/**
* Utility function which, given two pointers into disjoint circularly- linked lists, merges the
* two lists together into one circularly-linked list in O(1) time. Because the lists may be
* empty, the return value is the only pointer that's guaranteed to be to an element of the
* resulting list.
*
* <p>This function assumes that one and two are the minimum elements of the lists they are in,
* and returns a pointer to whichever is smaller. If this condition does not hold, the return
* value is some arbitrary pointer into the doubly-linked list.
*
* @param one A pointer into one of the two linked lists.
* @param two A pointer into the other of the two linked lists.
* @return A pointer to the smallest element of the resulting list.
*/
private static <T> Entry<T> mergeLists(Entry<T> one, Entry<T> two) {
/* There are four cases depending on whether the lists are null or not.
* We consider each separately.
*/
if (one == null && two == null) { // Both null, resulting list is null.
return null;
} else if (one != null && two == null) { // Two is null, result is one.
return one;
} else if (one == null && two != null) { // One is null, result is two.
return two;
} else { // Both non-null; actually do the splice.
/* This is actually not as easy as it seems. The idea is that we'll
* have two lists that look like this:
*
* +----+ +----+ +----+
* | |--N->|one |--N->| |
* | |<-P--| |<-P--| |
* +----+ +----+ +----+
*
*
* +----+ +----+ +----+
* | |--N->|two |--N->| |
* | |<-P--| |<-P--| |
* +----+ +----+ +----+
*
* And we want to relink everything to get
*
* +----+ +----+ +----+---+
* | |--N->|one | | | |
* | |<-P--| | | |<+ |
* +----+ +----+<-\ +----+ | |
* \ P | |
* N \ N |
* +----+ +----+ \->+----+ | |
* | |--N->|two | | | | |
* | |<-P--| | | | | P
* +----+ +----+ +----+ | |
* ^ | | |
* | +-------------+ |
* +-----------------+
*
*/
Entry<T> oneNext = one.mNext; // Cache this since we're about to overwrite it.
one.mNext = two.mNext;
one.mNext.mPrev = one;
two.mNext = oneNext;
two.mNext.mPrev = two;
/* Return a pointer to whichever's smaller. */
return one.mPriority < two.mPriority ? one : two;
}
}
/**
* Cuts a node from its parent. If the parent was already marked, recursively cuts that node from
* its parent as well.
*
* @param entry The node to cut from its parent.
*/
private void cutNode(Entry<T> entry) {
/* Begin by clearing the node's mark, since we just cut it. */
entry.mIsMarked = false;
/* Base case: If the node has no parent, we're done. */
if (entry.mParent == null) return;
/* Rewire the node's siblings around it, if it has any siblings. */
if (entry.mNext != entry) { // Has siblings
entry.mNext.mPrev = entry.mPrev;
entry.mPrev.mNext = entry.mNext;
}
/* If the node is the one identified by its parent as its child,
* we need to rewrite that pointer to point to some arbitrary other
* child.
*/
if (entry.mParent.mChild == entry) {
/* If there are any other children, pick one of them arbitrarily. */
if (entry.mNext != entry) {
entry.mParent.mChild = entry.mNext;
}
/* Otherwise, there aren't any children left and we should clear the
* pointer and drop the node's degree.
*/
else {
entry.mParent.mChild = null;
}
}
/* Decrease the degree of the parent, since it just lost a child. */
--entry.mParent.mDegree;
/* Splice this tree into the root list by converting it to a singleton
* and invoking the merge subroutine.
*/
entry.mPrev = entry.mNext = entry;
mMin = mergeLists(mMin, entry);
/* Mark the parent and recursively cut it if it's already been
* marked.
*/
if (entry.mParent.mIsMarked) cutNode(entry.mParent);
else entry.mParent.mIsMarked = true;
/* Clear the relocated node's parent; it's now a root. */
entry.mParent = null;
}
/**
* Dequeues and returns the minimum element of the Fibonacci heap. If the heap is empty, this
* throws a NoSuchElementException.
*
* @return The smallest element of the Fibonacci heap.
* @throws NoSuchElementException If the heap is empty.
*/
public Entry<T> dequeueMin() {
/* Check for whether we're empty. */
if (isEmpty()) throw new NoSuchElementException("Heap is empty.");
/* Otherwise, we're about to lose an element, so decrement the number of
* entries in this heap.
*/
--mSize;
/* Grab the minimum element so we know what to return. */
Entry<T> minElem = mMin;
/* Now, we need to get rid of this element from the list of roots. There
* are two cases to consider. First, if this is the only element in the
* list of roots, we set the list of roots to be null by clearing mMin.
* Otherwise, if it's not null, then we write the elements next to the
* min element around the min element to remove it, then arbitrarily
* reassign the min.
*/
if (mMin.mNext == mMin) { // Case one
mMin = null;
} else { // Case two
mMin.mPrev.mNext = mMin.mNext;
mMin.mNext.mPrev = mMin.mPrev;
mMin = mMin.mNext; // Arbitrary element of the root list.
}
/* Next, clear the parent fields of all of the min element's children,
* since they're about to become roots. Because the elements are
* stored in a circular list, the traversal is a bit complex.
*/
if (minElem.mChild != null) {
/* Keep track of the first visited node. */
Entry<?> curr = minElem.mChild;
do {
curr.mParent = null;
/* Walk to the next node, then stop if this is the node we
* started at.
*/
curr = curr.mNext;
} while (curr != minElem.mChild);
}
/* Next, splice the children of the root node into the topmost list,
* then set mMin to point somewhere in that list.
*/
mMin = mergeLists(mMin, minElem.mChild);
/* If there are no entries left, we're done. */
if (mMin == null) return minElem;
/* Next, we need to coalsce all of the roots so that there is only one
* tree of each degree. To track trees of each size, we allocate an
* ArrayList where the entry at position i is either null or the
* unique tree of degree i.
*/
List<Entry<T>> treeTable = new ArrayList<Entry<T>>();
/* We need to traverse the entire list, but since we're going to be
* messing around with it we have to be careful not to break our
* traversal order mid-stream. One major challenge is how to detect
* whether we're visiting the same node twice. To do this, we'll
* spent a bit of overhead adding all of the nodes to a list, and
* then will visit each element of this list in order.
*/
List<Entry<T>> toVisit = new ArrayList<Entry<T>>();
/* To add everything, we'll iterate across the elements until we
* find the first element twice. We check this by looping while the
* list is empty or while the current element isn't the first element
* of that list.
*/
for (Entry<T> curr = mMin; toVisit.isEmpty() || toVisit.get(0) != curr; curr = curr.mNext)
toVisit.add(curr);
/* Traverse this list and perform the appropriate unioning steps. */
for (Entry<T> curr : toVisit) {
/* Keep merging until a match arises. */
while (true) {
/* Ensure that the list is long enough to hold an element of this
* degree.
*/
while (curr.mDegree >= treeTable.size()) treeTable.add(null);
/* If nothing's here, we're can record that this tree has this size
* and are done processing.
*/
if (treeTable.get(curr.mDegree) == null) {
treeTable.set(curr.mDegree, curr);
break;
}
/* Otherwise, merge with what's there. */
Entry<T> other = treeTable.get(curr.mDegree);
treeTable.set(curr.mDegree, null); // Clear the slot
/* Determine which of the two trees has the smaller root, storing
* the two tree accordingly.
*/
Entry<T> min = (other.mPriority < curr.mPriority) ? other : curr;
Entry<T> max = (other.mPriority < curr.mPriority) ? curr : other;
/* Break max out of the root list, then merge it into min's child
* list.
*/
max.mNext.mPrev = max.mPrev;
max.mPrev.mNext = max.mNext;
/* Make it a singleton so that we can merge it. */
max.mNext = max.mPrev = max;
min.mChild = mergeLists(min.mChild, max);
/* Reparent max appropriately. */
max.mParent = min;
/* Clear max's mark, since it can now lose another child. */
max.mIsMarked = false;
/* Increase min's degree; it now has another child. */
++min.mDegree;
/* Continue merging this tree. */
curr = min;
}
/* Update the global min based on this node. Note that we compare
* for <= instead of < here. That's because if we just did a
* reparent operation that merged two different trees of equal
* priority, we need to make sure that the min pointer points to
* the root-level one.
*/
if (curr.mPriority <= mMin.mPriority) mMin = curr;
}
return minElem;
}
