### GlebsHP's blog

By GlebsHP, history, 7 years ago,
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 » 7 years ago, # |   0 Auto comment: topic has been updated by GlebsHP (previous revision, new revision, compare).
 » 7 years ago, # | ← Rev. 2 →   0 In the problem C I used a map to store previous queries to pass on the 100th test case. It worked, lol
•  » » 7 years ago, # ^ | ← Rev. 2 →   0 and main program?Are you using the is_sorted() function in stl?upd:sorry I'm bit naive
•  » » » 7 years ago, # ^ |   0 No, actually I precomputed the answer as said in the problem explanation.But my algorithm runs on O(k*m), Because it checks every column, it's pretty easy to repeat queries just to slow it down. Só I used a map to store previous queries. Here is a link: 24986527.
•  » » » » 7 years ago, # ^ |   +1 you are so clever.
•  » » » » 7 years ago, # ^ |   0 I saw your solution, the key here is using "pref" dynamic programming, not the map. No need to store the previous queries, I can pass all test without map and faster two times yours. 25166547
•  » » » » » 7 years ago, # ^ |   +1 What I meant was that I didn't need to quite do what the editorial said and it worked. Later I did what the editorial explained.cheers
•  » » » » » » 7 years ago, # ^ |   0 :) But they are similar!
•  » » » » 7 years ago, # ^ |   0 nice....haha...
•  » » 7 years ago, # ^ |   0 Seems distinct queries are very needed
 » 7 years ago, # | ← Rev. 3 →   +3 There may be a mistake in prob E's solution.ans[i] += opt[i].height; should be ans[i] += r[i].height;opt.back() should be opt.top()Is it right?
•  » » 7 years ago, # ^ |   0 Yes, fixed now. Thanks you!
 » 7 years ago, # |   0 for problem E can't we use modified longest increasing sub-sequence in O(nlgn) ?
•  » » 7 years ago, # ^ | ← Rev. 2 →   0 i did something similar to LIS, but in the end it's not so far from the solution proposed above #include #include #include using namespace std; int II(){int n;scanf("%d",&n);return n;} void II(int n){printf("%d\n",n);} typedef long long ll; void LL(ll n){printf("%lld\n",n);} typedef pair il; struct ring_t { int inner,outer,height; bool operator<(ring_t b) { if(b.outer==outer)return inner>b.inner; return outer>b.outer; } }; #define MAXN 100000 ring_t V[MAXN]; ring_t *vend=V; int main() { int N=II(); set dp; while(N--)*vend++={II(),II(),II()}; ll H=0; sort(V,vend); for(ring_t *x=V;xinner,x->outer,x->height); ll myH=0; { auto it=dp.lower_bound({x->outer,-1}); if(it!=dp.begin()) { it--; myH=it->second; } } myH+=x->height; H=max(H,myH); { auto r=dp.insert({x->inner,myH}); if(r.second) { auto it=r.first; auto prv=it;prv++; while(prv!=dp.end()&&it->second>=prv->second) { dp.erase(prv); prv=it;prv++; } } } //for(auto y:dp)printf("[%d|%d] ",y.first,y.second);putchar(10); } LL(H); }
 » 7 years ago, # | ← Rev. 2 →   0 In problem C, should be a[i, j] <= a[i + 1, j] instead of a[i, j] < a[i + 1, j].
 » 7 years ago, # |   +19 When CF round #403 begins?
 » 7 years ago, # |   0 I see that there is a binary search tag for problem C. Did anyone solve this problem using binary search? Please enlighten me on how to use binary search for this problem.
•  » » 7 years ago, # ^ |   0 If you just store the indexes which are greater than the next number in an adjacency list for each column. Then for each query you could apply binary search within that range and see if no such element exists even for one column, then Yes. But there seems to be some kind of optimization that I have missed. Plus, the approach is way slower than that in the editorial. This might helpPlease let me know if I'm wrong.
 » 7 years ago, # |   0 Here are two submissions for problem D.2634862726348633These two submissions contain exactly the same code. Yet, I didn't use the same compiler. The first one got TLE, and the second one got Accepted. I would be thankful if anybody could explain me why.
 » 6 years ago, # |   0 can someone provide me dp solution for B
•  » » 4 years ago, # ^ |   0 I have used hashing instead of dp and my solution works in O(n). 67647151
 » 5 years ago, # |   -7 DDangr coongj sanr vieetj nam muoon nawm >u<
 » 5 years ago, # |   0 Editorial for problem C should be:* $up(i,j) = up(i+1,j)$ if $a[i][j] \leq a[i+1][j]$* $up(i,j) = i$, otherwise.I think it is more suitable with your explanation.
 » 4 years ago, # | ← Rev. 3 →   0 I think I got E in O(n log n) time. Sort rings by outer radius first, then inner radius. Sorting should take O(n log n). From the end of the array (biggest outer radius), do the below for each ring: If the ring would fall through the ring that is on top of the current stack, get rid of ring on the top of the stack, repeat until fall is no longer detected. Then add the ring to the stack, since you can do so without the ring falling. Keep track of the sum of height of rings in the stack: when adding a ring, add to the sum, when removing from the stack, subtract from the sum. When adding a ring, update the "maxHeight" variable with the current sum of height, that is, maxHeight = max(maxHeight, currentSum), this variable will be the answer by the end. Knowing that each individual ring can only be added and removed to the stack one time each, this part of the program takes O(n) time.The whole algorithm thus takes O(n log n + n) = O(n log n) time.Feel free to try and prove me wrong.
 » 4 years ago, # |   0 Problem E can be solved using Fenwick tree and coordinate compression! First compress the values of a and b. Sort the rings first based on b,then on base of a in decreasing order. Fenwick tree will store the max height account so far till the value 'a'. We start iterating from 0 to n-1, and the for the max value in the fenwick tree till b[i]-1 call it qmax, and update fenwick tree at index a[i] by h[i]+qmax, and max height so far by h[i]+qmax!
 » 4 years ago, # | ← Rev. 2 →   0 I think that the editorial proof of D is unnecessarily involved. We can just prove by induction that at the $mth$ step, the length of string $s_t$, $t=n-m+1$ is maximum as produced by our algorithm. Can anyone prove me wrong?