Atcoder Beginner Contest 257 Discussion?

→ Обратите внимание

Соревнование идет
Codeforces Round 995 (Div. 3)
02:02:06
Зарегистрироваться »

→ Трансляции

Codeforces Round 995 Solution Discussion

aryanc403

До начала 02:07:06

Всё →

→ Лидеры (рейтинг)

№	Пользователь	Рейтинг
1	tourist	3985
2	jiangly	3814
3	jqdai0815	3682
4	Benq	3529
5	orzdevinwang	3526
6	ksun48	3517
7	Radewoosh	3410
8	hos.lyric	3399
9	ecnerwala	3392
9	Um_nik	3392

Страны | Города | Организации

Всё →

→ Лидеры (вклад)

№	Пользователь	Вклад
1	cry	169
2	maomao90	162
2	Um_nik	162
4	atcoder_official	161
5	djm03178	158
6	-is-this-fft-	157
7	adamant	155
8	awoo	154
8	Dominater069	154
10	luogu_official	150

Всё →

→ Найти пользователя

→ Прямой эфир

Детальнее →

Блог пользователя iLoveIOI

Atcoder Beginner Contest 257 Discussion?

Автор iLoveIOI, история, 2 года назад, По-английски

I couldn't find a discussion blog for this contest so here's one.

Does anyone know how to solve Ex?

iLoveIOI
2 года назад
36

Комментарии (31)

Показать архивные | Написать комментарий?

ustaritz

2 года назад, # |

For problem D, my solution with Binary Search + BFS works fine except for 01_random_03.txt. Any hint ?

→ Ответить

harpreet1237

2 года назад, # ^ |

← Rev. 2 →

+16

Did you binary search till 4e9 becauese I was facing same verdict till i realized cost could be 4e9.

→ Ответить

why_I

2 года назад, # ^ |

← Rev. 2 →

My solution also fails on the same test case. Is there anyone who could look into my code.

code

Your code here...

void myfunc()
{

            int n;
            cin >> n;
            vector<vector<int>>vv(n, vector<int>(3));
            for(int i=0; i<n; ++i)
                cin >> vv[i][0] >> vv[i][1] >> vv[i][2];
            
            vector<bool>vis(n);
            function<int(int, int)> dfs = [&](int cur, int s)
            {
                vis[cur] = true;
                int cnt = 1;
                int x = vv[cur][0], y = vv[cur][1], p = vv[cur][2];
               // cout << x << " " << y << endl;
                for(int i=0; i<n; ++i)
                {
                    if(vis[i])continue;
                    int x1 = vv[i][0], y1 = vv[i][1];
                    int dis = abs(x-x1)+abs(y-y1);
                    //cout << "SE " << x1 << " " << y1 << " " << dis << " " << p << endl;
                    if(dis<=s*p)
                    {
                        cnt+=dfs(i, s);
                    }
                }return cnt;
            };
            int ans = 2e15;
            for(int i=0; i<n; ++i)
            {
                int low = 0, high = 1e11;
                int res = 0;
                while(low<=high)
                {
                    int mid = (low+high)/2;
                    fill(all(vis), false);
                    if(dfs(i, mid)==n)
                    {
                        res = mid;
                        high = mid-1;
                    }else low = mid+1;
                }ans = min(ans, res);
            }
            cout << ans << endl;         

}

I have used #define int long long

→ Ответить

thatsramen

2 года назад, # ^ |

Overflow. P can be 10^9 10^11 * 10^9 = 10^20 which is too big even for long long.

→ Ответить

Kartik1508

2 года назад, # ^ |

-6

explain your solution

→ Ответить

bashkort

2 года назад, # ^ |

rude

→ Ответить

ustaritz

2 года назад, # ^ |

With 4e9 it works fine. thanks.

→ Ответить

rgnerdplayer

2 года назад, # |

+12

Submission In problem G why picking the longest transition (by prefix function) works?

→ Ответить

rgnerdplayer

2 года назад, # ^ |

Greedily removing the longest suffix of T that is a prefix of S seems to work as well.

→ Ответить

milind0110

2 года назад, # ^ |

Yeah did so after the contest and it passed but I am unable to prove it why does it work.

→ Ответить

Zain

2 года назад, # ^ |

Because ans[i]<=ans[j] for i<=j because you can always construct a solution for index i from the constructive words of solution j by either deleting some characters of the last word (giving the same ans) or deleting some words then some characters (giving smaller ans).

→ Ответить

sksshivam007

2 года назад, # |

D was a nice variation of Floyd Warshall algorithm

→ Ответить

Dan4Life

2 года назад, # ^ |

BFS is a nice variation of Floyd Warshall?

→ Ответить

harsh__h

2 года назад, # ^ |

← Rev. 3 →

You can also solve it using a slight modification of floyd warshall by defining

$$$ w[i][j] = \frac{|x_i-x_j|+|y_i-y_j|}{p[i]}$$$

and changing + operation with max

code

→ Ответить

Dan4Life

2 года назад, # ^ |

Interesting solution

→ Ответить

SummerSky

2 года назад, # ^ |

I used this idea as well, but I think that binary search solution is also very nice.

→ Ответить

lbntxdy

2 года назад, # |

I have failed on F because of a little mistake in my BFS :)

→ Ответить

Dan4Life

2 года назад, # ^ |

I got WA in one test case in F smh.

→ Ответить

neelakshsh

2 года назад, # |

Problem C? Can anyone mention the approach? Thanks!

→ Ответить

spookywooky

2 года назад, # ^ |

← Rev. 2 →

We can multiply all w[i]=w[i]*2, then check all w[i]+1 and w[i]-1 to be the possible best answer.

Each check can be done in O(logn) by sorting childs and adulds separate, then binary search the number of correct guessed ones for fixed x.

Edit: Or use prefix sums. Atfer sorting w[], we can split w[] at any position, then the correct guessed number is the number of childs left of the split position plus the number of adults right of the split position.

→ Ответить

llc5pg

2 года назад, # ^ |

Keep track of score0 and score1. Use map to iterate the weights and move the border line of guessing 0 vs guessing 1 from left to right. I submitted a simplified version of the code at: https://atcoder.jp/contests/abc257/submissions/32757902

I was unlucky to only solved it one minute after the contest ended.

→ Ответить

Dan4Life

2 года назад, # ^ |

← Rev. 2 →

check for all w[i], then also check for smallest weight-1 and largest weight+1(Solution)

→ Ответить

ninjamayank

2 года назад, # ^ |

just sort the children array and parent array and apply binary to get the ans. also check if we have only children or only adults

→ Ответить

adityagamer

2 года назад, # |

How to solve E?

→ Ответить

aspireboy007

2 года назад, # |

can any one help me with D , I think I miss understood the question . They are saying that : "Takahashi's objective is to become able to choose a starting trampoline such that he can reach any trampoline from the chosen one with some jumps."

some jumps means we can assume that he will be able to jump for each consecutive pair right ?

→ Ответить

adityagamer

2 года назад, # ^ |

No. Read sample test case explanation

→ Ответить

codermaniac

2 года назад, # |

← Rev. 3 →

Hi Everyone, Can anyone help to find the mistake in my code of F? Link to Solution

→ Ответить

Involution

2 года назад, # ^ |

← Rev. 2 →

Consider the case

3 2

0 1

0 3

Just add a line ans = min(ans, st + en + 2); and it should AC.

→ Ответить

codermaniac

2 года назад, # ^ |

Thanks!!!

→ Ответить

adityagamer

2 года назад, # |

← Rev. 3 →

For problem F, my solution is giving runtime error on 27 hidden cases. Why? https://atcoder.jp/contests/abc257/submissions/32791294

UPD: I have done the same thing as editorial except DFS instead of BFS, which I don't think should make any difference

→ Ответить

Spheniscine

2 года назад, # ^ |

← Rev. 2 →

It makes all the difference, because the graph isn't necessarily a tree/forest, so there may be more than one simple path between nodes of different lengths, and DFS doesn't necessarily find the shortest of them

→ Ответить

Соревнования по программированию 2.0

Время на сервере: 22.12.2024 17:47:54 (i1).

Десктопная версия, переключиться на мобильную.

При поддержке