RosettaCodeData/Task/Knapsack-problem-Bounded/Phix/knapsack-problem-bounded-2....

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sequence items = {
{"map", 9, 150, 1},
{"compass", 13, 35, 1},
{"water", 153, 200, 2},
{"sandwich", 50, 60, 2},
{"glucose", 15, 60, 2},
{"tin", 68, 45, 3},
{"banana", 27, 60, 3},
{"apple", 39, 40, 3},
{"cheese", 23, 30, 1},
{"beer", 52, 10, 3},
{"suntan cream", 11, 70, 1},
{"camera", 32, 30, 1},
{"T-shirt", 24, 15, 2},
{"trousers", 48, 10, 2},
{"umbrella", 73, 40, 1},
{"waterproof trousers", 42, 70, 1},
{"waterproof overclothes",43, 75, 1},
{"note-case", 22, 80, 1},
{"sunglasses", 7, 20, 1},
{"towel", 18, 12, 2},
{"socks", 4, 50, 1},
{"book", 30, 10, 2},
};
sequence {names,weights,points,counts} = columnize(items)
constant n = length(items)
function knapsack(int w)
int v
-- m is the achievable points matrix:
-- Note that Phix uses 1-based indexes, so m[1][1]
-- actually holds points for 0 items of weight 0,
-- and m[n+1][w+1] is for n items at weight w.
seq m = repeat(repeat(0,w+1),n+1)
for i=1 to n do
for j=1 to w+1 do -- (0 to w really)
m[i+1][j] = m[i][j]
for k=1 to counts[i] do
if k*weights[i]>j-1 then
exit
end if
v = m[i][j-k*weights[i]]+k*points[i]
if v>m[i+1][j] then
m[i+1][j] = v
end if
end for
end for
end for
seq s = repeat(0,n)
int j = w+1 -- (w -> 0 really)
for i=n+1 to 2 by -1 do -- (n to 1 really)
v = m[i][j]
int k = 0
while v!=m[i-1][j]+k*points[i-1] do
s[i-1] += 1
j -= weights[i-1]
k += 1
end while
end for
return s
end function
int tc = 0, tw = 0, tv = 0
seq s = knapsack(400)
for i=1 to n do
int si = s[i]
if si then
printf(1,"%-22s %5d %5d %5d\n", {names[i], si, si*weights[i], si*points[i]})
tc += si
tw += si*weights[i]
tv += si*points[i]
end if
end for
printf(1,"%-22s %5d %5d %5d\n", {"count, weight, points:", tc, tw, tv})