用c++编程设计3～4个并发进程共享系统的10个同类不可抢占的资源。各进程动态进行资源的申请和释放。

由于这是一个并发进程的问题，我们需要使用线程来实现。以下是一个示例程序，其中有3个进程，每个进程动态地申请和释放系统的10个资源。为了避免死锁，每个进程在申请资源时都会尝试申请全部资源，如果没有足够的资源，则会释放已经获得的资源并重新尝试申请。同时，每个进程在释放资源时也会检查是否还有其他进程在等待这些资源，如果有，则会通知等待的进程可以继续执行。

#include <iostream>
#include <thread>
#include <mutex>
#include <condition_variable>
#include <chrono>
#include <vector>

using namespace std;

const int NUM_RESOURCES = 10;

mutex mtx;
condition_variable cv;

vector<int> available(NUM_RESOURCES, 1);
vector<int> allocated(NUM_RESOURCES, 0);
vector<vector<int>> max_resources {
    {3, 1, 2, 1, 2, 1, 2, 1, 2, 1},
    {2, 1, 3, 1, 2, 1, 2, 1, 2, 1},
    {3, 2, 1, 1, 2, 1, 2, 1, 2, 1}
};

vector<vector<int>> need_resources(3, vector<int>(NUM_RESOURCES));

bool is_finished(vector<int> v) {
    for (int i = 0; i < NUM_RESOURCES; i++) {
        if (v[i] != 0) {
            return false;
        }
    }
    return true;
}

bool is_safe() {
    vector<int> work = available;
    vector<bool> finish(3, false);

    while (!is_finished(finish)) {
        bool found = false;
        for (int i = 0; i < 3; i++) {
            if (!finish[i]) {
                bool can_allocate = true;
                for (int j = 0; j < NUM_RESOURCES; j++) {
                    if (need_resources[i][j] > work[j]) {
                        can_allocate = false;
                        break;
                    }
                }
                if (can_allocate) {
                    found = true;
                    finish[i] = true;
                    for (int j = 0; j < NUM_RESOURCES; j++) {
                        work[j] += allocated[i][j];
                    }
                }
            }
        }
        if (!found) {
            return false;
        }
    }
    return true;
}

void request_resources(int process_id, vector<int> request) {
    unique_lock<mutex> lck(mtx);
    cout << "Process " << process_id << " is requesting resources." << endl;

    while (true) {
        bool can_allocate = true;
        for (int i = 0; i < NUM_RESOURCES; i++) {
            if (request[i] > available[i]) {
                can_allocate = false;
                break;
            }
        }
        if (can_allocate) {
            for (int i = 0; i < NUM_RESOURCES; i++) {
                available[i] -= request[i];
                allocated[process_id][i] += request[i];
                need_resources[process_id][i] = max_resources[process_id][i] - allocated[process_id][i];
            }
            if (is_safe()) {
                cout << "Process " << process_id << " has been granted resources." << endl;
                break;
            } else {
                for (int i = 0; i < NUM_RESOURCES; i++) {
                    available[i] += request[i];
                    allocated[process_id][i] -= request[i];
                    need_resources[process_id][i] = max_resources[process_id][i] - allocated[process_id][i];
                }
            }
        }
        cv.wait(lck);
    }
}

void release_resources(int process_id, vector<int> release) {
    unique_lock<mutex> lck(mtx);
    for (int i = 0; i < NUM_RESOURCES; i++) {
        available[i] += release[i];
        allocated[process_id][i] -= release[i];
        need_resources[process_id][i] = max_resources[process_id][i] - allocated[process_id][i];
    }
    cout << "Process " << process_id << " has released resources." << endl;
    cv.notify_all();
}

void process_1() {
    vector<int> request(NUM_RESOURCES, 0);
    vector<int> release(NUM_RESOURCES, 0);

    request[0] = 2;
    request[1] = 1;
    request[2] = 1;

    release[0] = 1;
    release[1] = 1;

    request_resources(0, request);
    this_thread::sleep_for(chrono::seconds(2));
    release_resources(0, release);
}

void process_2() {
    vector<int> request(NUM_RESOURCES, 0);
    vector<int> release(NUM_RESOURCES, 0);

    request[0] = 1;
    request[2] = 2;

    release[0] = 1;
    release[2] = 1;

    request_resources(1, request);
    this_thread::sleep_for(chrono::seconds(2));
    release_resources(1, release);
}

void process_3() {
    vector<int> request(NUM_RESOURCES, 0);
    vector<int> release(NUM_RESOURCES, 0);

    request[1] = 1;
    request[3] = 1;
    request[4] = 1;

    release[1] = 1;

    request_resources(2, request);
    this_thread::sleep_for(chrono::seconds(2));
    release_resources(2, release);
}

int main() {
    thread t1(process_1);
    thread t2(process_2);
    thread t3(process_3);

    t1.join();
    t2.join();
    t3.join();

    return 0;
}

在上面的示例程序中，我们使用了3个线程来模拟3个进程。每个进程都会动态地申请和释放系统的10个资源。在申请资源时，每个进程都会尝试申请全部资源，如果没有足够的资源，则会释放已经获得的资源并重新尝试申请。同时，在释放资源时，每个进程也会检查是否还有其他进程在等待这些资源，如果有，则会通知等待的进程可以继续执行。这样，我们就实现了一个简单的并发进程共享资源的例子