FILLMTR - Editorial

editorial
, , , ,
21

The problem is not yet added to the practice portal

22

It is already added dear.

23

I tried to solve it using bicoloring(dfs) but getting TLE on subtask 2 of task 2. Can someone help me ? Here is my code:
https://www.codechef.com/viewsolution/15367330

24

The editorial seems actually very long. But its good as it is detailed.
I have summarized the tutorial in my short ~10 min video here : https://www.youtube.com/watch?v=6qWg7-O4Fmg

You can refer this for complete list of video tutorials for SEP17 including WEASTELX.

1 Like
25

The editorial seems actually very long. But its good as it is detailed.
I have summarized the tutorial in my short ~10 min video here : https://www.youtube.com/watch?v=6qWg7-O4Fmg

You can refer this for complete list of video tutorials for SEP17 including WEASTELX.

26

Yeah, but was not visible at that moment. Fixed now.

27

I used bipartite graph for 1 difference edges and disjoint sets for 0 difference edges. I am getting WA. Please, take a look at the code :


[1]


  [1]: https://www.codechef.com/viewsolution/15414051
28

I have used the same approach

Explanation https://discuss.codechef.com/questions/109357/fillmtr-editorial/110997

A link to code is given alongwith…

Feel free to ask anything…
Upvote…

29

@taran_1407
Sorry I don’t know how to reply to your comment.
I used the same approach. Can you find out why my code is getting TLE in the 2nd task of the 2nd subtask ?

[My Code][1]
[1]: https://www.codechef.com/viewsolution/15379751

30

` cananyone tell me why my logic isnt working
#include
#include
using namespace std;

        /*
         * 
         */
        const int MAX=2e6+9;
        int main() {
            int i  ,t ,n ,q;
            cin>>t;
            while(t--){
                cin>>n>>q;
                int v[MAX]={0} ,x ,y ,d , a[MAX]={0};
                for(i=1;i<=n;i++){
                    a[i]=0;
                    v[i]=0;
                }
               int f=0;
                while(q--){
                    cin>>x>>y>>d;
                    if(x==y){
                        if(d!=0){
                            f=1;
                        }
                        v[x]=1;
                        a[x]=0;
                        
                    }
                    else  if(v[x]==0&&v[y]==0){
                  
                        a[x]=0;
                        a[y]=d+a[x];
                        v[x]=1;
                        v[y]=1;
                    }
                    else  if(v[x]==0&&v[y]==1){
                        a[x]=a[y]+d;
                        v[x]=1;
                    }
                    else  if(v[x]==1&&v[y]==0){
                        a[y]=a[x]+d;
                        v[y]=1;
                    }
                    else  if(v[x]==1&&v[y]==1){
                        if(abs(a[x]-a[y])!=d){
                            f=1;
                        }
                    }
                }
                if(f==1){
                    cout<<"no"<<"\n";
                }
                
                else{
                    cout<<"yes"<<"\n";
                }
                
            }
            
        }
         
           `can anyone tell me why my logic is not working

first i declared two array a[] and other array v[] array v will keep the record of all the integers which
are visited now during each query 4 condition arises

  1. both 1st and 2nd are unvisited
  2. 1st is visited and 2nd is unvisited
    3 1st is unvisited and 2nd is visited
    4 both are visited
    now
    for the 1st condition
    i marked both the nodes as visited
    and assigned a[i]=d;
    a[j]=a[i]+d;
    for the 2nd condition
    marked 2nd node as visited
    a[j]=a[i]+d;
    for 3rd condition
    marked 1st node as visited
    a[i]=a[j]+d;
    for 4 th condition
    since both nodes are visited and are assigned some values
    therefore
    if(abs(a[i]-a[j])!=d)
    ans=no and exit
    else
    ans=yes;
31

Will anyone please tell me whats the problem with my solution …
Thnks in advance…

i tried checking if graph is bipartite or not…
https://www.ideone.com/OGDY8I

PLease do check …

32

can anyone give me test case where my code fails ?
here is the link - https://www.codechef.com/viewsolution/15423175

33

Well, You ought to go for vertex 2-coloring instead of bipartite graph testing…

You might like to have a look at my approach with explanation

Feel free to ask anything…

Please upvote

34

Well, i dont code in C language, so i can’t understand your code…

But, You might like to have a look at my approach with explanation

Feel free to ask anything…

Please upvote…

35

can anyone tell me what is wrong in this logic

  1. Initially all nodes are uncolored.
  2. As I get values of i,j and val, if both a[i] and a[j] are uncolored then if val is 0 then give them same color or if val is 1 then give different colors.
  3. If anyone of a[j] or a[i] is already colored then color other one accordingly if val is 1 then different color or same color if val is 0.
  4. If both a[i] and a[j] are colored then if val is 0 then they should have same color if not then matrix is wrong similarly if val is 1 then they should have different color if not the matrix is wrong.
  5. In all other cases matrix will be correct.
36

Hello! can anyone pls tell me why the solution on going according to author’s logic is getting WA? I have written code according to the logic explained in the “Author’s Solution” section of the editorial which seems to be absolutely correct. Any help is much appreciated. Thanks!! U can find my code here.

37

Simple solution can found only by sorting the initial m commands and then trying to assign the values to the array. This takes O( m*log(m) ) time, which is the time for sorting. U can see the solution here.