#include<stdio.h>
#include<conio.h>
#include<iostream.h>
typedef struct Max1
{
	int m_a;
	int m_b;
	int m_c;
}Max;
typedef struct Allocation1
{
	int a_a;
	int a_b;
	int a_c;
}Allocation;
typedef struct Need1
{
	int n_a;
	int n_b;
	int n_c;
}Need;
struct Available1
{
	int av_a;
	int av_b;
	int av_c;
} q;
struct pr
{
	char name;
	Max max;
	Allocation allocation;
	Need need;
	int finishflag;
}p[5];
char na[5];
//*****************************

void init()//讀入
{
	printf("各進程的NEED：\n");
	FILE *fp;
	fp=fopen("1.txt","r+");
	for(int i=0;i<5;i++)
	{

		fscanf(fp,"%c,%d,%d,%d,%d,%d,%d\n",&p[i].name,&p[i].max.m_a,&p[i].max.m_b,&p[i].max.m_c,&p[i].allocation.a_a,&p[i].allocation.a_b,&p[i].allocation.a_c  );
		p[i].need.n_a=p[i].max.m_a-p[i].allocation.a_a;
		p[i].need.n_b=p[i].max.m_b-p[i].allocation.a_b;
		p[i].need.n_c=p[i].max.m_c-p[i].allocation.a_c;
		
		printf("%c:%d,%d,%d\n",p[i].name,p[i].need.n_a,p[i].need.n_b,p[i].need.n_c);

		
	}
	fclose(fp);
}

//*****************************
int fenpei()//分配
{
	printf("Available:\n");
	printf("%d,%d,%d\n",q.av_a,q.av_b,q.av_c);
	int finishcnt=0,k=0,count=0;
	for(int j=0;j<5;j++)
		p[j].finishflag=0;

	while(finishcnt<5)
	{
		for(int i=0;i<5;i++)
		{
			if(p[i].finishflag==0&&q.av_a>=p[i].need.n_a&&q.av_b>=p[i].need.n_b&&q.av_c>=p[i].need.n_c)
			{
				q.av_a+=p[i].allocation.a_a;
				q.av_b+=p[i].allocation.a_b;
				q.av_c+=p[i].allocation.a_c;
				p[i].finishflag=1;
				finishcnt++;
				na[k++]=p[i].name;
				
				break;
			}
			
		}
		count++;//禁止循環(huán)過多
		if(count>5)return 0;
	}
return 1;

}

//*****************************

int shq()
{
	int m,i,j,k;
	printf("請輸入進程號和請求資源！如：0   2 0 2\n");
	scanf("%d%d%d%d",&m,&i,&j,&k);
	if(i<=p[m].need.n_a&&j<=p[m].need.n_b &&k<=p[m].need.n_c)
	{
		if(i<=q.av_a&&j<=q.av_b&&k<=q.av_c)
		{
			p[m].allocation.a_a+=i;
			p[m].allocation.a_b+=j;
			p[m].allocation.a_c+=k;
			
			p[m].need.n_a=p[m].max.m_a-p[m].allocation.a_a;
			p[m].need.n_b=p[m].max.m_b-p[m].allocation.a_b;
			p[m].need.n_c=p[m].max.m_c-p[m].allocation.a_c;
	
			printf("各進程的NEED:\n");
			for(int w=0;w<5;w++)
			printf("%c:%d,%d,%d\n",p[w].name,p[w].need.n_a,p[w].need.n_b,p[w].need.n_c);
			return 1;
		}
		else
			printf("Request>Available\n讓%c等待......\n",p[m].name);
	}
	else printf("Request>Need\n讓%c等待......\n",p[m].name);
	return 0;

}


//*****************************

void main()
{
	int flag;
	char c;
	cout<<"\n\n\n\n\n\n\n\n\n\t           ";
	for(int i=0;i<30;i++)
		cout<<"*";
	cout<<endl<<"\t"<<"           * 銀  行  家  算  法  演  示 *"<<endl;
	cout<<endl<<"\t"<<"           * 信管031班  呂燕  20034940104 *"<<endl;
	printf("\t           ");
	for( i=0;i<30;i++)
		printf("*");
	printf("\n\n\n\n\n");
	getch();
	printf("\t    請確認(rèn)已經(jīng)在\"1.txt\"中正確輸入各進程的有關(guān)信息\n");
	getch();
	init();
	q.av_a=10;
	q.av_b=5;
	q.av_c=7;
	
	while(flag)
	{
		for(i=0;i<5;i++)
		{
			q.av_a-= p[i].allocation.a_a;
			q.av_b-= p[i].allocation.a_b;
			q.av_c-= p[i].allocation.a_c;
		}
		if(fenpei())
		{
			printf("\n這樣配置資源是安全的！\001_\001\n");
			printf("其安全序列是：  ");
			for(int k=0;k<5;k++)
				printf("-->%c",na[k]);
			printf("\n");
			printf("有進程發(fā)出Request請求向量嗎？？？(Enter y or Y)\n");
			c=getch();
			if(c=='y'||c=='Y')
			{
				if(shq())continue;
				else break;
			}else flag=0;
			
		}
		else {flag=0;printf("不安全\n");}
		

	}
}