Given a list of N arbitrarily permutated integers from set {1..N}.
Need to find the ordering numbers of each integer in the LONGEST
increasing sequence to which this number belongs. Sample:
List:
[4, 5, 6, 1, 2, 7, 3]
Corresponding ordering numbers:
[1, 2, 3, 1, 2, 4, 3]
Details:
e.g. number 7 belongs to increasing sequence 1, 2, 7;
but this sequence is not the LONGEST sequence for "7".
The longest sequence for the 7 is 4, 5, 6, 7.
So, the 7's ordering number in this sequence is 4.
The salt of the thing is to do this with an O(n*log(n))
algorithm!
The straightforward O(n^2) algorithm is toooo slooow.
Any ideas?
Sep 7 '05
39  3121 
n00m wrote: Oh!
Seems you misunderstand me!
See how the last block in your code should look:
for tc in range(10):
_ = stdin.readline( )
sequence = [int(ch) for ch in stdin.readline( ).split()]
supers = supernumbers(se quence)
print len(supers)
for i in supers:
print i,
Ah, I misunderstood the input. I thought they'd run it once for
each of their test cases. Handling exactly ten inputs sucks, so
how about :
while True:
if not stdin.readline( ).strip():
break
sequence = [int(ch) for ch in stdin.readline( ).split()]
supers = supernumbers(se quence)
print len(supers)
for i in supers:
print i,
print
--
--Bryan
n00m wrote: It also timed out:(
Could be. Yet you did write:
It's incredibly fast!
I never intended to submit this program for competition. The
contest ranks in speed order, and there is no way Python can
compete with truly-compiled languages on such low-level code.
I'd bet money that the algorithm I used (coded in C) can run
with the winners. I also think I'd wager that the Python version
outright trumps them on code size.
My first version bombed for the zero-length sequence. That was a
mistake, sorry, but it may not be one of their test-cases. I
wonder how many of the accepted entries would perform properly.
--
--Bryan
Bryan Olson wrote: Could be. Yet you did write: > It's incredibly fast!
I just was obliged to exclaim "It's incredibly fast!"
because I THOUGHT your first version handled ALL TEN
testcases from the input. But the code read from the
*20-lines* input *ONLY 2* its first lines.
Usually they place heavy data testcase(s) at the end
of the (whole) input. Like this:
3
2 3 1
7
4 5 6 1 2 7 3
....
....
....
100000
456 2 6789 ... ... ... ... ... 55444 1 ... 234
Surely producing an answer for list [2, 3, 1] will
be "incredibly fast" for ANY language and for ANY
algorithm.
My first version bombed for the zero-length sequence. That was a
mistake, sorry, but it may not be one of their test-cases.
In my turn I can bet there's not an empty sequence testcase in the
input.
I
wonder how many of the accepted entries would perform properly.
Info of such kind they keep in secret (along with what the input
data are).
One more thing.
They (the e-judge's admins) are not gods and they don't warrant
that if they put 9 sec timelimit for a problem then this problem
can be "solved" in all accepted languages (e.g. in Python).
I never intended to submit this program for competition.
"Competitio n" is not quite relevant word here. It just LOOKS as
if it is a "regular" competetion. There nobody blames anybody.
Moreover, judging on my own experience, there nobody is even
interested in anybody. It's just a fun (but very useful fun).
Bryan;
My own version also timed out.
And now I can tell: it's incredibly SLOW.
Nevertheless it would be interesting to compare
speed of my code against yours. I can't do it myself
because my Python is of 2.3.4 version.
Just uncomment "your" part.
import bisect
def oops(w,a,b):
for m in w:
j=bisect.bisect _left(a,m)
a.insert(j,m)
b.insert(j,max( b[:j]+[0])+1)
def n00m(n,w):
a,b=[],[]
oops(w,a,b)
v=map(lambda x: -x, w[::-1])
c,d=[],[]
oops(v,c,d)
e=map(sum, zip(b, d[::-1]))
mx=max(e)
f=[]
for i in xrange(n):
if e[i]==mx:
f.append(i+1)
print len(f)
def one_way(seq):
n = len(seq)
dominators = [n + 1] * (n * 1)
score = [None] * n
end = 0
for (i, x) in enumerate(seq):
low, high = 0, end
while high - low > 10:
mid = (low + high) >> 1
if dominators[mid] < x:
low = mid + 1
else:
high = mid + 1
while dominators[low] < x:
low += 1
dominators[low] = x
score[i] = low
end = max(end, low + 1)
return score
def supernumbers(se q):
forscore = one_way(seq)
opposite = [len(seq) - x for x in reversed(seq)]
backscore = reversed(one_wa y(opposite))
score = map(sum, zip(forscore, backscore))
winner = max(score + [0])
return sorted([seq[i] for i in range(len(seq)) if score[i] ==
winner])
def b_olson(sequenc e):
supers = supernumbers(se quence)
print len(supers)
import random, time
n=50000
w=range(1,n+1)
random.shuffle( w)
t=time.time()
n00m(n,w)
print 'n00m:',time.ti me()-t
"""
t=time.time()
b_olson(w)
print 'b_olson:',time .time()-t
"""
[Bryan Olson, on the problem at http://spoj.sphere.pl/problems/SUPPER/
] I never intended to submit this program for competition. The
contest ranks in speed order, and there is no way Python can
compete with truly-compiled languages on such low-level code.
I'd bet money that the algorithm I used (coded in C) can run
with the winners. I also think I'd wager that the Python version
outright trumps them on code size.
Oh, it's not that bad <wink>. I took a stab at a Python program for
this, and it passed (3.44 seconds). It just barely made it onto the
list of "best" solutions, which I also guess is ranked by elapsed
runtime. The Java program right above it took 2.91 seconds, but ate
more than 27x as much RAM ;-)
I didn't make any effort to speed this, beyond picking a reasonable
algorithm, so maybe someone else can slash the runtime (while I
usually enjoy such silliness, I can't make time for it at present).
I'll include the code below.
Alas, without access to the input data they use, it's hard to guess
what might be important in their data. On my home box, chewing over
random 100,000-element permutations took less than a second each
(including the time to generate them); I'm pretty sure they're using
slower HW than mine (3.4 GHz P5).
My first version bombed for the zero-length sequence. That was a
mistake, sorry, but it may not be one of their test-cases. I
wonder how many of the accepted entries would perform properly.
No idea here, and didn't even think about it.
Notes: the `all` returned by crack() is a list such that all[i] is
list of all (index, value) pairs such that the longest increasing
subsequence ending with `value` is of length i+1; `value` is at index
`index` in the input permutation. The maximal LISs thus end with the
values in all[-1]. findall() iterates backwards over `all`, to
accumulate all the values that appear in _some_ maximal LIS. There's
clearly wasted work in findall() (if someone is looking for an
algorithmic point to attack). Curiously, no use is made of that
values are integers, outside of input and output; any values with a
total ordering would work fine in crack() and findall().
"""
# http://spoj.sphere.pl/problems/SUPPER/
def crack(xs):
from bisect import bisect_right as find
smallest = []
all = []
n = 0
for index, x in enumerate(xs):
i = find(smallest, x)
if i == n:
smallest.append (x)
all.append([(index, x)])
n += 1
else:
all[i].append((index, x))
if x < smallest[i]:
smallest[i] = x
return all
def findall(all):
constraints = all[-1]
allints = [pair[1] for pair in constraints]
for i in xrange(len(all) - 2, -1, -1):
survivors = []
for pair in all[i]:
index, value = pair
for index_limit, value_limit in constraints:
if index < index_limit and value < value_limit:
survivors.appen d(pair)
allints.append( value)
break
constraints = survivors
return sorted(allints)
def main():
import sys
while 1:
n = sys.stdin.readl ine()
if not n:
break
n = int(n)
perm = map(int, sys.stdin.readl ine().split())
assert n == len(perm)
supers = findall(crack(p erm))
perm = None # just to free memory
print len(supers)
print " ".join(map( str, supers))
if __name__ == "__main__":
main()
"""
Tim Peters;
INCREDIBLE~ 241433 2005-09-11 04:23:40 Tim Peters accepted 3.44 7096 PYTH
BRAVO!
I just wonder have I grey cells enough for to understand how your
algo works... and hopefully it's not your last solved problem on
the contester.
I'm pretty sure they're using
slower HW than mine (3.4 GHz P5).
As I mentioned before their 4 identical machines are PIII Xeon 700MHz.
PS:
each accepted solution automatically gets into "Best Solutions" list.
[Tim Peters, on the problem at http://spoj.sphere.pl/problems/SUPPER/
] ...
[n0**@narod.ru] INCREDIBLE~
241433 2005-09-11 04:23:40 Tim Peters accepted 3.44 7096 PYTH
BRAVO!
It's different now ;-) I added the two lines
import psyco
psyco.full()
and time dropped to 2.29, while memory consumption zoomed:
2005-09-11 18:44:57 Tim Peters accepted 2.29 42512 PYTH
That surprised me: my own test cases on Windows using Python 2.4.1
enjoyed the same order of speedup, but barely increased RAM usage.
Perhaps they installed an older (or newer <0.9 wink>) version of
psyco.
I just wonder have I grey cells enough for to understand how your
algo works...
You do! Work out some small examples by hand, and it will become
clear; be sure to read the comments before the code, because they
explain it.
and hopefully it's not your last solved problem on the contester.
I have to stop now, else I wouldn't do anything else <0.3 wink>.
...
Tim Peters wrote: [Bryan Olson, on the problem at
http://spoj.sphere.pl/problems/SUPPER/
]
I never intended to submit this program for competition. The
contest ranks in speed order, and there is no way Python can
compete with truly-compiled languages on such low-level code.
I'd bet money that the algorithm I used (coded in C) can run
with the winners. I also think I'd wager that the Python version
outright trumps them on code size.
Oh, it's not that bad <wink>. I took a stab at a Python program for
this, and it passed (3.44 seconds).
[...] I didn't make any effort to speed this, beyond picking a reasonable
algorithm, so maybe someone else can slash the runtime
Hmmm ... I used the Theta(n lg n) algorithm ... how the heck...
Aha! The 'bisect' module changed since last I looked. It still
has the Python implementation, but then the last few lines say:
# Overwrite above definitions with a fast C implementation
try:
from _bisect import bisect_right, bisect_left, insort_left,
insort_right, insort, bisect
except ImportError:
pass
Binary-search is the inner-loop in this algorithm. I wrote my
own binary-search, so I was executing Theta(n lg n) Python
statements. Tim's use of bisect means that his inner-loop is
in C, so he does Theta(n) Python statements and Theta(n lg n) C
statement.
The key to fast Python: use a good algorithm, and keep the inner
loop in C.
--
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