Panasonic Programming Contest 2024（AtCoder Beginner Contest 375） Announcement

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Panasonic Programming Contest 2024（AtCoder Beginner Contest 375） Announcement

Автор atcoder_official, история, 7 недель назад, По-английски

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

Contest URL: https://atcoder.jp/contests/abc375
Start Time: http://www.timeanddate.com/worldclock/fixedtime.html?iso=20241012T2100&p1=248
Duration: 100 minutes
Writer: cn449, kyopro_friends, ynymxiaolongbao
Tester: MtSaka, sounansya
Rated range: ~ 1999
The point values: 100-150-400-400-450-550-575

We are looking forward to your participation!

atcoder

atcoder_official
7 недель назад
40

Комментарии (40)

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Lunatic_Frisbee

7 недель назад, # |

Hoping to get more active on AtCoder starting from this ABC

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mustard_with_fries69420

7 недель назад, # |

Wow, the gap from B to C is huge o.O

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SNX03

7 недель назад, # ^ |

But gap btw D and E is low

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yvmr_33

6 недель назад, # ^ |

I think C is not difficult to get the logic , understanding and implementation is little difficult.

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xiaozha

7 недель назад, # |

← Rev. 2 →

C 400pts？That's so amazing.

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13579abc

6 недель назад, # ^ |

D 400pts too, first time

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LHLeisus

7 недель назад, # |

What is the Distribution of Problem Statements on the top page for?

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iamfreeezing

6 недель назад, # ^ |

Same question

→ Ответить

Physics212303

6 недель назад, # ^ |

← Rev. 2 →

+14

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.

→ Ответить

caojh07

6 недель назад, # |

Wow, 400 points problem C!

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llc5pg

6 недель назад, # |

375 is 3/8 times 1000.

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vector1024

6 недель назад, # |

It seems that problem C will be very hard this time.

GL!

→ Ответить

allenxing

6 недель назад, # |

I hope we can all get a high score in the competition. Good luck!

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yuanzihan

6 недель назад, # |

Wow, C and D have the same score this time, both are 400 points

GL&&HF

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lmz121107

6 недель назад, # |

400 points problem C? It must be difficult!

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harvey_wzy

6 недель назад, # |

As a participant, I confirm that problems are problems

→ Ответить

hjiaming

6 недель назад, # |

Hoping to get 0.

→ Ответить

Wayne13770834134_

6 недель назад, # |

Hoping it is easy enough......

→ Ответить

ThomasWang

6 недель назад, # |

-9

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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mustard_with_fries69420

6 недель назад, # |

← Rev. 2 →

G is almost the same as 567E

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then00bprogrammer

6 недель назад, # |

Can someone explain C and G?

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letsgetgood

6 недель назад, # ^ |

← Rev. 2 →

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.

Code

import sys

def main():
    n = int(input())
    mat = list(map(lambda s: list(s.strip()), sys.stdin.readlines()))
    mat2 = [[0] * n for _ in range(n)]
    
    for i in range(1, n+1):
        for j in range(1, n+1):
            edge_dist = min(i-1, n-i, j-1, n-j)
            times = (edge_dist + 1) % 4 # how many rotations
            target_i, target_j = i, j
            for k in range(times):
                target_i, target_j = target_j, n + 1 - target_i
            mat2[target_i-1][target_j-1] = mat[i-1][j-1]
    
    for row in mat2:
        print("".join(map(str, row)))

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infinitix

6 недель назад, # ^ |

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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Stark0509

6 недель назад, # ^ |

← Rev. 3 →

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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gabriel88766

6 недель назад, # ^ |

← Rev. 2 →

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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then00bprogrammer

6 недель назад, # ^ |

Thank you, can you please also explain how to find edges which are part of all shortest paths?

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gabriel88766

6 недель назад, # ^ |

← Rev. 2 →

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

Code


vector<ll> adj[N], adjw[N], par[N], adj2[N];
ll dist[N];


void dijkstra(int src){
    memset(dist, 63, sizeof(dist));
    dist[src] = 0;
    priority_queue<pair<ll,ll>, vector<pair<ll,ll>>, greater<pair<ll,ll>>> pq;
    pq.push({0, src});
    while(!pq.empty()){
        auto d = pq.top().first;
        auto u = pq.top().second;
        pq.pop();
        if(d > dist[u]) continue;
        for(int i=0;i<adj[u].size();i++){
            ll v = adj[u][i], w = adjw[u][i];
            if(dist[v] > dist[u] + w){
                par[v].clear();
                par[v].push_back(u);
                dist[v] = dist[u] + w;
                pq.push({dist[v], v});
            }else if(dist[v] == dist[u] + w){
                par[v].push_back(u);
            }
        }
    }
}

//in main
queue<int> q;
vector<bool> vis(n+1, false);
q.push(n);
vis[n] =true;
while(!q.empty()){
    auto u = q.front();
    q.pop();
    for(auto v : par[u]){
        adj2[u].push_back(v);
        adj2[v].push_back(u);
        if(!vis[v]){
            vis[v] = true;
            q.push(v);
        }
    }
}

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brave-kid

6 недель назад, # ^ |

← Rev. 2 →

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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gabriel88766

6 недель назад, # ^ |

← Rev. 4 →

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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brave-kid

6 недель назад, # ^ |

thanks

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tronaldDump45

6 недель назад, # |

where will the editorial for the problems be published?

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xiaoping3418

6 недель назад, # |

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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is_this_greedy

6 недель назад, # |

-19

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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Astroflexx

6 недель назад, # |

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.

→ Ответить

mzibrahmi

6 недель назад, # ^ |

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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Madhav_1608

6 недель назад, # |

← Rev. 4 →

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.

→ Ответить

oj1

6 недель назад, # |

$$$C$$$ $$$\gt\gt$$$ $$$E$$$

→ Ответить

wxy2010

6 недель назад, # |

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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wxy2010

6 недель назад, # ^ |

← Rev. 2 →

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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ArmeriaLeap

6 недель назад, # |

I think this contest's F is quite interesting, despite I haven't got it during the contest.

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