
alloc[claims.user][i]=alloc[claims.user][i]-claims.num[i];
need[claims.user][i]=need[claims.user][i]+claims.num[i];
}
return 0;
}
else
{

printf("safety_chk's result %d \n",1);
}
return 1;
}
void main()
{
int alloc[5][3]={{0,1,0},{2,0,0},{3,0,2},{2,1,1},{0,0,2}};
int need[5][3]={{7,4,3},{1,2,2},{6,0,0},{0,1,1},{4,3,1}};
int avail[3]={3,3,2};
if(process(alloc,need,avail)==0)
printf("sorry,we cannot help you!\n");
else printf("operation complete!\n");
return;
}