/*
銀行家算法

銀行家可以把一定數(shù)量的資金供多個用戶周轉(zhuǎn)使用，為保證資金的安全銀行家規(guī)定：

1.當(dāng)一個用戶對資金的最大需求量不超過銀行家現(xiàn)有的資金就要接納該用戶;
2.用戶可以分期貸款,但垡的總數(shù)不能超過最大需求量;
3.當(dāng)銀行家現(xiàn)有的資金不能滿足用戶的沿需垡數(shù)時,對用戶的貸款可推遲支付,但總能使用戶在有限的時間里得到貸款;
4當(dāng)用戶得到所需的全部資金后,一定能在有限的時間里歸還所有的資金.

假設(shè)共有3類資源A B C,分別為10,5,7個
進程列表存在文件中,格式為

進程名 對A類資源需求數(shù)量  對B類資源需求數(shù)量  對c資源需求數(shù)量 

輸出每個進程的處理結(jié)果


本程序包括:銀行家算法


  VC++調(diào)試通過

  (C)copyright by Neo

  歡迎大家測試 請問題請Email:sony006@163.com
*/

#include<stdio.h>
#include<string.h>
#include<iostream.h>
#include<stdlib.h>
#include<iomanip.h>
#include<conio.h>

const int MAX_P=20;
const int MAXA=10;  //定義A類資源的數(shù)量
const int MAXB=5;
const int MAXC=7;


typedef struct node{
int a;
int b;
int c;
int remain_a;
int remain_b;
int remain_c;
}bank;

typedef struct node1{
char name[20];
int a;
int b;
int c;
int need_a;
int need_b;
int need_c;
}process;

bank banker;
process processes[MAX_P];
int quantity;

//初始化函數(shù)
void initial()
{
int i;

banker.a=MAXA;
banker.b=MAXB;
banker.c=MAXC;
banker.remain_a=MAXA;
banker.remain_b=MAXB;
banker.remain_c=MAXC;

for(i=0;i<MAX_P;i++){
  strcpy(processes[i].name,"");
  processes[i].a=0;
  processes[i].b=0;
  processes[i].c=0;
  processes[i].need_a=0;
  processes[i].need_b=0;
  processes[i].need_c=0;
}

}

//新加作業(yè)
void add()
{
char name[20];
int flag=0;
int t;
int need_a,need_b,need_c;
int i;

cout<<endl;
cout<<"新加作業(yè)"<<endl;
cout<<"━━━━━━━━━━━━━━━━━━━━━━━━━━━━"<<endl;
cout<<"請輸入新加作業(yè)名:";
cin>>name;

for(i=0;i<quantity;i++){
  if(!strcmp(processes[i].name,name)){
   flag=1;
   break;
  }
}

if(flag){
  cout<<"錯誤,作業(yè)已存在"<<endl;
}
else{
  cout<<"本作業(yè)所需A類資源:";
  cin>>need_a;
  cout<<"本作業(yè)所需B類資源:";
  cin>>need_b;
  cout<<"本作業(yè)所需C類資源:";
  cin>>need_c;

  t=1;
  cout<<need_a<<banker.remain_a;
  if(need_a>banker.remain_a){
   cout<<"錯誤,所需A類資源大于銀行家所剩A類資源"<<endl;
   t=0;
  }
  if(need_b>banker.remain_b){
   cout<<"錯誤,所需B類資源大于銀行家所剩B類資源"<<endl;
   t=0;
  }
  if(need_c>banker.remain_c){
   cout<<"錯誤,所需C類資源大于銀行家所剩C類資源"<<endl;
   t=0;
  }
  if(t){
   strcpy(processes[quantity].name,name);
   processes[quantity].need_a=need_a;
   processes[quantity].need_b=need_b;
   processes[quantity].need_c=need_c;

   quantity++;
   cout<<"新加作業(yè)成功"<<endl;
  }
  else{
   cout<<"新加作業(yè)失敗"<<endl;
  }
}
}

//為作業(yè)申請資源
void bid()
{
char name[20];
int i,p;
int a,b,c;
int flag;

cout<<endl<<"為作業(yè)申請資源"<<endl;
cout<<"━━━━━━━━━━━━━━━━━━━━━━━━━━━━"<<endl;
cout<<"要申請資源的作業(yè)名:";
cin>>name;

p=-1;
for(i=0;i<quantity;i++){
  if(!strcmp(processes[i].name,name)){
   p=i;
   break;
  }
}
if(p!=-1){
  cout<<"該作業(yè)要申請A類資源數(shù)量:";
  cin>>a;
  cout<<"該作業(yè)要申請B類資源數(shù)量:";
  cin>>b;
  cout<<"該作業(yè)要申請C類資源數(shù)量:";
  cin>>c;

  flag=1;
  if((a>banker.remain_a)||(a>processes[p].need_a-processes[p].a)){
   cout<<"錯誤,所申請A類資源大于銀行家所剩A類資源或該進程還需數(shù)量"<<endl;
   flag=0;
  }
  if((b>banker.remain_b)||(b>processes[p].need_b-processes[p].b)){
   cout<<"錯誤,所申請B類資源大于銀行家所剩B類資源或該進程還需數(shù)量"<<endl;
   flag=0;
  }
  if((c>banker.remain_c)||(c>processes[p].need_c-processes[p].c)){
   cout<<"錯誤,所申請C類資源大于銀行家所剩C類資源或該進程還需數(shù)量"<<endl;
   flag=0;
  }
  if(flag){
   banker.remain_a-=a;
   banker.remain_b-=b;
   banker.remain_c-=c;

   processes[p].a+=a;
   processes[p].b+=b;
   processes[p].c+=c;

   cout<<"為作業(yè)申請資源成功"<<endl;
  }
  else{
   cout<<"為作業(yè)申請資源失敗"<<endl;
  }
}
else{
  cout<<"該作業(yè)不存在"<<endl;
}
}

//撤消作業(yè)
void finished()
{
char name[20];
int i,p;

cout<<endl<<"撤消作業(yè)"<<endl;
cout<<"━━━━━━━━━━━━━━━━━━━━━━━━━━━━"<<endl;
cout<<"要撤消作業(yè)名:";
cin>>name;

p=-1;
for(i=0;i<quantity;i++){
  if(!strcmp(processes[i].name,name)){
   p=i;
   break;
  }
}
if(p!=-1){
  banker.remain_a+=processes[p].a;
  banker.remain_b+=processes[p].b;
  banker.remain_c+=processes[p].c;
  for(i=p;i<quantity-1;i++){
   processes[i]=processes[i+1];
  }
  strcpy(processes[quantity-1].name,"");
  processes[quantity-1].a=0;
  processes[quantity-1].b=0;
  processes[quantity-1].c=0;
  processes[quantity-1].need_a=0;
  processes[quantity-1].need_b=0;
  processes[quantity-1].need_c=0;

  quantity--;

  cout<<"撤消作業(yè)成功"<<endl;
}
else{
  cout<<"撤消作業(yè)失敗"<<endl;
}
}

//查看資源情況
void view()
{
int i;

cout<<endl<<"查看資源情況"<<endl;
cout<<"━━━━━━━━━━━━━━━━━━━━━━━━━━━━"<<endl;
cout<<"銀行家所剩資源(剩余資源/總共資源)"<<endl;
cout<<"A類:"<<banker.remain_a<<"/"<<banker.a;
cout<<"   B類:"<<banker.remain_b<<"/"<<banker.b;
cout<<"   C類:"<<banker.remain_c<<"/"<<banker.c;

cout<<endl<<endl<<"作業(yè)占用情況(已占用資源/所需資源)"<<endl<<endl;
if(quantity>0){
  for(i=0;i<quantity;i++){
   cout<<"作業(yè)名:"<<processes[i].name<<endl;
   cout<<"A類:"<<processes[i].a<<"/"<<processes[i].need_a;
   cout<<"   B類:"<<processes[i].b<<"/"<<processes[i].need_b;
   cout<<"   C類:"<<processes[i].c<<"/"<<processes[i].need_c;
   cout<<endl;
  }
}
else{
  cout<<"當(dāng)前沒有作業(yè)"<<endl;
}
}

//顯示版權(quán)信息函數(shù)
void version()
{
cout<<endl<<endl;

cout<<" ┏━━━━━━━━━━━━━━━━━━━━━━━┓"<<endl;
cout<<" ┃　　　　　　 　銀 行 家 算 法　 　　　　　　　┃"<<endl;
cout<<" ┠───────────────────────┨"<<endl;
cout<<" ┃　   　(c)All Right Reserved  Neo　　　 　　　┃"<<endl;
cout<<" ┃　　　　　　sony006@163.com　　 　　　　　　　┃"<<endl;
cout<<" ┃　　　　　version 2004  build 1122　　　　　　┃"<<endl;
cout<<" ┗━━━━━━━━━━━━━━━━━━━━━━━┛"<<endl;
cout<<endl<<endl;
}

void main()
{
int chioce;
int flag=1;


initial();

version();

while(flag){
  cout<<"━━━━━━━━━━━━━━━━━━━━━━━━━━━━"<<endl;
  cout<<"1.新加作業(yè)  2.為作業(yè)申請資源  3.撤消作業(yè)"<<endl;
  cout<<"4.查看資源情況   0.退出系統(tǒng)"<<endl;
  cout<<"請選擇:";
  cin>>chioce;

  switch(chioce){
  case 1:
   add();
   break;
  case 2:
   bid();
   break;
  case 3:
   finished();
   break;
  case 4:
   view();
   break;
  case 0:
   flag=0;
   break;
  default:
   cout<<"選擇錯誤"<<endl<<endl;
  }
}
}