c++实现银行家算法-蒲公英云

大三上学期的时候自己写的银行家算法的代码，复制粘贴到编译器即可运行，希望帮助到其他人！
/* Financier algorithm.(银行家算法) */ 
//=====<head file>==============================================================
#include<stdio.h>
#include<windows.h>
#include<conio.h>
//=====<data structure>=========================================================
struct systemInformation
{
    int proceeding; //number of preceedings.
    int resource;   //number of resources.
    int available[6];//six kinds of resources.
    int maxResource[6];//each max element stand for the max number of one kind of resources .
    int allocation[10][6];//allocation[i][j]=k stand for proceeding[i] has been allocated k resource[j].
    int need[10][6];//need[i][j]=k stands for proceeding[i] need k resource[j] for finishing.
    int finish[10];//finish[i]=1 => proceeding[i] have been finished .
}; 
struct currentProceedingInformation
{
    int proceedingID;
    int need[6];
};
//=====<static variable>========================================================
struct systemInformation information;
struct currentProceedingInformation informationForCheck;
int haveBeenInitializedTag = 0;
//=====<function declare>=======================================================
char menu();//show menu UI.
void drawTable();//to show users the current system information.
void defaultInput();//use default information to run this algorithm.
void input();
void financier();//financier algorithm
int  securityCheck();//a subordinate function of financier .
//=====<mian function>========================================================== 
int main()
{
    do
    {
        char option = menu();
        switch(option)
        {
            case'1':
            {
                input();
                drawTable();
                break;
            }
            case'2':
            {
                defaultInput();
                system("cls");
                drawTable();
                printf("\n[Enter]继续\n");
                getch();
                break;
            }
            case'3':
            {
                ;//nothing
                break;
            }
            case'4':
            {
                system("cls");
                drawTable();
                printf("\n[Enter]继续\n");
                getch();
                goto sign;              
            }
        }
        system("cls");
        financier();
        sign:;
    }while(1);
    return 0;
}
//=====<functions>============================================================== 
char menu()
{
    do
    {
        system("cls");
        printf(" 银行家算法\n");
        printf(" ----------------------------\n");
        printf("\n");
        printf(" (1)---输入数据\n");
        printf("\n");
        printf(" (2)---使用默认数据\n");
        printf("\n");
        if(haveBeenInitializedTag == 1)
        {
            printf(" (3)--继续\n");
            printf("\n");
            printf(" (4)--显示系统状态\n");
            printf("\n"); 
        }
        printf(" (0)---退出程序\n");
        printf("\n");
        printf(" ----------------------------\n");
        printf("\n请选择:");
        char optionTag = getch();
        if(optionTag == '0')
        {
            system("cls");
            printf("\n\n\t\t<欢迎下次使用>\n\n\t\t");
            Sleep(1000);
            exit(0);
        }
        else if(optionTag == '1' || optionTag == '2')
            return optionTag; 
        else if((optionTag == '3' ||optionTag == '4') && haveBeenInitializedTag == 1)
            return optionTag;
        else 
        { 
            printf("\n#输入错误#");
            Sleep(500); 
        } 
    }while(1);
}
void defaultInput()
{
    information.proceeding = 5;
    information.resource = 3;
    information.available[0] = 3;
    information.available[1] = 3;
    information.available[2] = 2;
    information.maxResource[0] = 10;
    information.maxResource[1] = 5;
    information.maxResource[2] = 7;
    information.allocation[0][0] = 0;
    information.allocation[0][1] = 1;
    information.allocation[0][2] = 0;
    information.allocation[1][0] = 2;
    information.allocation[1][1] = 0;
    information.allocation[1][2] = 0;
    information.allocation[2][0] = 3;
    information.allocation[2][1] = 0;
    information.allocation[2][2] = 2;
    information.allocation[3][0] = 2;
    information.allocation[3][1] = 1;
    information.allocation[3][2] = 1;
    information.allocation[4][0] = 0;
    information.allocation[4][1] = 0;
    information.allocation[4][2] = 2;
    information.need[0][0] = 7;
    information.need[0][1] = 4;
    information.need[0][2] = 3;
    information.need[1][0] = 1;
    information.need[1][1] = 2;
    information.need[1][2] = 2;
    information.need[2][0] = 6;
    information.need[2][1] = 0;
    information.need[2][2] = 0;
    information.need[3][0] = 0;
    information.need[3][1] = 1;
    information.need[3][2] = 1;
    information.need[4][0] = 4;
    information.need[4][1] = 3;
    information.need[4][2] = 1;
    //finish[]=0;
    for(int i=0;i<5;i++)
    {
        information.finish[i] = 0;
    }
    haveBeenInitializedTag = 1;
}
void input()
{
    system("cls");
    printf("输入信息\n");
    printf("---------------------------\n");
    printf("\n");
    //input proceeding 
    printf("\n请输入进程的个数:"); 
    int temporary=0;
    scanf("%d",&temporary);
    if(temporary<=0||temporary>10)
    {
        printf("\n#输入错误,再见#\n");
        getch();
        exit(0); 
    }
    else
        information.proceeding = temporary;
    printf("\n#以下用 P0-p%d 来表示\n",temporary);
    //input resource
    printf("\n请输入资源的个数:");
    scanf("%d",&temporary);
    if(temporary<=0 ||temporary>6)
    {
        printf("\n#输入错误,再见#\n");
        getch();
        exit(0); 
    }
    else
        information.resource = temporary;
    printf("\n#以下用 A-%c 来表示",temporary+65); 
    //input maxResource
    for(int i=0;i<information.resource;i++)
    {
        printf("\n请输入%c类资源总数:",i+65);
        scanf("%d",&temporary);
        if(temporary<=0)
        {
            printf("\n#输入错误,再见#");
            getch();
            exit(0); 
        }
        else
            information.maxResource[i] = temporary;
    } 
    //input allocation
    for(int i=0;i<information.proceeding;i++)
    {
        for(int j=0;j<information.resource;j++)
        {
            printf("\n请输入P%d 当前已占有%c 类资源的数目:",i,j+65);
            if(temporary<=0)
            {
                printf("\n#输入错误,再见#");
                getch();
                exit(0); 
            }
            else
                information.allocation[i][j] = temporary;
        }
    }
    //input need
    for(int i=0;i<information.proceeding;i++)
    {
        for(int j=0;j<information.resource;j++)
        {
            printf("\n请输入P%d 当前尚需要%c 类资源的数目:",i,j+65);
            if(temporary<=0)
            {
                printf("\n#输入错误,再见#");
                getch();
                exit(0); 
            }
            else
                information.allocation[i][j] = temporary;
        }
    }
    //finish[]=0;
    for(int i=0;i<information.proceeding;i++)
    {
        information.finish[i] = 0;
    }
    haveBeenInitializedTag = 1;
}
void drawTable()
{
    printf("\n\n当前系统状态:\n\n");
    printf("P :A allication/need ...\n");
    for(int i=0;i<information.proceeding;i++)
    {
        printf("\n");
        printf("P%d: ",i);
        if(information.finish[i] == 0)
        {
            //
            for(int j=0;j<information.resource;j++)
            {
                printf("%c %d/%d ",j+65,information.allocation[i][j],information.need[i][j]);
            } 
            printf(";");
        }
        else
            printf("已完成"); 
    }
    printf("\n系统剩余: ");
    for(int i=0;i<information.resource;i++) 
    {
        printf("%c:%d ",i+65,information.available[i]);
    }
    printf("\n");
}
void financier()
{
    system("cls");
    printf("银行家算法\n");
    printf("----------------------------\n\n");
    fflush(stdin);//empty input block.
    printf("请输入要申请资源的程序(如 P1): ");
    int temporary = -1;
    char p;
    scanf("%c%d",&p,&temporary);
    if(temporary<0||temporary>=information.proceeding)
    {
        printf("\n#输入错误,再见#");
        Sleep(1000);
        exit(0); 
    }
    else
        informationForCheck.proceedingID = temporary;
    printf("\n");
    for(int i=0;i<information.resource;i++)
    {
        printf("请输入 P%d 要申请 %c 类资源的数目:",informationForCheck.proceedingID,i+65);
        scanf("%d",&temporary);
        if(temporary < 0)
        {
        printf("\n#输入错误,再见#");
        Sleep(1000);
        exit(0);            
        } 
        else if(temporary > information.need[informationForCheck.proceedingID][i])
        {
            printf("\n#输入错误#\n错误信息: 申请资源量大于该进程需求量\n");
            Sleep(1000);
            exit(0);                
        }
        else
            informationForCheck.need[i] = temporary;        
    }
    if(1 == securityCheck())//the request can be satisfied .
    {
        for(int i=0;i<information.resource;i++)//allocate
        {
            //refresh need
            information.need[informationForCheck.proceedingID][i] = information.need[informationForCheck.proceedingID][i] - informationForCheck.need[i];
            //refresh allocation
            information.allocation[informationForCheck.proceedingID][i] = information.allocation[informationForCheck.proceedingID][i]+informationForCheck.need[i];
            //refresh available
            information.available[i] = information.available[i] - informationForCheck.need[i]; 
        }
    }
    else
    {
        ;//nothing.
    }   
}
int securityCheck()
{
    system("cls");
    printf("\n安全性检查\n");
    printf("----------------------------\n");
    struct systemInformation copyInformation = information,temporaryInformation;//to protect original data.
    //STEP ONE:experimental allocate
    for(int i=0;i<copyInformation.resource;i++)
    {
        //refresh need
        copyInformation.need[informationForCheck.proceedingID][i] = copyInformation.need[informationForCheck.proceedingID][i] - informationForCheck.need[i];
        //refresh allocation
        copyInformation.allocation[informationForCheck.proceedingID][i] = copyInformation.allocation[informationForCheck.proceedingID][i]+informationForCheck.need[i];
        //refresh available
        copyInformation.available[i] = copyInformation.available[i] - informationForCheck.need[i]; 
    }
    printf("\n\n第一步:试探性分配");
    //temporaryInformation <- information <- copyInformation --- to drawTable
    temporaryInformation = information;
    information = copyInformation;
    drawTable();
    //copyInformation <- information <- temporaryInformation -- return to the original condition.
    copyInformation = information;
    information = temporaryInformation;
    printf("\n[Enter]下一步\n"); 
    getch();
    //STEP TWO:to check whether all of proceedings can finish .
    system("cls");
    printf("\n安全性检查\n");
    printf("----------------------------\n");
    printf("\n\n第二步:检测是否所有进程都能够完成.\n"); 
    int haveUnfinishedProceedingTag = 0;//Are there unfinished proceedings ?
    int changeTag = 0;//whether finishTag has changed.
    int eachSatisfiedTag = 0;//eachSatisfiedTag = resource => changeTag = 1; 
    for(int i=0;i<copyInformation.proceeding;i++)//scan proceedings for those times.
    {
        //set variables in order
        changeTag = 0;
        eachSatisfiedTag = 0;
        for(int j=0;j<copyInformation.proceeding;j++)//scan all proceedings each time. 
        {
            if(copyInformation.finish[j] == 0)//this proceeding is unfinished now .
            {
                eachSatisfiedTag = 0;
                //if(need:<=available)
                for(int k=0;k<copyInformation.resource;k++)
                {
                    if(copyInformation.need[j][k] <= copyInformation.available[k])
                        eachSatisfiedTag++; 
                }
                if(eachSatisfiedTag == copyInformation.resource)//this proceeding can finish.
                {
                    copyInformation.finish[j] = 1;//to record its finish .
                    changeTag = 1;//finish[] was changed .
                    //then refresh copyInformation
                    for(int k=0;k<copyInformation.resource;k++)//retrieve its resource .
                    {
                        copyInformation.available[k] = copyInformation.available[k] + copyInformation.allocation[j][k];
                    }
                    break;
                }//if-finish
            }//if
        }//for-eachTime.
        //whether are all proceedings finished ?
        haveUnfinishedProceedingTag = 1;
        for(int k=0;k<copyInformation.proceeding;k++)//have unfinished proceedings ?
        {
            if(copyInformation.finish[k] == 0)
            {
                haveUnfinishedProceedingTag = 0;//yes
                break;
            }
        } 
        if(haveUnfinishedProceedingTag == 0 && changeTag == 0)//have no way to continue running.
        {
            printf("\n无法完成分配!\n");
            getch();
            return 0;
        }
        else
        {
            //temporaryInformation <- information <- copyInformation --- to drawTable
            temporaryInformation = information;
            information = copyInformation;
            drawTable();
            //copyInformation <- information <- temporaryInformation -- return to the original condition.
            copyInformation = information;
            information = temporaryInformation;
            printf("\n[Enter]继续\n");
            getch();    
        }
    }//for-scan
    system("cls");
    printf("\n所有程序都可以运行完成!\n");
    Sleep(500); 
    printf("\n#系统为安全状态,可以分配#");
    Sleep(500); 
    printf("\n\n[Enter]完成"); 
    getch();
    return 1;
}