Good afternoon everyone. Today, I’m going to talk about a unique sleeping schedule named Uberman sleep schedule. It’s a  kind of polyphasic sleep. Polyphasic sleep is a practice of sleeping multiple times in a 24-hour-period. Usually more than two, on contrast to normal biphasic sleep that we usually do. Uberman sleep schedule, which is named by Dr. Claudio Stampi, a chronobiology researcher, is adopted by many animals sleep behaviour. Instead of sleeping in long chunks of time, such as the eight hours a night recommended to humans, certain animals sleep for short periods of time throughout the day and night.

Uberman itself propose 6 times of sleep in a day of 24-hour-period. Each of them is 20 minutes long and separated 4 hours after another. In total, people doing Uberman schedule only sleep 2 hours in a day. If you do a searching on google about it, you will find that todays more people doing this sleep schedule. Anyway, every type of sleep schedule has its own pros and cons.

There are at least four pros why people doing Uberman sleep schedule. First, you have a lot of time to do everything you want. With 22 hours a day you may feel like superman since more things can be done compare to normal people who only have 15 up to 17 hours. Secondly, you will not lose every phase of your day, because you will be ready every time. You willl see the morning, the afternoon, the evening, and even the dawn. Thirdly, for those who want to have a diet, this kind of sleep schedule perhaps will help you much. Lastly, some resources say that many geniuses have this type of sleep. For instance, Leonardo Da Vinci and Thomas Alva Edison are said to be success having Uberman sleep schedule. So, maybe you will also become genius, if you do so.

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This is terrible, they will convert 30% of all rooms into triple rooms. I think my room now is not big enough to accommodate 3 people. Hopefully I still get the double room next semester.

The First Goal : ACM ICPC World Final 2009!

The name “NTUsiasm” is given by Wendy, it’s better than the first one he proposed, hehe…

Actually February problem-set is not so hard.

Bronze Division:

-diningb -> straightforward DP problem, which you should precompute the cost to change to card 1 for all length and cost to change to card 2 as well, once you’ve done that you can combine it and get the result, O(N).

-racing -> just a typical simulation problem, all you can do is run trough all the track, O(N).

-cowmult -> using an easy observation of mathematics, we can solve it very easy, O(1).

Silver Division:

-lines -> mathematis + sorting = full score, the only thing that made my code fail to get full score is because my array wasn’t big enough, I forgot that the number of gradiens will probably reach N^2, once I change the array [1..210] to [1..40010] I will get full score, hiks, one more time I hate bug! O(N^2).

-meteor -> a tricky flood fill problem, all you have to do just do a flood fill and simulate the meteors chronologically (you have to sort them first), don’t forget to code your solution clearly, my code didn’t get full score coz I did some bug (arrrghhh, I hate bug!), O(max(T,X,Y)^2).

-egroup -> similar to diningb in bronze division, but this one is a little bit more difficult since we also have card 3, the keyword is only “precompute” and then you can combine it and get the result, another no-hard DP, I got full score for this, O(N).

Gold Division:

-cgame -> when I saw this problem I think about knapsack algorithm, but it’s impossible to have a normal knapsack within that constraint, all we have to do is hack the normal knapsack algorithm (as RK said in the analysis), the fact that we have to hack the normal way make this problem considered as a output-only problem, I only got 5-first-testcases.

-hotel -> problem with advance data structure, I realize what to do to solve it but until now I can’t code binary indexed tree or something like that, that means I have to practice it more, I only got the small testcases since I only submitted my kind of brute force code, the expected solution is run for O(MlogN).

-grading -> a typical DP-push problem, but before that you have to convince yourself with a lemma that no need to change a height into particular height that doesn’t exist in the height set, in other words you only have N possible height to try that makes this can be done in O(N^2), I got full score.

Want to know further about PINTU? Just visit http://pintusingapura.org!

Finally, I start my career on TopCoder.Com with a not-yet-good JAVA language. My first rating is 1416 (blue), not bad for the first time competitor. Wish a good progress forward for me.