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Panasonic Programming Contest 2024（AtCoder Beginner Contest 375） Announcement
By atcoder_official, history, 6 weeks ago, In English

We will hold Panasonic Programming Contest 2024（AtCoder Beginner Contest 375）.

We are looking forward to your participation!

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6 weeks ago, # |
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Hoping to get more active on AtCoder starting from this ABC

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6 weeks ago, # |
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Wow, the gap from B to C is huge o.O

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    6 weeks ago, # ^ |
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    But gap btw D and E is low

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    6 weeks ago, # ^ |
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    I think C is not difficult to get the logic , understanding and implementation is little difficult.

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6 weeks ago, # |
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C 400pts？That's so amazing.

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6 weeks ago, # |
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What is the Distribution of Problem Statements on the top page for?

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    6 weeks ago, # ^ |
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    Same question

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    6 weeks ago, # ^ |
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    AtCoder suffered DDoS attacks during some contests last year. During that period, Panasonic (also the sponsor of this contest) sponsored a contest, ABC301. To avoid the effect of the DDoS attack, they distributed the problem statements in PDF format during the contest.

    Later, when Panasonic hosted contests again, they might have directly copied the problem statements from previous contests without any modifications, so the PDF version of the problem statements was retained—even though the statement was still from ABC301.

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6 weeks ago, # |
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Wow, 400 points problem C!

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6 weeks ago, # |
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375 is 3/8 times 1000.

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6 weeks ago, # |
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It seems that problem C will be very hard this time.

GL!

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6 weeks ago, # |
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I hope we can all get a high score in the competition. Good luck!

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6 weeks ago, # |
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Wow, C and D have the same score this time, both are 400 points

GL&&HF

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6 weeks ago, # |
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400 points problem C? It must be difficult!

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6 weeks ago, # |
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As a participant, I confirm that problems are problems

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6 weeks ago, # |
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Hoping to get 0.

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6 weeks ago, # |
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Hoping it is easy enough......

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6 weeks ago, # |
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Is there a bit of water in the data of question E, I can still AC if I write the DP transfer wrongly.

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6 weeks ago, # |
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G is almost the same as 567E

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6 weeks ago, # |
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Can someone explain C and G?

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    6 weeks ago, # ^ |
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    C is just implementation, do what the question is saying to do. Note that the operation rotates a square submatrix each time. The size of this submatrix goes from n to 1. For every index, figure out how many times it will get rotated(distance from edge % 4), and just do the operation that many times.

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      6 weeks ago, # ^ |
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      Nice implementation. I ended up writing a complex code for performing complete rotation on each outer strip of (n-i)*(n-i) square.

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    6 weeks ago, # ^ |
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    To solve C, is helpful to visualize each operation as a 90 degrees clock-wise rotation in the square. Then, the border of the complete square N*N will be rotated once, the border of the centered square of N-2*N-2 will be rotated two times and so on. You can only implement a function that performs rotation, and then have in count that each 4 rotations you get the original values in the border of the square.

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    6 weeks ago, # ^ |
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    For G I ran dijkstra and made a new graph based in all edges for all shortest path. an edge is part of a shortest path if it make the distance lower or keep the distance equal from 1 to some vertex when running dijkstra. Then I made a new graph based on this edges and found all bridges. bridges can make the new graph disconnected and then there won't be a path from 1 to N using that graph.

    For C, I divided the grid in layers (like a onion). in layer 1 you will rotate 1, and in general in layer i you will rotate each cell i times. find i%4 and rotate i%4 times, because after 4 rotations the cell will be in the same place.

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      6 weeks ago, # ^ |
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      Thank you, can you please also explain how to find edges which are part of all shortest paths?

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        6 weeks ago, # ^ |
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        Let dist[x] be the distance from 1 to x, and w[a][b] = weight of the edge ab

        if the distance from "1" to "b" is X, and dist[a] + w[a][b] = X, and b is in some shortest path, then ab is one of these edges.

        First I ran dijkstra and keep a "parent" vector, then I ran a bfs from N to 1 to make the new graph

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      6 weeks ago, # ^ |
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      in problem $$$C$$$ I have to rotate for each $$$i$$$ upto $$$N/2$$$?Also how did you reach this observation?What was your observation?

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        6 weeks ago, # ^ |
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        This way

        image

        Need to rotate all cells

        This is just what I got after read the statement. The main observation to solve the problem is you don't need to rotate i times, only i%4 times. and the other observation is how to get the layer. the layer is the minimum distance from i to 0, from i to n+1, from j to 1, from j to n+1

        int layer = min(min(i, n+1-i), min(j, n+1-j));
layer %= 4;
int ri = i, rj = j;
for(int r=1;r<=layer;r++){
    swap(ri, rj);
    rj = n + 1 - rj;
}
ans[ri][rj] = gr[i][j];
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6 weeks ago, # |
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where will the editorial for the problems be published?

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6 weeks ago, # |
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A little confusion on C. It took me a while to realize that: all pairs {x, y} between a and b means: a << x <= b and a <= y <= b.

Otherwise, the problem setting is excellent as usual.

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6 weeks ago, # |
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For G , consider all edges which are part of smallest path from 1 to N . Now we need to find all the bridges in the final graph . Its implementation heavy.

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6 weeks ago, # |
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Parkinson's disease, Crohn's disease, Sickle cell anemia, Amyotrophic lateral sclerosis (ALS), Alzheimer's disease, Multiple sclerosis (MS), Cystic fibrosis, Lupus, Tuberculosis, Hepatitis C, Rheumatoid arthritis, Diabetes mellitus, Hypertension, Asthma, Ebola, Malaria, Lyme disease, Dengue fever, Psoriasis, Influenza.

These are the diseases I suffered while trying to solve problem C.

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    6 weeks ago, # ^ |
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    I left C after reading the statement cause C and D gave me same points but D's statement was much simpler.

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6 weeks ago, # |
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Can somebody please help me understand my mistake in the submission for G? Submission Link-> https://atcoder.jp/contests/abc375/submissions/58739004

Edit: I got my error, my infinity value was not enough in dijkstra implementation.

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6 weeks ago, # |
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$$$C$$$ $$$\gt\gt$$$ $$$E$$$

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6 weeks ago, # |
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Well, G is weaker than a well-known problem in China called Bridge.

Link: https://www.luogu.com.cn/problem/P2685

The solution is to calculate any shortest path of the graph first, and get the point for each point v where the shortest path from 1 or n to v part from the shortest path from 1 to n. For each edge (u,v) which is not on the path, we can know the range it can influence. So that we can use a segtree to maintain the shortest path from 1 to n when we delete an edge on the former path.

BTW, I tried to modify the code I wrote for that problem to pass G but failed :(

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    6 weeks ago, # ^ |
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    I see the official editorial and it is easier than the problem I just mentioned so it have a simplier implementation. But I believe many of the people must have solved this problem before.

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6 weeks ago, # |
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I think this contest's F is quite interesting, despite I haven't got it during the contest.

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