o-timberland

can someone give me the third test case for this task...cant find any mistakes in the code but it still gives 10/26

Hm I had some problems with input / output . I changed something and it works at my computer .

Here are the lines I changed :

char s[5];

scanf("%d %d",&n,&q);
scanf(" %s",&s);
scanf("%d %d",&a,&b);
scanf("%d %d",&a,&b);
scanf("%d",&a);

hmm tried changing that but it seems thats not the problem.i have 10/26 so its not like i dont pass any tests...

oduleodule just to remaind u about the 3.test case:D

try to remove "\n" from scanfs :D

still 10/26:D

I have 25/26..can someone give me a hint to speed up my brute force

Imas opis resenja ovde pa pogledaj : http://www.yuoi.nis.edu.rs/takmicenja/2011.0.kv1/5.o-timberlend/o-timberlend.solution.pdf

ma pogledao sam resenje ali mene interesuje vise sto ovaj moj kod ne radi...

meni nije jasno kako je moj jadan brute force dobio 96 bodova?

na yuoi kaze brute force oko 60 poena:D

eh...jbg to se onda zove sreća


btw. kada će medalje

a daj za ovaj racun dali test primere a za ovaj najtezi nisu...

ok, mislim da sam ti nasao gresku...

kada je niz[i][i1] == 1, onda stavis sum = 0, medjutim tu treba da stavis i

for (it = asd.begin(); it != asd.end(); ++it) niz1[i1][(*it).first] = 0

da video sam...resio sam...ne mogu da verujem sta nisam video:@...hvala u svakom slucaju:D

Postoji i bolje rjesenje, slozenosti N^2 po queryu :)

jel možeš objasnit kako si ti radio na natjecanju??

slozenosti O(n^2) po svakoj povrsini?kako to postizes??