i converted editorialist’s solution to java Code… but in java same code is giving TLE… Here is link:- https://www.codechef.com/viewsolution/18880190
I have a method to share but,the problem with my method was that it only fetched me 40 points,it did not pass the last test cases. But, I tried my best to optimize my code as much possible.
Method :- During input I counted frequency of every elements in matrix. Then I created 2 maps xyz & ank. xyz map stored the cumulative count of pair combination and ank is for marking visited cell for particular pair. Example take pair of (1,2) and let frequency of 1 be 10 and frequency of 2 be 20. So when first time path of (1,2) will occur xyz will store the count of (1,2). Suppose in first path (1,2) contains 15 elements. So, when next time (1,2) path is possible then it will check frequency(1+2) - xyz(1,2) > tot.Here, tot is maximum count variable which will keep getting updated throughout the execution.If condition satisfies then only it will do DFS traversal.
In above @akamoha test case freq of 1=50 and other elements =1. So, count for (1,2)=51. But now it will not do DFS traversal for any other pair as frequency condition won’t satisfy.
It’s a type of brute force but I will highly appreciate if anyone can tell where to optimize my code.
My method = https://www.codechef.com/viewsolution/18869954
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You may want to check out my code xD
You can optimise the code. Obvious things are reducing function calls, replacing Math.max() etc. by ternary operator. Also, remember that usually, constant associated with hashing is high. You can have a look at my code for reference.
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I did not use BFS/DFS - just DSU for the set of all the cells in the field along with subset sizes. First, join all same type pairs of neighbors and save a copy of the resulting data. Then do the following - break the set of all the (unordered) pairs of different neighbor types into subsets in such a way that no two pairs in a subset have a common type. For each subset maintain a hash set of all the types in it and a list of index pairs of neighbors. Maintain a hash map mapping a pair of types into a subset. For each pair of different type neighbors add their indices to the corresponding list. The first time a pair of types is encountered find a subset that contains neither of the two types. If no such subset exists create a new one.
For each of the resulting lists perform union of all the pairs in it restoring to the saved state before processing each list. Maintain the maximal size when performing unions.
Here is python (PyPy) solution.
UPDATE.
This solution got AC, but it will be slow for a case when there is a type with a large number of different neighbor types. Instead we can simply maintain a hash map of separate lists for each pair of neighbor types, and at each DSU iteration restore only those elements that have been changed, as described in the editorial.
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I used just dfs and it fetched me 80 points…i was getting tle in one of the test cases of 2nd subtask.
In dfs , what i did was…i just fixed one number and counted the connecting components of all other numbers ansd the fixed number connected to it which are greater (maintaining each ones count in a map) …thus i was able to count each pair’s connected component in just one dfs run . Also i have used many optimisations to prevent tles … but still i was getting one tle .
However , finally using some conditions (which i thought could have been the reason for the tle )…i was able to get 100 points (though i am not satisfied by my approach).
But still i want to know why i was getting tle in the 2nd subtask’s 3rd test case …if anyone could help 
Here is my solution .
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I formed the components as in the author’s solution and then used the edge base graph traversal instead of dsu.The edges here are the edges of the component graph.I feel in this way we can guarantee the bound to be O(8*N*M) what was tried to achieve in the editorialist’s solution.
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https://www.codechef.com/viewsolution/18897826
I need help. Actually I don’t know it is due to StackOverflow and I can’t resist to avoid it.
@likecs: I think there should be a correction in the pseudo code given for the first subtask. I think it should be:
if vis[i][j] or (a[i][j] != c1 and a[i][j] != c2):
Correct me if I’m wrong. But I think this is correct.
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Can anyone please explain how did we calculate the time complexity of “DSU + sorting map” solution since it has loops and we have to reset values evertime after we perform union find on components?
TIA!
I was talking about akamoha’s test case, the one which is mentioned in the comment above.
We perform one DSU per edge and reset only those values that have been affected when processing a set of edges.
Yeah, you are right.
for (auto &vec : ed) {
vector < int > changed;
for (auto &it : vec.second) {
int x = it.first, y = it.second;
changed.push_back(x);
changed.push_back(y);
int xx = getWho(x), yy = getWho(y);
if (xx == yy) {
continue;
}
if (rand() & 1) {
swap(xx, yy);
}
who[xx] = yy;
sz[yy] += sz[xx];
ans = max(ans, sz[yy]);
}
for (auto &it : changed) {
sz[it] = sizes[it];
who[it] = it;
}
}
**can someone explain how this DSU part works in author’s solution??
In short , what is use of getWho() function and who[] array??**Being new to competitive programming, and knowing nothing about the methods described here (or at least not by the abbreviations used), I developed a completely different method based on my previous experience of flood-filling. This scored me 100 points, completing in 1.66 seconds. If anyone is interested please see my submitted solution (in C#), or ask me to explain it further here.
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@codeislifenluv: Have a look here: https://discuss.codechef.com/questions/129644/need-some-help-in-twofl
Corrected.
@smartnj, you are correct. My implementation would give TLE, because I traverse each component the number of elements it has in it. Thus the test case provided above have similar idea where a lot of 1's are clustered close to each other.
getWho is the “find” while who is the “parent” array compared to the usual DSU implementation.