请用python帮我写一个代码，设计有 n 个进程共享 1 个系统资源的系统，基于银行家算法和安全性算法，进 程可动态的申请和释放资源，系统按各进程的申请动态的分配资源。要求打印资源动态分配矩阵

在Python中，我们可以使用`multiprocessing`库模拟多进程环境，并结合一些数据结构来实现基于银行家算法的安全性检查。以下是简化的示例代码：

import heapq
from multiprocessing import Process, Manager

# 定义资源列表
resources = [0] * 4  # 假设有4种资源，每种初始都是0
limits = [10, 8, 6, 4]  # 每种资源的最大可用量
processes = []

def allocate_resources(process_id, needed):
    global resources, limits
    if can_allocate(process_id, needed):
        for i in range(len(needed)):
            resources[i] += needed[i]
            print(f"Process {process_id}: Requested {i+1} resource(s) ({needed[i]}), current allocation: {resources[i]}")
        return True
    else:
        print(f"Process {process_id}: Allocation denied due to insufficient resources.")
        return False

def release_resources(process_id, released):
    global resources
    for i in range(len(released)):
        resources[i] -= released[i]
        print(f"Process {process_id}: Released {i+1} resource(s) ({released[i]}), new allocation: {resources[i]}")

def can_allocate(process_id, needed):
    global resources, limits
    available = []
    for i, limit in enumerate(limits):
        if resources[i] + needed[i] <= limit:
            available.append(limit - resources[i])
    if not available:
        return False
    # 使用堆来查找最有利的资源组合
    min_heap = [-value for value in available]  # 负数表示最小值，因为我们需要最大可用资源
    heapq.heapify(min_heap)
    
    while len(min_heap) > 0 and sum(needed) > 0:
        # 如果当前分配的资源可以满足所有请求，则接受分配
        if -heapq.heappop(min_heap) >= needed[0]:
            for i in range(len(needed)):
                if needed[i] > 0:
                    min_heap.append(-(available[i] - needed[i]))
                    break
            processes[process_id].running = False
            sum_needed = sum(needed)
        else:
            # 否则，回退并尝试下一个最小可用资源
            pass
    
    return len(min_heap) == 0 or processes[process_id].running  # 返回是否分配成功

if __name__ == "__main__":
    manager = Manager()
    processes_data = manager.list()  # 存储进程信息

    for i in range(5):  # 假设有5个进程
        process = Process(target=run_process, args=(i, processes_data))
        process.start()
        processes.append(process)

    # 等待所有进程结束
    for process in processes:
        process.join()

def run_process(process_id, processes_data):
    running = True  # 标记进程是否还在运行
    while running:
        request = input(f"Enter resources needed for Process {process_id} (separated by comma, leave blank to finish): ")
        if request == "":
            running = False
        else:
            requested = list(map(int, request.split(",")))
            if allocate_resources(process_id, requested):
                processes_data.append((process_id, requested))
            else:
                break

    release_resources(process_id, requested)  # 无论分配成功与否，都需要释放已申请的资源