@蕾蕾老师 补交作业，之前忘记提交了
#include <iostream>
#include<stdio.h>
#include<stdlib.h>
using namespace std;
# define m 50
int no1;//进程数
int no2;//资源数
int r;
int allocation[m][m];//分配矩阵
int need[m][m];//需求矩阵
int available[m];//可用资源数目
int max[m][m]; //最大需求矩阵
char name1[m], name2[m];
void main()
{
void check();
void print();
int i, j, p = 0, q = 0;
char c;
int request[m], allocation[m][m],need1[m][m],available[m];
printf("银行家算法实现\n");
printf("请输入进程数：\n");
scanf("%d",&no2);
printf("请输入Max矩阵：\n");
for (int i = 0; i<no1; i++)
{
for (int j = 0; j<no2; j++)
{
scanf("%d",&max[i][j]);//输入已知进程最大资源需求量
};
};
printf("请输入allocation 矩阵：\n");
for (int i = 0; i =no1)
{
printf("输入错误，请重新输入：\n");
continue;
}
else break;
};
printf("\n请输入该进程所请求的资源数request[j]:\n");
for (int j = 0; j need[i][j]) p=1;
//判断请求是否超过该进程所需要的资源数
if (p)
{
printf("请求资源超过该进程资源需求量，请求失败！\n");
}
else
{
for (int j = 0; j<no2; j++)
{
if(request[j]>available[j]) q=1;
//判断请求是否超过可用资源数
if (q)
{
printf("没有做够的资源分配，请求失败！\n");
}
else
{
for (int j = 0; j<no2; j++)
{
available[j]=available[j];
allocation[i][j]=allocation[i][j];
need[i][j]=need[i][j];
};
printf("返回分配前资源数\n");
print();
};
};
}
printf("\n你还要继续分配吗？ Y or N ?\n");
//判断是否继续进行资源分配
c=getche();}
}while (c=='y'||c=='Y');
}
};
void check() //安全算法函数
{
int k, f, v = 0, i, j;
int work[m], a[m];
bool finish[m];
r = 1;
for (i = 0; i < no1; i++)
{
finish[i] = false;//初始化京城均没得到足够资源数并完成
};
for (int i = 0; i < no2; i++)
{
work[i] = available[i];//work 表示课提供进程继续运行的各资源数
};
k = no1;
do
{
for (i = 0; i < no1; i++)
{
if (finish[i] == false)
{
f = 1;
for (j = 0; j < no2; j++)
{
if (need[i][j] > work[j])
{
f = 0;
}
if (f == 1)
{
finish[i] = true;
a[v++] = i;
for (int j = 0; j < no2; j++)
{
work[j] += allocation[i][j];
}
}
}
}
}
k--;
} while (k > 0);
f = 1;
for (int i = 0; i < no1; i++)
{
if (finish[i] == false)
{
f = 0;
break;
}
}
if (f == 0)
{
printf("系统处在不安全状态！");
r = 0;
}
else
{
printf("\n系统当前为安全状态，安全序列为：\n");
for (int i = 0; i < no1; i++)
{
printf("p%d", a[i]);
}
}
};
void print()
{
printf("\n");
printf("此时刻资源分配情况");
printf("进程名/号 | Max | Allocation | Need |\n");
for (int i = 0; i < no1; i++)
{
printf("p%d/%d", i, i);
for (int j = 0; j < no2; j++)
{
printf("%d", max[i][j]);
};
for (int j = 0; j < no2; j++)
{
printf("%d", allocation[i][j]);
};
for (int j = 0; j < no2; j++)
{
printf("%d", need[i][j]);
}
printf("\n");
}
printf("各类资源可利用的资源数为：");
for (int j = 0; j < no2; j++)
{
printf("%d", available[j]);
}
printf("\n");
}
