用c语言数据结构写长包房

在C语言中，我们可以使用链表数据结构来模拟长包房的概念，因为链表可以动态地存储和管理房间的分配情况。下面是一个简单的例子，我们将创建一个`Room`节点结构体来表示每个房间，并使用链表维护所有可用的房间。

#include <stdio.h>
#include <stdlib.h>

// 定义一个Room结构体，包含房间ID、是否已被预订的标志
typedef struct Room {
    int roomID;
    int isBooked; // 0 表示空闲，1 表示已预订
    struct Room* next; // 指向下一个房间
} Room;

// 创建新房间并将其添加到链表尾部
Room* createRoom(int id) {
    Room* newRoom = (Room*)malloc(sizeof(Room));
    if (!newRoom) {
        printf("Memory allocation failed.\n");
        return NULL;
    }
    newRoom->roomID = id;
    newRoom->isBooked = 0;
    newRoom->next = NULL;
    return newRoom;
}

// 验证并设置房间是否预订
void bookRoom(Room** head, int roomID) {
    Room* current = *head;
    while (current != NULL) {
        if (current->roomID == roomID && !current->isBooked) {
            current->isBooked = 1;
            printf("Room %d booked successfully.\n", roomID);
            break;
        }
        current = current->next;
    }
}

// 打印当前所有房间的状态
void displayRooms(Room* head) {
    Room* current = head;
    while (current != NULL) {
        printf("Room %d: %s\n", current->roomID, current->isBooked ? "Occupied" : "Available");
        current = current->next;
    }
}

int main() {
    // 初始化链表为空
    Room* roomList = NULL;

    // 添加一些房间
    roomList = createRoom(1); // 空闲
    roomList = createRoom(2); // 空闲
    roomList = createRoom(3); // 空闲

    // 预订房间
    bookRoom(&roomList, 2);

    // 显示当前房间状态
    displayRooms(roomList);

    return 0;
}