设计一种数据结构,以便在以下约束条件下在一台计算机上保留将来的作业。
1)每个作业都需要机器的k个时间单位。
2)机器一次只能执行一项工作。
3)时间是系统的一部分。未来的工作会在不同的时间出现。仅当在k个时间段内(之后和之前)没有任何保留的情况下,才保留对未来作业的保留
4)每当作业完成时(或其保留时间加k等于当前时间),就会将其从系统中删除。
例子:
Let time taken by a job (or k) be = 4
At time 0: Reservation request for a job at time 2 in
future comes in, reservation is done as machine
will be available (no conflicting reservations)
Reservations {2}
At time 3: Reservation requests at times 15, 7, 20 and 3.
Job at 7, 15 and 20 can be reserved, but at 3
cannot be reserved as it conflicts with a
reserved at 2.
Reservations {2, 7, 15, 20}
At time 6: Reservation requests at times 30, 17, 35 and 45
Jobs at 30, 35 and 45 are reserved, but at 17
cannot be reserved as it conflicts with a reserved
at 15.
Reservations {7, 15, 30, 35, 45}.
Note that job at 2 is removed as it must be finished by 6.
让我们考虑此任务的不同数据结构。
一种解决方案是使所有将来的保留按数组排序。可以使用二进制搜索在O(Logn)中完成检查冲突的时间复杂度,但是插入和删除操作需要O(n)时间。
此处不能使用散列,因为搜索不是精确搜索,而是在k个时间范围内的搜索。
这个想法是使用二进制搜索树来维护一组保留的作业。对于每个预订请求,仅在没有冲突的预订时才插入。插入作业时,请执行“在k个时间范围内检查”。如果从根插入路径上有ak远的节点,则拒绝保留请求,否则进行保留。
// A BST node to store future reservations
struct node
{
int time; // reservation time
struct node *left, *right;
};
// A utility function to create a new BST node
struct node *newNode(int item)
{
struct node *temp =
(struct node *)malloc(sizeof(struct node));
temp->time = item;
temp->left = temp->right = NULL;
return temp;
}
/* BST insert to process a new reservation request at
a given time (future time). This function does
reservation only if there is no existing job within
k time frame of new job */
struct node* insert(struct node* root, int time, int k)
{
/* If the tree is empty, return a new node */
if (root == NULL) return newNode(time);
// Check if this job conflicts with existing
// reservations
if ((time-k < root->time) && (time+k > root->time))
return root;
/* Otherwise, recur down the tree */
if (time < root->time)
root->left = insert(root->left, time, k);
else
root->right = insert(root->right, time, k);
/* return the (unchanged) node pointer */
return root;
}
删除作业是简单的BST删除操作。
正常的BST花费O(h)时间进行插入和删除操作。我们可以使用自平衡二进制搜索树(例如AVL,Red-Black,..)在O(Log n)时间内进行这两种操作。
这个问题是从麻省理工学院的讲座中采纳的。