C语言实现银行家算法

#define _CRT_SECURE_NO_WARNINGS
#include<stdio.h>
#include<malloc.h>
#include<string.h>
#include<windows.h>

void Init(int **arr,int row, int col)         //赋值
{       
	//int** p= (int **)malloc(sizeof(int *)*row);//申请一组一维指针空间。
	for (int i = 0; i < row; i++)
		arr[i] = (int *)malloc(sizeof(int)* col); //对于每个一维指针，申请一行数据的空间。
	for (int i = 0; i < col; ++i)
		printf("   %c", 65 + i);
	printf("\n");
	for (int i = 0; i < row; i++)
	{
		printf("P%d ", i);
		for (int j = 0; j < col; j++)
		{
			scanf("%d", &arr[i][j]);
		}
	}
	
}

void Print_double(int **arr, int row, int col)       //打印矩阵
{
	for (int i = 0; i < col; ++i)
	{
		printf("   %c", 65 + i);
	}
	printf("\n");
	for (int i = 0; i < row; i++)
	{
		printf("P%d", i);
		for (int j = 0; j < col; j++)//这个循环可以输出一行元素。
			printf(" %d ", arr[i][j]);//输出i行j列的元素。
		printf("\n");//每行输入结束加换行符。
	}
}

void Print_sing(int* arr, int num)         //打印单表
{
	for (int i = 0; i < num; ++i)
	{
		printf(" %d ", arr[i]);
	}
}
int ID(int* arr, int procsee)    //1 成功        -1失败   //输出序列号
{
	int count = 0;
	int sum = 0;
	int ret = -1;
	for (int i = 0; i < procsee; ++i)
	{
		sum += i;
	}
	for (int k = 0; k < procsee; ++k)
	{
		++count;
	}
	if (count == sum)
	{
		ret = 1;
		printf("安全序列号为：\n");
		for (int i = 0; i < procsee; ++i)
		{
			if (i == procsee - 1)
				printf("P%d", arr[i]);
			else
			printf(" P%d ->", arr[i]);
		}
		printf("\n");
		return ret;
	}
	return ret;
}
         //Allocation  //Need  //Work ==Avaliable
void Safe(int **arr1, int **arr2,int* arr3,int *safe,int process,int resource)    //安全序列
{
		int true[10] = { 0 };
		int Work[10] = { 0 };
		for (int i = 0; i < resource; ++i)
		{
			Work[i] = arr3[i];
		}
	int safe_num = 0;
	int i = 0;
	int n = 0;
	while (safe_num != process)
	{
		if (true[i] == 0)
		{
			int finish = 0;
			for (int j = 0; j < resource; ++j)
			{
				if (arr2[i][j] <= Work[j])
				{
					++finish;
				}
			}
			if (finish == resource)
			{
				for (int k = 0; k < resource; ++k)
				{
					Work[k] += arr1[i][k];
				}
				true[i] = 1;
				safe[n] = i;
				++safe_num;
				i = -1;
				++n;
			}
		}
		++i;
	}
	ID(safe, process);
}
           //AL          //NE      //AV
void Dis(int** arr1, int** arr2, int* arr3, int* safe, int process, int resource)  //分配
{
	int p1 = process;
	int p2 = resource;
	int Request[10] = { 0 };
			int num = 0;
			int r2 = 0, r3 = 0;
			printf("请输入想要分配的进程序号(0 -- %d):\n", process - 1);
			scanf("%d", &num);
			printf("请输入请求的资源数目: \n");                 //获得Request
			for (int i = 0; i < resource; ++i)
			{
				scanf("%d", &Request[i]);
			}
			printf("\n");
			for (int i = 0; i < resource; ++i)        // Avaliable = Avaliable - Request  Allocation = Allocation + Request
			{                                             // Need = Need -Request
				arr1[num][i] += Request[i];
				arr2[num][i] -= Request[i];
				arr3[i] -= Request[i];
				if (arr1[num][i] < 0 || arr2[num][i] < 0 || arr3[i] < 0)
					r3 = 1;
			}
			if (r3 == 0)
			{
				for (int i = 0; i < process; ++i)            // 把Avaliable 和 Need 进行比较    
				{
					int r1 = 0;
					for (int j = 0; j < resource; ++j)
					{
						if (arr3[j] >= arr2[i][j])             //若Avaliable  >= Need        ++r1
						{
							++r1;
						}
					}
					if (r1 == resource)                           //若r1 == resource  则说明有一个Need 能够被全部满足 则++r2
					{
						++r2;
					}
				}
				if (r2 != 0)       
				{
					printf("已占用的资源矩阵为：\n");
					Print_double(arr1, process, resource);
					printf("需要的资源矩阵为：\n");
					Print_double(arr2, process, resource);
					printf("\n");
					printf("可利用的资源为：\n");
					Print_sing(arr3, resource);
					printf("\n");
					Safe(arr1, arr2, arr3, safe, p1, p2);
				}
				else
				{
					printf("可利用资源小于系统所需资源，分配失败，展示原本的资源\n");
					for (int i = 0; i < resource; ++i)
					{
						arr1[num][i] -= Request[i];
						arr2[num][i] += Request[i];
						arr3[i] += Request[i];
					}
					printf("已占用的资源矩阵为：\n");
					Print_double(arr1, process, resource);
					printf("需要的资源矩阵为：\n");
					Print_double(arr2, process, resource);
					printf("\n");
					printf("可利用的资源为：\n");
					Print_sing(arr3, resource);
				}
			}
			else
				printf("可利用资源小于系统所需资源，分配失败\n");
			for (int i = 0; i < resource; ++i)       
			{
				arr1[num][i] -= Request[i];
				arr2[num][i] += Request[i];
				arr3[i] += Request[i];
			}

}


void menu()
{
	printf("*************************1. 输入数据 ************************\n");
	printf("*************************2. 展示数据 ************************\n");
	printf("*************************3. 安全序列 ************************\n");
	printf("*************************4. 资源分配 ************************\n");
	printf("*************************5.  退出    *************************\n");
}

int main()
{
	int process = 0;
	int resource = 0;
	int safe[10] = { 0 };
	int input;
	int Work[10] = { 0 };
	int Avaliable[10] = { 0 };
	int** Allocation = (int**)malloc(sizeof(int *)* process);
	int** Need = (int**)malloc(sizeof(int *)* process);
	while (1)
	{
		menu();
		printf("请输入你的选项：\n");
		scanf("%d", &input);
		switch (input)
		{
		case 1:	printf("请输入进程数量和资源数量：");
			scanf("%d%d", &process, &resource);
			if (process <= 0 || resource <= 0)
				printf("输入错误，请重新输入!\n");
			printf("请输入已占用的资源数组\n");
			Init(Allocation, process, resource);
			printf("请输入需要的资源数组\n");
			Init(Need, process, resource);
			printf("请输入可利用资源数目: \n");
			for (int i = 0; i < resource; ++i)
			{
				scanf("%d", &Avaliable[i]);
			}
			for (int i = 0; i < resource; ++i)
			{
				Work[i] = Avaliable[i];
			}
			break;
		case 2: printf("已占用的资源为:\n");
			Print_double(Allocation, process, resource);
			printf("\n");
			printf("系统所需的资源为:\n");
			Print_double(Need, process, resource);
			printf("\n");
			printf("你输入的可利用资源数目为：\n");
			for (int i = 0; i < resource; ++i)
			{
				printf("%d ", Avaliable[i]);
			}
			printf("\n");
			break;
		case 3:Safe(Allocation, Need, Avaliable, safe, process, resource);
			break;
		case 4:	Dis(Allocation, Need, Avaliable, safe, process, resource);
			break;
		case 5: exit(0);
		default:
			break;
		}
	}
	return 0;
}

结果如下：

来源：CSDN

作者：丶独醒

链接：https://blog.csdn.net/m1059247324/article/details/103848199