Panasonic Programming Contest 2024（AtCoder Beginner Contest 375） Announcement

Codeforces и Polygon могут быть недоступны в период с 6 декабря, 22:00 (МСК) по 7 декабря, 00:00 (МСК) в связи с проведением технических работ. ×

→ Обратите внимание

До соревнования
Codeforces Round 991 (Div. 3)
19:25:10
Зарегистрироваться »

→ Лидеры (рейтинг)

№	Пользователь	Рейтинг
1	tourist	3985
2	jiangly	3741
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3489
7	Radewoosh	3483
8	Kevin114514	3443
9	ecnerwala	3392
9	Um_nik	3392

Страны | Города | Организации

Всё →

→ Лидеры (вклад)

№	Пользователь	Вклад
1	cry	167
2	Um_nik	163
3	maomao90	162
3	atcoder_official	162
5	adamant	158
5	-is-this-fft-	158
7	awoo	156
8	djm03178	155
9	TheScrasse	154
10	Dominater069	153

Всё →

→ Найти пользователя

→ Прямой эфир

Детальнее →

Блог пользователя atcoder_official

Panasonic Programming Contest 2024（AtCoder Beginner Contest 375） Announcement

Автор atcoder_official, история, 2 месяца 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
2 месяца назад
41

Комментарии (41)

Написать комментарий?

Lunatic_Frisbee

2 месяца назад, # |

Hoping to get more active on AtCoder starting from this ABC

→ Ответить

mustard_with_fries69420

2 месяца назад, # |

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

→ Ответить

SNX03

2 месяца назад, # ^ |

But gap btw D and E is low

→ Ответить

yvmr_33

2 месяца назад, # ^ |

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

→ Ответить

xiaozha

2 месяца назад, # |

← Rev. 2 →

C 400pts？That's so amazing.

→ Ответить

13579abc

2 месяца назад, # ^ |

D 400pts too, first time

→ Ответить

LHLeisus

2 месяца назад, # |

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

→ Ответить

iamfreeezing

2 месяца назад, # ^ |

Same question

→ Ответить

Physics212303

2 месяца 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

2 месяца назад, # |

Wow, 400 points problem C!

→ Ответить

llc5pg

2 месяца назад, # |

375 is 3/8 times 1000.

→ Ответить

vector1024

2 месяца назад, # |

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

GL!

→ Ответить

allenxing

2 месяца назад, # |

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

→ Ответить

yuanzihan

2 месяца назад, # |

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

GL&&HF

→ Ответить

lmz121107

2 месяца назад, # |

400 points problem C? It must be difficult!

→ Ответить

harvey_wzy

2 месяца назад, # |

As a participant, I confirm that problems are problems

→ Ответить

hjiaming

2 месяца назад, # |

Hoping to get 0.

→ Ответить

Wayne13770834134_

2 месяца назад, # |

Hoping it is easy enough......

→ Ответить

ThomasWang

2 месяца 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.

→ Ответить

mustard_with_fries69420

2 месяца назад, # |

← Rev. 2 →

G is almost the same as 567E

→ Ответить

then00bprogrammer

2 месяца назад, # |

Can someone explain C and G?

→ Ответить

letsgetgood

2 месяца 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)))

→ Ответить

infinitix

2 месяца назад, # ^ |

Nice implementation. I ended up writing a complex code for performing complete rotation on each outer strip of (n-i)*(n-i) square.

→ Ответить

Stark0509

2 месяца 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.

→ Ответить

gabriel88766

2 месяца 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.

→ Ответить

then00bprogrammer

2 месяца назад, # ^ |

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

→ Ответить

gabriel88766

2 месяца 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);
        }
    }
}

→ Ответить

brave-kid

2 месяца 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?

→ Ответить

gabriel88766

2 месяца 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];

→ Ответить

brave-kid

2 месяца назад, # ^ |

thanks

→ Ответить

tronaldDump45

2 месяца назад, # |

where will the editorial for the problems be published?

→ Ответить

xiaoping3418

2 месяца 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.

→ Ответить

is_this_greedy

2 месяца 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.

→ Ответить

Astroflexx

2 месяца 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

2 месяца назад, # ^ |

I left C after reading the statement cause C and D gave me same points but D's statement was much simpler.

→ Ответить

Madhav_1608

2 месяца 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

2 месяца назад, # |

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

→ Ответить

wxy2010

2 месяца 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 :(

→ Ответить

wxy2010

2 месяца 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.

→ Ответить

ArmeriaLeap

7 недель назад, # |

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

→ Ответить

shinchankosen

26 часов назад, # |

According to https://atcoder.jp/users/ynymxiaolongbao, his name at CF is YoshikaMiyafuji, not ynymxiaolongbao. If you just write the name on atcoder, wouldn't it be better not to connect it to the user's link like kyopro_friends?

→ Ответить

Соревнования по программированию 2.0

Время на сервере: 04.12.2024 22:09:52 (l1).

Десктопная версия, переключиться на мобильную.

При поддержке