UNIQUE VISION Programming Contest 2024 Summer (AtCoder Beginner Contest 359) Announcement

6 months ago, # |

← Rev. 2 →

Idk how I passed C after 4 WAs, just did some +/- math. Idea was to move diagonally as much as possible and align the x coordinate or y coordinate with the target node.

Code

#include <bits/stdc++.h>
using namespace std;
#define fast ios::sync_with_stdio(0); cin.tie(0); cout.tie(0);
#define ll long long
#define all(v) v.begin(),v.end()
#define F first
#define S second
#define pb push_back
#define mp make_pair
#define pi pair<int,int>
#define pii pair<int,pi>
#define rep(i,n) for(int i=0;i<n;i++)
#define rrep(i,n) for(int i=n-1;i>=0;--i)
#define cusrep(i,a,b,j) for(int i=a;i<b;i+=j)
#define rng(i,a,b) for(int i=a;i<b;i++)
#define rrng(i,a,b) for(int i=a;i>b;--i)
#define ar array

void run_case() {
  ll sx, sy, tx, ty;
  cin >> sx >> sy >> tx >> ty;
  ll alpha = min(abs(sx-tx), abs(sy-ty));
  ll nx = (sx>=tx) ? sx - alpha : sx + alpha;
  ll ny = (sy>=ty) ? sy - alpha : sy + alpha;

  if(ty%2==0){
      if(abs(tx)%2!=0){
        --tx;
      }
    } else {
      if(abs(tx)%2==0){
        --tx;
      }
  }
  if(sy!=ty){
    if(ty%2==0){
      if(abs(nx)%2!=0){
        --nx;
      }
    } else {
      if(abs(nx)%2==0){
        --nx;
      }
    }
  }
  ll ans = alpha + (1LL*abs(nx - (tx))/2LL) + 1LL*abs(ny - ty);
  cout << ans << "\n";
}
 
int main() {
  fast;
  int t = 1;
  while(t--){
    run_case();
  }
}

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_trie_again

6 months ago, # |

+25

what the fck was C

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6 months ago, # ^ |

Idk man, I just figured some +/- math after moving diagonally as much as possible and aligning resultant x or y with target node. After that it was tricky to move in the same direction where I did some guessing.

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6 months ago, # ^ |

I felt like leaving the contest .

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ExplodingFreeze

6 months ago, # |

Is there a simpler solution to G than Small to Large merging? Or did people just manage to implement that approach in like 5 mins @_@

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Pirate_King

6 months ago, # ^ |

-8

small to large merging is simple ??

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andrewtam

6 months ago, # ^ |

+18

It's a classic euler tour + monostack problem.

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qiuzx

6 months ago, # ^ |

There are indeed simpler solutions, but small to large merging is also simple enough. It's only about 30 lines and 5 mins or so is enough to implement it (my solution in contest was small to large merging).

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kelzzin2_fan

6 months ago, # ^ |

Could you roughly explain your approach using small-to-large merging?

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ok12

6 months ago, # |

can someone please explain the idea behind the 3rd question import math a,b=map(int,input().split()) c,d=map(int,input().split()) t=math.ceil(1.5*abs(b-d))

if t<=abs(a-c) and t!=0: print(int(abs(a-c) -1.5*abs(b-d))//2 + abs(b-d)+1) else: a=abs(b-d) print(a) please explain me why my code is incorrect please

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TrendBattles

6 months ago, # ^ |

you can try to move the initial and the end points to adjacent ones (where (x + y) mod 2 = 0) such that each one is still in its tile and then draft some examples

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ok12

6 months ago, # ^ |

thanks

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udta_kouwa

6 months ago, # |

← Rev. 3 →

My solution for problem-D is to store last K characters as state. AC

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Ishaan_23

6 months ago, # ^ |

Same here, I used a bitset and converted it using to_ulong

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skill_Issued

6 months ago, # ^ |

nice mate. I used bits instead of using strings. Submission

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Dkboss3103

6 months ago, # ^ |

this solution is really good.

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then00bprogrammer

6 months ago, # |

Great contest. D was a really good problem, wish I would've solved E a bit quicker though. Also, How to do C?

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Solved ABDEFG. So strange.

6 months ago, # ^ |

Idea is to move diagonally as much as possible and then align target with the resultant x or y coordinate. Tricky part was to move in the same direction in x. In y its just that difference of y coordinates as height of tiles are 1.

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someanth

6 months ago, # |

how can we get atcoder contest testcase at which our solution got failed...

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Parzival18s

6 months ago, # |

Anybody who found C,D tougher than E,F or just me?

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FFTotoro

6 months ago, # |

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FFTotoro

6 months ago, # ^ |

And G is the easy version of CF1725E.

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shove_it

6 months ago, # ^ |

Can you explain the complexity of your E solution?

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6 months ago, # |

Anyone can give hints on how to solve D?.

My approach to count all the possibilities such there is atleast 1 subarray of K length that is a palindrome and then the subtract that from 2^q and get the answer, but I couldn't calculate it properly(due to overcounting).

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6 months ago, # ^ |

bitmask dp

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6 months ago, # ^ |

Damn bro!! Pease explain a bit further, what will you store using bitmasking? and time complexity of your solution

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https://atcoder.jp/contests/abc359/submissions/54847829

6 months ago, # ^ |

← Rev. 2 →

in mask I will store values of last $$$K$$$ question marks

$$$O(2^K * N * K)$$$

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gurnam771

6 months ago, # ^ |

i think for each index we only need last k-1 elements and check if they are forming palindrome or not

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6 months ago, # ^ |

same thing

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6 months ago, # ^ |

thanks

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ExplodingFreeze

6 months ago, # ^ |

Let $$$dp[i][m]$$$ (for $$$k - 1 \leq i \lt n$$$) represent the number of ways of constructing a good prefix of the first $$$i$$$ characters ending with mask $$$m$$$.

For $$$i = k - 1$$$, this is fairly direct, check all $$$2^k$$$ possible masks and see if they work (are possible and don't contain any palindromes). If they do, set $$$dp[k - 1][m] = 1$$$, otherwise set it to $$$0$$$.

Now notice that the masks for position $$$i$$$ and $$$i - 1$$$ share $$$k - 1$$$ characters. So for any $$$i \geq k$$$, we can check each of the $$$2^k$$$ masks again. If a mask $$$m$$$ is valid, we simply calculate $$$dp[i][m] = dp[i - 1][m'] + dp[i - 1][m' + 1]$$$ where $$$m'$$$ is $$$m$$$ with its highest ($$$(k - 1)$$$-th) bit disabled.

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6 months ago, # ^ |

thanks a lot

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comingsoon.cpp

6 months ago, # |

I think problem B's statement could have been made a lot easier to understand.

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shviv1

6 months ago, # |

My code failed for last three testcases in C.Can someone point out where the mistake lies?

Code

ll sx,sy;
cin>>sx>>sy;

ll tx,ty;
cin>>tx>>ty;



ll ans=(abs(ty-sy));

ans+=(max(0LL,((abs(sx-tx)-(abs(ty-sy))))/2));


cout<<ans<<endl;

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6 months ago, # ^ |

answer for

4 0
-1 0

should be 3 and not 2.

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shviv1

6 months ago, # ^ |

But -1 is invalid input

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6 months ago, # ^ |

2 1
200 98

Expected is 148 but your code gives 147

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Kannayya

6 months ago, # |

← Rev. 2 →

"Can anyone explain C, please? I should have at least tried d, as I've spent so much time with C, actually, not so much time but significant time. I'm feeling bad about it now."

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6 months ago, # |

D is good.

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polosatic

6 months ago, # |

← Rev. 4 →

I used small to large in G. But I can't understand what's wrong in my solution. mp[v][c] contains two numbers: the number of vertices of color c in v's subtree and the total length of paths to them.

Code

#include <bits/stdc++.h>
using namespace std;
#define ll long long
#define ld long double
#define fi first
#define se second
#define pb push_back
#pragma GCC optimize("unroll-loops,O3")
#pragma GCC target("avx2,bmi,bmi2,lzcnt,popcnt")
#define cok cout << (ok ? "YES\n" : "NO\n");
#define dbg(x) cout << (#x) << ": " << x << endl;
#define dbga(x,l,r) cout << (#x) << ": "; for (int ii=l;ii<r;ii++) cout << x[ii] << " "; cout << endl;
#define int long long
#define pi pair<int, int>
const int N = 1e6+9;
int n, ans, a[N], sz[N];
vector<int> vec[N];
map<int, pi> *mp[N];
int add[N];
void dfs(int v, int p = -1) {
    sz[v] = 1;
    pi mx = {-1, -1};
    for (int u: vec[v]) {
        if (u == p) continue;
        dfs(u, v);
        sz[v] += sz[u];
        mx = max(mx, pi{sz[u], u});
    }
    if (mx.se == -1) {
        mp[v] = new map<int, pi>;
        (*mp[v])[a[v]] = {1, 0};
        return;
    }
    mp[v] = mp[mx.se];
    add[v] = add[mx.se] + 1;
    ans += (*mp[v])[a[v]].se + (*mp[v])[a[v]].fi * add[v];
    (*mp[v])[a[v]].fi++;
    (*mp[v])[a[v]].se -= add[v];
    for (int u: vec[v]) {
        if (u == mx.se || u == p) continue;
        for (auto [color, para]: *mp[u]) {
            ans += ((*mp[v])[color].se + (*mp[v])[color].fi * add[v]) * para.fi + (para.se + para.fi * (add[u] + 1)) * (*mp[v])[a[v]].fi;
        }
        for (auto [color, para]: *mp[u]) {
            (*mp[v])[color].fi += para.fi;
            (*mp[v])[color].se += para.se + para.fi * add[u] - para.fi * add[v];
        }
    }
}
signed main() {
    cin.tie(0); ios_base::sync_with_stdio(0);
    cin >> n;
    for (int i = 1; i < n; i++) {
        int a, b;
        cin >> a >> b;
        a--;b--;
        vec[a].pb(b);
        vec[b].pb(a);
    }
    for (int i = 0; i < n; i++) cin >> a[i];
    dfs(0);
    cout << ans;
}

UPD: submission

UPD2: I found issue, AC

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Chimpanzee

6 months ago, # |

+63

If C has a million haters I'm one of them. If C has a hundred haters I'm one of them. If C has one hater it's me. If C has zero haters I have died. If the world is against C I am with the world, if the world is for C I am against the world.

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6 months ago, # ^ |

D E easier than C

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Noobish_Monk

6 months ago, # ^ |

The whole contest was easier

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nichke

6 months ago, # ^ |

+25

Tbh C was cute problem. I didn't like it either, but it's really simple and solution is pretty concise. I feel like CP turned into mindlessly rewriting same algos over and over, especially on AtCoder ABC. Problems like this humiliate you in a good way.

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https://www.luogu.com.cn/problem/P7840

6 months ago, # ^ |

I was helllbent on solving C, finally did and missed on other problems D,E,...

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Chimpanzee

6 months ago, # ^ |

I don't particularly enjoy writing math casework

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pkhaustov

6 months ago, # ^ |

+33

I also don't like problems like this one. But since ABC is supposed to be educational, I try to make use of such problems. For example, I've learned the following lessons (in this particular order):

Skip nasty non-standard grid problems whenever it's possible. Especially if there are problems with more points to get.
Try to find out "simple observation" (as they will call it in editorial for sure) before considering shit tons of edge cases.
Test your solution like crazy. After the first WA, write stress test immediately, don't wait for luck.
Solving more and more similar problems make your more resistant and prepared to them.
Competitive programming is 100% pain.

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Chimpanzee

6 months ago, # ^ |

Last point is real

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harvey_wzy

6 months ago, # ^ |

If C has n haters ($$$ 3 \le n $$$) I'm one of them. If C has two haters that's you and me. If C has one hater I have AFOed. If the world is for C, I'm with you.

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lazy-guy-1618

6 months ago, # |

E was stack and $$$\text{O(N)}$$$ ? (Asking cuz constraints had 2e5)

Code

void solve() {
    int n;
    cin >> n;
    vector<int> h(n), dif(n);
    inc(i, n) cin >> h[i];
    stack<pair<int, int> > st;
    st.push({1e9 + 1, 0});
    inc(i, n){
        auto val = st.top().first;
        if(val < h[i]){
            while(val < h[i]){
                st.pop();
                val = st.top().first;
            }
        }
        dif[i] = st.top().second;
        st.push({h[i], i+1});
    }

    vector<ll> final(n + 1);
    for(int i = 1; i<=n; ++i){
        final[i] = final[dif[i-1]] + (1LL*h[i-1]*(i - dif[i-1]));
    }

    for(int i = 1; i<=n; ++i) cout << final[i] + 1 << " ";
    cout << "\n";
    return;
}

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0x0002

6 months ago, # |

So typical problem G. Maybe it will be better to swap some orders of the problem.

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Misaka16172

6 months ago, # |

Could you guys just STOP making "greedy in grid" problems such as C plz? :(

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someanth

6 months ago, # |

how can we get atcoder contest testcase at which our solution got failed...

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gza

6 months ago, # |

+113

this is the original problem of F?

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6 months ago, # ^ |

Why ABC usually has an original problem?

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incra

6 months ago, # ^ |

My code which get AC in AT also get AC in LG!

But I suppose it wasn't on purpose, as there weren't many easy ideas:(

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CF_xyy

6 months ago, # ^ |

But I think ABC is a good way to let us find the same problem so that get AC.

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gza

6 months ago, # ^ |

so that get AC is wrong in grammar,you should modify it to be "so that you can get AC on the original problem."

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CF_xyy

6 months ago, # ^ |

Yes,you are right.

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6 months ago, # |

This problem is same as F!

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DukhiAatma420

6 months ago, # |

← Rev. 3 →

+21

C was exactly same as https://mirror.codeforces.com/contest/1421/problem/D. Just consider the right point of tiles (moving between left and right didn't encounter any cost), and those right points forms the hexagon.

It can be solved via linear algebra, no need to go into greedy

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Coder2020HCY

6 months ago, # |

+25

Problem F has appeared here previously : https://www.luogu.com.cn/problem/P7840

However, Luogu is a website that currently mainly used in China. So I think it is a coincidence. I only point out for notice.

Here is the translation by Copilot:

There are $$$n$$$ servers.
Each server has a busyness level denoted by $$$v_i$$$.
The servers are connected using $$$n−1$$$ network links.
Each server has a connection count $$$d_i$$$.

The total running time of the server network is given by the expression:

$$$ \sum_{i = 1}^n d_i^2 v_i $$$

The goal is to minimize this value.

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6 months ago, # ^ |

I think so.

I wonder who finds this original problem?

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ziyistudy

6 months ago, # |

-29

Didn't Atcoder check for if there have been the same problem？？？

This problem from Luogu same as today's G.

And this problem is from the year 2021.

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Coder2020HCY

6 months ago, # ^ |

+26

It is a Chinese website and it only support Chinese... It is extremely difficult to check EVERY CP website and search whether there is a same problem. So It can be only coincidence. BTW it didn't cause any significant effect, because it is just an ABC contest.

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Coder2020HCY

6 months ago, # ^ |

+29

They do nothing wrong. Try not to be too aggressive.

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gsczl71

6 months ago, # ^ |

Is it hard to check weight? Inexplicably! Only a fool thinks so. The questions are exactly the same, and luogu is a great site for brushing up on questions, and the bottom line as a question writer is to not come up with exactly the same questions. Even though the questions are in Chinese, can't you learn Chinese? You can't read Chinese, can you? And ziyistudy is stating facts and you're defending a shit group of question writers. I think you are very wrong.

Translated with www.DeepL.com/Translator (free version)

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