vovuh's blog

Codeforces Round #550 (Div. 3) Editorial
By vovuh, history, 5 years ago, In English

1144A - Diverse Strings

Idea: MikeMirzayanov

Tutorial
Solution

1144B - Parity Alternated Deletions

Idea: MikeMirzayanov

Tutorial
Solution

1144C - Two Shuffled Sequences

Idea: MikeMirzayanov

Tutorial
Solution

1144D - Equalize Them All

Idea: vovuh

Tutorial
Solution

1144E - Median String

Idea: vovuh

Tutorial
Solution

1144F - Graph Without Long Directed Paths

Idea: MikeMirzayanov

Tutorial
Solution

1144G - Two Merged Sequences

There is different solution for the problem, it is pretty interesting! Thanks, Roundgod!

Idea: vovuh

Tutorial
Solution
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Comments Comments (41)
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5 years ago, # |
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G Tutorial is not available?

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5 years ago, # |
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In Problem G, I'm getting WA on test 12. Please help me out. 52149636

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5 years ago, # |
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Why is problem D tagged as dfs and similar?

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5 years ago, # |
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Interesting idea on E, very cool problem!

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5 years ago, # |
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Solving E problem using Python gave TLE, don't know why. I stored the Number as int(base 26). Shouldn't it work?

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    5 years ago, # ^ |
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    Nobody can see your submitted solutions. You're probably getting TLE because the operations you are doing on the numbers are too slow.

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      5 years ago, # ^ |
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      No problem, I think I got the Reason why it is giving TLE.

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5 years ago, # |
← Rev. 3   Vote: I like it 0 Vote: I do not like it

I don't understand E's editorial,can anyone please explain?What are we doing? I thought of seeing the string as base 26,so we can convert it into decimal and take the mean. For example: az = 26^0*25 + 26^1*0 = 25, bf = 26^0*5 + 26^1*1 = 31 now taking the mean we get 28,converting into base 26 we get 12 (26^0*2+26^1*1 = 28),which is bc. But for large string lengths we may have to do 26^100000,so I thought it is not feasible. What are we doing in this editorial?

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    5 years ago, # ^ |
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    You misunderstood the meaning.We can represent the string to a number, with base 26. And then we can figure out that if string s is lexicographically smaller than t, the "number" which s represents is smaller than t's "number". So what we need to calculate is a "number" which is the median of [s's number, t's number]. We can Easily calculate it by writing "BigIntegar", to storing big number and doing operations. Here's my Code Link: Here

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      5 years ago, # ^ |
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      Thanks a lot for replying,in author's code,I understand the string is stored as int with base 26,but what are the 2 operations done? Is he adding the 2 vectors?

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        5 years ago, # ^ |
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        2 vectors represents 2 BigInt(base 26). so, now he add the BigInt and then divide it by 2. (Same as we did addition and division with paper and pencil in childhood :) )

        Then print the vector as characters.

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5 years ago, # |
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This is a real Div.3 :)

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5 years ago, # |
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Another solution to G by considering whether the max.element in the array is in the decreasing or increasing subsequence:52159992

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5 years ago, # |
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anyone please help. In problem F, why my code give me Runtime Error on test 15?.

My Submission: 52160039

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5 years ago, # |
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Runtime error in F: can someone help me debug: https://codeforces.com/contest/1144/submission/52208937

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    5 years ago, # ^ |
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    maybe it's the for-loop for finding the largest degree

    you iterate it for i = [0, m), while vector "graph" has the size of n. RTE when m > n.

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5 years ago, # |
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for E:someone plzz explain how division by 2 is done in 26 base

for (int i = 0; i <= k; ++i)

{

    int rem = a[i] % 2;

    a[i] /= 2;

    if (i + 1 <= k) {

       a[i + 1] += rem * 26;

    } else {

       assert(rem == 0);

}

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +3 Vote: I do not like it

    If the digit in the pos[i] is not divisible by 2, you can just pass a one to the next digit, a 1 from $$$1*26^i$$$ is equivalent to 26 in $$$26^\left( i-1 \right)$$$, if it is divisible by 2 then you just need to divide it by 2

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5 years ago, # |
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problem ID:1144F

problem F is also done by BFS

problem F code(happy coding)

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5 years ago, # |
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Can I solve E using big integer?

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5 years ago, # |
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for problem g is this correct approach? let say i have two array X(increasing),Y(decreasing) and A is Merge(X,Y). Find the longest increasing sequence of A this sequence will contain X and atmost one element of Y and after removing that element from Y Y will still remain decreasing.

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5 years ago, # |
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Why is this solution for E giving TLE? https://codeforces.com/contest/1144/submission/52223063

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    5 years ago, # ^ |
    ← Rev. 2   Vote: I like it +5 Vote: I do not like it

    You are doing ans = ans + (char)(median[i] + 'a'); This takes more time as compared to ans+=(char)(median[i] + 'a'); I am not sure of its reason. But i too have faced similar issues before.

    Can someone explain why this removes TLE?

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5 years ago, # |
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there is a another way to solve problem G
first we found the LIS of the array and then there is a fact : if we can split the given sequence a into one increasing sequence and one decreasing sequence then at most one of the elements of the LIS could not appear in x ( x is the increasing sequence ) otherwise in the y ( the decreasing sequence ) we have two numbers A , B which a[A] < a[B] and A < B and its impossible so if we fix which index i is not gonna appear in x we can solve the problem

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5 years ago, # |
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In problem E, why is it only allowed to do a[i] %= 26 when i > 0?

For this test case, for example:

6

nijfvj

tvqhwp

The sum in the base 26 is: 33 3 25 13 17 24

Why is it allowed to have this "33" in the first position?

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    5 years ago, # ^ |
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    In my program，its allow a[i] % 26 in i = 0

    but if u do that, you need to deal with leading zeros，it's a little bit more trouble, but it doesn't really matter.

    For example

    33 3 25 13 17 24 -> 1 7 3 25 13 17 24

    then u div 2

    0 16 14 25 19 21 25

    now,u need cheak the zero,if u dont do that ,the answer will be an extra a

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5 years ago, # |
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Help me! I dont understand print problem F in tutorial: cout << (color[e[i].first] < color[e[i].second]); ???????????? please

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    5 years ago, # ^ |
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    when you split vertices in to two part . you want obtain direct of edges such that every edge in part one direct to the second pare or you can direct all edges from second part to the first part . and we know that every edges has two end point such that one of them from first and one of them from second (because its bipartite) so if color[first endpoint] < color[second endpoint] that means color[first endpoint] = 0 so that condition return 1 and otherwise return 0

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      5 years ago, # ^ |
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      ^_^ yeah, thanks U somuch

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5 years ago, # |
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In editorial 'G',what is the meaning of 'maximum possible minimal'?

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    5 years ago, # ^ |
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    minimal element from the sequence is a const value , this const value be maximum as possible means .

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5 years ago, # |
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Somebody please tell me why this submission 52323233 works but this submission 52321524 does not ??

The only difference in both the codes is that in accepted one, I've accepted the input as strings while in wrong answer I've accepted them in character arrays

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5 years ago, # |
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I can not understand the tutorial of problem G.

Please anyone Explain the tutorial.

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5 years ago, # |
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In problem F,the solution does not include the visited array.So,won't it end in a infinite loop and give tle as it will keep on checking two vertices again and again?

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    5 years ago, # ^ |
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    color acts as a visited array. If any node has color other than -1 means it is already visited.

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5 years ago, # |
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But I my opinion , the statue may be dp[0][i] represent that "The i-th element is in increasing sequence and the max possible element in the decreasing sequence",and dp[1][i] represent that "The i-th element is in decreasing sequence and the min possible element in the increasing sequence" May be I am Wrong ,But I hope you can correct my mistake , Thank you

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7 months ago, # |
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Problem F
Why does this give TLE??

#include <bits/stdc++.h>
using namespace std;

#pragma GCC optimize "trapv"
#define ll long long
#define nline cout << "\n"
#define nl "\n"
#define all(x) x.begin(), x.end()
#define sz(x) (int)x.size()
#define pb(x) push_back(x)
#define pii pair<int, int>
#define db(str, x) cout << str << " = " << x << nl;
#define dbv(a)        for(auto it: a)cout<<it<<" "; cout << nl;
#define db1(a)        cout<<a<<"\n";
#define db2(a,b)      cout<<a<<" "<<b<<"\n";
#define db3(a,b,c)    cout<<a<<" "<<b<<" "<<c<<"\n";
#define fastio() ios_base::sync_with_stdio(false);cin.tie(0);

const int M = 1e9+7;
const int N = 1e5;


int dfs(int u, vector<int>& color, set<pii>& edges, vector<vector<int>> adj){
  for(int child : adj[u]){
    if(color[child]){
      if(color[child] == color[u])
        return true;
    }
    else{
      if(color[u] == 1){
        color[child] = 2;
        edges.insert({u, child});
      }
      else{ 
        color[child] = 1;
        edges.insert({child, u});
      }
      if(dfs(child, color, edges, adj))
        return true;
    }
  }
  return false;
}

void solve(){
   int n, m;
   cin >> n >> m;
   vector<vector<int>> adj(n);
   vector<pii> tmp;
   int x, y;
   for(int i=0; i<m; i++){
      cin >> x >> y;
      x--,y--;
      adj[x].pb(y);
      adj[y].pb(x);
      tmp.push_back({x, y});
   }
   vector<int> color(n);
   set<pii> edges;
   color[0] = 1;
   
   if(dfs(0, color, edges, adj)){
    db1("NO");
   }
   else{
    db1("YES");
    for(auto pr : tmp){
      if(edges.count(pr)) cout << 0;
      else cout << 1;
    }
    nline;
   }
  
}

int main(){
  fastio()
  #ifndef ONLINE_JUDGE
  freopen("Input.txt", "r", stdin);
  freopen("Output.txt", "w", stdout);
  #endif

  int t=1;
  // cin >> t;
  for(int tc=1; tc<=t; tc++){
    solve();
  }
  
  return 0;
}
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