Priority queue: Difference between revisions
Added XPL0 example.
SqrtNegInf (talk | contribs) m (→{{header|Perl}}: future-proof for 5.36) |
(Added XPL0 example.) |
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[Tax return, 2]
[Solve RC tasks, 1]
</pre>
=={{header|XPL0}}==
The highest priority item is the one with the minimum number, as in 1st priority.
<syntaxhighlight lang "XPL0">def PQSize = 10; \Maximum number of items priority queue can hold
int PQ(PQSize*2), PQI;
func Remove; \Remove and return item with highest priority
int Min, I, MinI, Item;
[if PQI <= 0 then return 0;
Min:= -1>>1; I:= PQI;
while I > 0 do
[I:= I-2;
if PQ(I) < Min then
[Min:= PQ(I);
MinI:= I;
];
];
Item:= PQ(MinI+1); \get highest priority Item
PQI:= PQI-2;
PQ(MinI):= PQ(PQI); \replace that Item with last item
PQ(MinI+1):= PQ(PQI+1);
return Item;
];
proc Insert(Priority, Item); \Insert item into priority queue
int Priority, Item;
[if PQI >= PQSize*2 then return;
PQ(PQI):= Priority;
PQ(PQI+1):= Item;
PQI:= PQI+2;
];
int Items, I;
[Items:= [
[3, "Clear drains"],
[4, "Feed cat"],
[5, "Make tea"],
[1, "Solve RC tasks"],
[2, "Tax return"] ];
PQI:= 0;
for I:= 0 to 5-1 do
Insert(Items(I,0), Items(I,1));
for I:= 0 to 5-1 do
[Text(0, Remove); CrLf(0)];
]</syntaxhighlight>
{{out}}
<pre>
Solve RC tasks
Tax return
Clear drains
Feed cat
Make tea
</pre>
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