....that involve LCA, MST, Treap, Sparse Table, Binary Manipulations and other advanced techniques.
Trying to build a good foundation. Thanks!
# | User | Rating |
---|---|---|
1 | tourist | 4009 |
2 | jiangly | 3773 |
3 | Radewoosh | 3646 |
4 | ecnerwala | 3624 |
5 | jqdai0815 | 3620 |
5 | Benq | 3620 |
7 | orzdevinwang | 3612 |
8 | Geothermal | 3569 |
8 | cnnfls_csy | 3569 |
10 | gyh20 | 3447 |
# | User | Contrib. |
---|---|---|
1 | cry | 161 |
2 | awoo | 160 |
2 | maomao90 | 160 |
4 | atcoder_official | 157 |
5 | -is-this-fft- | 155 |
5 | nor | 155 |
7 | adamant | 153 |
8 | maroonrk | 152 |
8 | Um_nik | 152 |
10 | djm03178 | 146 |
....that involve LCA, MST, Treap, Sparse Table, Binary Manipulations and other advanced techniques.
Trying to build a good foundation. Thanks!
Name |
---|
I recently tried this interesting problem. Would surely recommend it.
https://codeforces.net/problemset/problem/1615/D
Thank you!
if u have an explination to the solution can you please provide it ? i wrote a blog once that the tutorial wasnt clear and also kept trying with it from time to time and never got it
The important observation here is that we are only concerned with the parity of no. of bits on any path, as mentioned in the editorial. Thus, we need to only look at the no. of set bits in a number, and not which bits are set.
Then we can impose some restrictions on the parity of bits on each path in the input. Any edge with non negative value can also be viewed as a path of size 1. Hence, we can try to impose all these restrictions and see if these give us a valid combination. Then, we will work with the negative weighted edges and obtain their values in the same way.
To store the parity of bits in any path, we are using DSU (but it is probably not required if we root our tree beforehand). To obtain the parity between any two nodes, we can xor the parity of each path from the root of the tree. Since, if any edge occurs in both the paths, it will not be on the simple path between the two nodes and hence get cancelled. If at any point, the conditions are not satisfied or are contradictory, we know that the required tree is not possible. Finally to obtain the weights of the negative edges, we can check for the required parity between the two nodes, and set our answer as 0 or 1. Here is my solution, feel free to ask any queries.
https://codeforces.net/contest/1615/submission/269522430
The solution can also be viewed as a variant of bipartite matching. Each node is colored either 0 or 1, depending on the parity of path between the node and root.
https://codeforces.net/contest/809/problem/D
https://codeforces.net/contest/963/problem/D
https://codeforces.net/contest/911/problem/G
https://codeforces.net/contest/1977/problem/D
https://codeforces.net/contest/1975/problem/E
ty good sir!
609E - Minimum spanning tree for each edge
using MST and LCA
i tried this interesting problem, although it might be a bit simple for the techniques you are talking about: 1305G - Kuroni and Antihype
UPDATE: CodingPokemon said that these problems are wayyyyyy to easy for your level
aw hell nah
i call cap on the update :skul:
Here is another problem, taht uses treap: 4A - Watermelon
learn binary search
.....good tip for newbies and pupils trying to reach specialist/expert
are you not tryna reach expert lmao
im trying to reach master :)
believe it or not but masters still need to know binary search
suggest me some mst problems
This one is nice: uses lca and mst -- [NOIP2013] — 货物运输
Why are you asking for more problems when you haven't finished the ones given to you by smax and RobeZH?
I'm in China bru, and I got permission from RobeZH to stop doing the assigned problems
then ask robezh for more problems smh
surely he can offer you some 603 probs
I forgot to specify I'm in Hangzhou
no way you are at xyd camp wtf strong
blud u think I would be solving 2100-2400 CF problems rn otherwise? (tho most are from luogu)
ok but still
learn binary search
we did, in the first day + ternary search
surely learn adhoc
ummmm i believe chinese people aren't russian enough, therefore we aren't learning adhoc :D
usaco still has ad hoc fyi
yea i know :pray:
yes and that is precisely the reason you remain a specialist
at least im not afraid to comment on my main, and I don't cheat on cf just for rating.... reminds me of someone i know...
I've been solving problems from this list, it's a very good list!
https://codeforces.net/blog/entry/91600
It would be great if you could recommend the problems recommended to you during your training camp :)
https://codeforces.net/blog/entry/117352