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Knapsack problem/Continuous: Difference between revisions
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<br>Some amount of code was added to format the output better.
<lang rexx>/*REXX pgm solves the continuous burglar's knapsack problem; items with weight and value*/
@.= /*═══════ name weight value ══════*/
@.1 = 'flitch 4 30 '
@.2 = 'beef 3.8 36 '
@.3 = 'pork 5.4 43 '
@.4 = 'greaves 2.4 45 '
@.5 = 'brawn 2.5 56 '
@.6 = 'welt 3.7 67 '
@.7 = 'ham 3.6 90 '
@.8 = 'salami 3 95 '
@.9 = 'sausage 5.9 98 '
parse arg maxW d . /*get possible arguments from the C.L. */
if maxW=='' | maxW=="," then maxW=15 /*the burglar's knapsack maximum weight*/
if d=='' | d=="," then d= 3 /*
wL=d+length('weight'); nL=d+length("total weight");
do #=1 while @.#\=='';
▲ end /*#*/ /* [↑] assign item to separate lists. */
call show 'unsorted item list' /*display the header and the @ list.*/
call sortD /*invoke descemdomg sort
call hdr "burglar's knapsack contents"
do j=1 for # while totW<maxW; f=1 /*process the items. */
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totW=totW+w.j*f; totV=totV+v.j*f /*add it ───► totals. */
call syf left(word('{all}',1+(f\==1)),5) n.j, w.j*f, v.j*f
end /*j*/
call sep; say
call sy left('total weight', nL, "─"), $(format(totW,,d))
call sy left('total value', nL, "─"), , $(format(totV,,d))
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do k=s-1 by -1 to 1 while v.k/w.k<u/!; ?=k+1; n.?=n.k; w.?=w.k;v.?=v.k;end
?=k+1; n.?=a; w.?=!; v.?=u
end /*s*/; return /* ↑↑↑ algorithm is OK for small arrays*/
/*──────────────────────────────────────────────────────────────────────────────────────*/
hdr: say; say; say center(arg(1),50,'─'); say; call title; call sep; return
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