1.c.txt

这是< 计算机操作系统>>中的一个例子

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#define N 10
#define M 10
typedef struct{
 int ip;
 int able[N];
 int remain[M];
 int visited[N];
}WorkNode;

int ass[N][M],need[N][M],r[M],pnum,rnum,top=-1;
WorkNode current,stack[N];

void add(int [],int []);
void init();
void output();
int small(int [],int []);
int bankalgo();

int main(){
 int ct=0;
 clrscr();
 init();
 ct=bankalgo();
 if(ct)
  printf("\nit is safe,and it has %d solutions\n",ct);
 else
  printf("\nit is unsafe\n");
 return 0;
}

void init(){
 int i,j;
 printf("process number:");
 scanf("%d",&pnum);
 printf("resource number:");
 scanf("%d",&rnum);
 printf("resource series:");
 for(i=0;i<rnum;i++)scanf("%d",r+i);
 printf("assigned matrix:\n");
 for(i=0;i<pnum;i++){
  printf("p%d:",i);
  for(j=0;j<rnum;j++)scanf("%d",&ass[i][j]);
 }
 printf("needed matrix:\n");
 for(i=0;i<pnum;i++){
  printf("p%d:",i);
  for(j=0;j<rnum;j++)scanf("%d",&need[i][j]);
 }
}

void add(int x[],int y[]){
 int i;
 for(i=0;i<rnum;i++)
  x[i]+=y[i];
}

int small(int x[],int y[]){
 int i;
 for(i=0;i<rnum;i++)
  if(x[i]>y[i])break;
 if(i==rnum)return 1;
 return 0;
}

void output(){
 int i;
 for(i=0;i<top;i++)
  printf("p%d-->",stack[i].able[stack[i].ip]);
 printf("p%d\n",stack[i].able[stack[i].ip]);
}


int bankalgo(){
 int i,j,sum=0,ct=0;
 current.ip=-1;
 memset(current.visited,sizeof(current.visited),0);
 for(i=0;i<rnum;i++){
  for(j=0;j<pnum;j++)
   sum+=ass[j][i];
  current.remain[i]=r[i]-sum;
  sum=0;
 }
 for(i=0;i<pnum;i++){
  if(small(need[i],current.remain) )
   current.able[++current.ip]=i;
 }
 if(current.ip==-1)return 0;
 while(1){
  stack[++top]=current;
  current.visited[current.able[current.ip]]=1;
  add(current.remain,ass[current.able[current.ip]]);
  current.ip=-1;
  for(i=0;i<pnum;i++){
   if(!current.visited[i] && small(need[i],current.remain))
    current.able[++current.ip]=i;
  }
  if(current.ip==-1){
   if(top==pnum-1){
    output();
    ct++;
   }
   current=stack[top--];
   while(current.ip==0 && top>-1)current=stack[top--];
   if(top==-1 && current.ip==0)break;
   else current.ip--;
  }
 }
 return ct;
}
测试结果如下(Andrews S.Tanenbaum的《现代操统系统》179页)

process number:5
resource number:4
resource series:6 3 4 2
assined matrix:
p0:3 0 1 1
p1:0 1 0 0
p2:1 1 1 0
p3:1 1 0 1
p4:0 0 0 0
needed matrix:
p0:1 1 0 0
p1:0 1 1 2
p2:3 1 0 0
p3:0 0 1 0
p4:2 1 1 0
p3-->p4-->p0-->p2-->p1
p3-->p4-->p0-->p1-->p2
p3-->p0-->p4-->p2-->p1
p3-->p0-->p4-->p1-->p2
p3-->p0-->p2-->p4-->p1
p3-->p0-->p2-->p1-->p4
p3-->p0-->p1-->p4-->p2
p3-->p0-->p1-->p2-->p4
it is safe,and it has 8 solutions