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Priority queue: Difference between revisions
Moved to correct location by language name
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1, Solve RC tasks
</pre>
=={{header|Clojure}}==
<syntaxhighlight lang="clojure">user=> (use 'clojure.data.priority-map)
; priority-map can be used as a priority queue
user=> (def p (priority-map "Clear drains" 3, "Feed cat" 4, "Make tea" 5, "Solve RC tasks" 1))
#'user/p
user=> p
{"Solve RC tasks" 1, "Clear drains" 3, "Feed cat" 4, "Make tea" 5}
; You can use assoc or conj to add items
user=> (assoc p "Tax return" 2)
{"Solve RC tasks" 1, "Tax return" 2, "Clear drains" 3, "Feed cat" 4, "Make tea" 5}
; peek to get first item, pop to give you back the priority-map with the first item removed
user=> (peek p)
["Solve RC tasks" 1]
; Merge priority-maps together
user=> (into p [["Wax Car" 4]["Paint Fence" 1]["Sand Floor" 3]])
{"Solve RC tasks" 1, "Paint Fence" 1, "Clear drains" 3, "Sand Floor" 3, "Wax Car" 4, "Feed cat" 4, "Make tea" 5}</syntaxhighlight>
=={{header|CLU}}==
This is a priority queue based on a binary heap. It uses CLU's dynamic
array to store the data.
There are no intrinsic limits on what kind of data can be used for the
priority or the values themselves, except that the priority datatype
must support the less-than operator.
<syntaxhighlight lang="clu">prio_queue = cluster [P, T: type] is new, empty, push, pop
where P has lt: proctype (P,P) returns (bool)
item = struct[prio: P, val: T]
rep = array[item]
new = proc () returns (cvt)
return (rep$create(0))
end new
empty = proc (pq: cvt) returns (bool)
return (rep$empty(pq))
end empty
parent = proc (k: int) returns (int)
return ((k-1)/2)
end parent
left = proc (k: int) returns (int)
return (2*k + 1)
end left
right = proc (k: int) returns (int)
return (2*k + 2)
end right
swap = proc (pq: rep, a: int, b: int)
temp: item := pq[a]
pq[a] := pq[b]
pq[b] := temp
end swap
min_heapify = proc (pq: rep, k: int)
l: int := left(k)
r: int := right(k)
smallest: int := k
if l < rep$size(pq) cand pq[l].prio < pq[smallest].prio then
smallest := l
end
if r < rep$size(pq) cand pq[r].prio < pq[smallest].prio then
smallest := r
end
if smallest ~= k then
swap(pq, k, smallest)
min_heapify(pq, smallest)
end
end min_heapify
push = proc (pq: cvt, prio: P, val: T)
rep$addh(pq, item${prio: prio, val: val})
i: int := rep$high(pq)
while i ~= 0 cand pq[i].prio < pq[parent(i)].prio do
swap(pq, i, parent(i))
i := parent(i)
end
end push
pop = proc (pq: cvt) returns (P, T) signals (empty)
if empty(up(pq)) then signal empty end
if rep$size(pq) = 1 then
i: item := rep$remh(pq)
return (i.prio, i.val)
end
root: item := pq[0]
pq[0] := rep$remh(pq)
min_heapify(pq, 0)
return (root.prio, root.val)
end pop
end prio_queue
start_up = proc ()
% use ints for priority and strings for data
prioq = prio_queue[int,string]
% make the priority queue
pq: prioq := prioq$new()
% add some tasks
prioq$push(pq, 3, "Clear drains")
prioq$push(pq, 4, "Feed cat")
prioq$push(pq, 5, "Make tea")
prioq$push(pq, 1, "Solve RC tasks")
prioq$push(pq, 2, "Tax return")
% print them all out in order
po: stream := stream$primary_output()
while ~prioq$empty(pq) do
prio: int task: string
prio, task := prioq$pop(pq)
stream$putl(po, int$unparse(prio) || ": " || task)
end
end start_up</syntaxhighlight>
{{out}}
<pre>1: Solve RC tasks
2: Tax return
3: Clear drains
4: Feed cat
5: Make tea</pre>
=={{header|COBOL}}==
Line 2,583 ⟶ 2,715:
+0000000006 EAT SCONES.
</pre>
=={{header|CoffeeScript}}==
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