银行家算法

模拟实现银行家算法，用银行家算法实现资源分配和安全性检查。

通过本次实验，加深对死锁概念的理解和掌握。

#include <iostream>
using namespace std;

void progressarrangement(int progressnumber,int resourcenumber,int Need[10][10],int Available[10],int Allocation[10][10],int state[10])
{
        int pnumber=0;
	    for(int j=0;j<progressnumber;j++)
		{
			for(int g=0;g<progressnumber;g++)
			{
				int k=0;
		    	for(int i=0;i<resourcenumber;i++)
			    {	
				     if(Need[g][i]<=Available[i])
				     k++;
			    }
			
                if(k==resourcenumber&&state[g]==0)
			    {
				    cout<<"正在被执行的进程是P"<<g<<endl;
			        for(int i=0;i<resourcenumber;i++)
				        Available[i]=Available[i]+Allocation[g][i];
				    state[g]=1;;
				    pnumber++;
			    }
			}
		}

		if(pnumber<progressnumber)
			 cout<<"找不到安全序列"<<endl;

        for(int i=0;i<progressnumber;i++)
		    state[i]=0;
	    
        for(int j=0;j<progressnumber;j++)
		    for(int i=0;i<resourcenumber;i++)
		        Available[i]=Available[i]-Allocation[j][i];
}

void dealrequstion(int progressnumber,int resourcenumber,int Need[10][10],int Available[10],int Allocation[10][10],int requstion[10][10],int state[10])
{
    int n;
	cout<<"请输入申请资源的进程号Pn"<<endl;
	cin>>n;
	cout<<"请输入申请的各类资源个数"<<endl;
	for(int i=0;i<resourcenumber;i++)
		cin>>requstion[n][i];
	int number=0;
	for(int i=0;i<resourcenumber;i++)
	if(requstion[n][i]>Available[i]||requstion[n][i]>Need[n][i])
		cout<<"申请的资源不能得到分配"<<endl; 
	else {
        Allocation[n][i]=Allocation[n][i]+requstion[n][i];
	    Need[n][i]-=requstion[n][i];
		Available[i]=Available[i]-requstion[n][i];
		number++;
	}

	if(number==resourcenumber)
        progressarrangement(progressnumber,resourcenumber,Need,Available,Allocation,state);
}

void inputdata(int &progressnumber,int &resourcenumber,int Need[10][10],int Available[10],int Allocation[10][10],int requstion[10][10])
{
    cout<<"请输入系统总的资源种类数："<<endl;
	cin>>resourcenumber;
	cout<<"请输入系统总的进程数："<<endl;
	cin>>progressnumber;
	cout<<"请输入每类资源的个数"<<endl;
	for(int i=0;i<resourcenumber;i++)
	cin>>Available[i];
	for(int j=0;j<progressnumber;j++)
	{
		cout<<"请输入进程P"<<j<<"已占有的各类资源数"<<endl;
		for(int i=0;i<resourcenumber;i++)
			cin>>Allocation[j][i];
	}

	for(int j=0;j<progressnumber;j++)
	{
		cout<<"请输入进程P"<<j<<"还需要的各类资源数"<<endl;
		for(int i=0;i<resourcenumber;i++)
		cin>>Need[j][i];
	}

	for(int j=0;j<progressnumber;j++)
		for(int i=0;i<resourcenumber;i++)
			Available[i]=Available[i]-Allocation[j][i];
			cout<<" MAX       Allocation       Need    "<<endl;
		for(int j=0;j<progressnumber;j++)
		{
		    for(int i=0;i<resourcenumber;i++)
			    cout<<Allocation[j][i]+Need[j][i]<<"  ";
		    cout<<"     ";
		
            for(int i=0;i<resourcenumber;i++)
			    cout<<Allocation[j][i]<<"  ";
		
            cout<<"     ";
		
            for(int i=0;i<resourcenumber;i++)
			    cout<<Need[j][i]<<"  ";
	
           	cout<<endl;
		}

		cout<<"可获得的资源数为"<<endl;
		for(int i=0;i<resourcenumber;i++)
		    cout<<Available[i]<<"  ";
		cout<<endl;
}

void mainmenu(int progressnumber,int resourcenumber,int Need[10][10],int Available[10],int Allocation[10][10],int requstion[10][10],int state[10])
{
	int number;
	int h=0;
	while(h!=1)
	{
	    cout<<"欢迎****银行家算法模拟程序"<<endl;
		cout<<"1.请录入相关数据"<<endl;
		cout<<"2.根据银行家算法，写出安全序列"<<endl;
		cout<<"3.申请资源"<<endl;
		cout<<"4.结束程序"<<endl;
		cout<<"输入你的选择："<<endl;
		cin>>number;
		switch(number)
		{
		case 1: inputdata(progressnumber,resourcenumber,Need,Available,Allocation,requstion);
			    break;
		case 2: progressarrangement(progressnumber,resourcenumber,Need,Available,Allocation,state);
			    break;
		case 3: dealrequstion(progressnumber,resourcenumber,Need,Available,Allocation,requstion,state);
				break;
		case 4:h=1;
		}
	}
} 

int resourcenumber,progressnumber;

int main()
{
    int Allocation[10][10],Need[10][10];
    int Available[10],requstion[10][10];
	int state[10]={0};
	
	//int  Allocation[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}};
     mainmenu(progressnumber,resourcenumber,Need,Available,Allocation,requstion,state);
	return 0;
}