|
|
вернуться в форумTo all of you who cannot overcome #3 !!!
All the suggestions about redundant '\n' and about that the dictionary may have repeated words have nothing to do with WA #3 (read condition carefully).
If you use any kind of backward tracking, be it recursion or DFS, check the "no luck" condition attentively. First i used checking the next letter to be '\0' at the beginning of my recursion function, and if so returned 'true'. This is wrong! You must make word-end check BEFORE you make the next step, otherwise you may get "deadlock" "there is no direction to go" though there is no need to go anymore!!!
Test word for the example table
rabadaabracabrac: YES
(starting from [3,1] in C-like indices). How did you get rabadaabracabrac: YES
In the condition there is said,that words could not have self-intersections. But if you don't intersect you will not get this word.
My program writes NO on this test Thank you, melkiy! I made the same mistake :( My program got YES on this test, by have WA#3 It's not a test case for WA#3:
rabadaabracabrac: YES and it's not working. My reason for WA #3 was that I tried to use 1d char array to store the table and used i -> i - 1 to go left and i -> i + 1 to go right, however, if you are in the first column you can't go left and if you're on the 4th row you can't go right |
|
|
