### arham_doshi's blog

By arham_doshi, history, 4 months ago, I got bored with solving and wanted to do something which is related to cp and also very fun so i decided to write this tutorial.bare me for my bad English .

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## 19) Planet queries I

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Hope this is helpfull to you . I will try to add more as I solve furthure. Comments (58)
 » Auto comment: topic has been updated by arham_doshi (previous revision, new revision, compare).
•  » » https://cses.fi/problemset/First register to create an account then you will be able to submit your codes.
•  » » » Thanks :)
 » Nice initiative.
 » Thanks arham_doshi
 » all hail ......editorialist arham_doshi
 » This may also help. https://github.com/ankitpriyarup/CSES_ProblemSet_Solution
 » Hi, thanks for the editorials. I am unable to understand the problem $High$ $Scores$. It says we can use a tunnel several times, that means if there is an edge with positive weight can't we use it several times(infinite times i.e. $a$ -> $b$ then $b$ -> $a$ ... ) and our answer will be -1? and i think in the editorial you meant Bellman ford instead of Floyd Warshall.
•  » » Yes, we can use it hence you have to print -1 as said in the problem statement.
•  » » » If that is the case why our answer is 5 for sample? We can use the edge with positive weight from 1 and use it infinite times and our answer will be -1. Sorry if I misunderstood you:(
•  » » » » All tunnels are one-way tunnels...means it's a directed graph.
•  » » » » » Oops. Sorry my bad. :(
•  » » » » yeah may be i got confused sed between bellman ford and floyd warshal.talking about sample it is a directed edge so after going from a to b you canot go back. " the tunnel starts at room a, ends at room b "
 » how to solve grid problems Here is an additional trick for grid problems. Specifically for problems where the input is some kind of maze with # for "blocked" cells and . for "free" cells. I like the approach of using the dx and dy arrays but the possible function feels clunky to me in some problems.Instead I put a "frame" made out of # around the whole grid. That is: when the input is ..#..#. .#..#.. #.#..#. .#...#. instead I consider the input ######### #..#..#.# #.#..#..# ##.#..#.# #.#...#.# ######### It is really easy to implement. Initially fill the whole grid with #, then read the input.
•  » » 4 months ago, # ^ | ← Rev. 2 →   Nice one, i like it. I use possible at many places like when when solveing a dp probelm
 » Auto comment: topic has been updated by arham_doshi (previous revision, new revision, compare).
 » Auto comment: topic has been updated by arham_doshi (previous revision, new revision, compare).
 » I am weak at graph. I think, it will help me so much..Thank you so much Sir !!!
 » 4 months ago, # | ← Rev. 2 →   I'm having trouble with Building Roads problem. For some reason, I am getting TLE for test cases 6-11, although my solution is similar in idea to ones that have been accepted. How can I change my code so I don't get TLE? #include #include using namespace __gnu_pbds; using namespace std; typedef long long ll; typedef long double ld; typedef unsigned int ui; typedef unsigned long ul; typedef pair pi; typedef pair pl; typedef vector vi; typedef vector vl; typedef vector vpi; typedef vector vpl; typedef priority_queue pq; typedef priority_queue,greater> pqg; typedef tree,rb_tree_tag,tree_order_statistics_node_update> indexed_set; #define mp make_pair #define pb push_back #define f first #define s second #define lb lower_bound #define ub upper_bound void dfs(int start, vector adj, vector& visited){ if(visited[start]) return; visited[start]=true; for(auto u:adj[start]) dfs(u,adj,visited); } int main() { ios::sync_with_stdio(0); cin.tie(0); int n,m; cin>>n>>m; vector adj(n+1,vi()); for(int i=0; i>a>>b; adj[a].pb(b); adj[b].pb(a); } vector visited(n+1,false); vi add; for(int i=1; i<=n; i++){ if(!visited[i]){ dfs(i,adj, visited); add.pb(i); } } cout<1){ for(int i=1; i
•  » » you can do it using dsu its way more simpler and easier to code,you wont get TLE at all.Here's my code https://pastebin.com/Xx0UnNtq
•  » » I used a similar solution, but much simpler and without DSU.https://ideone.com/q5NCal
 » 4 months ago, # | ← Rev. 2 →   I'm currently making Video editorials for tree section of CSES. Interested people can check : CSES (DP ON) TREE ALGORITHMSI'm hoping to finish it in around a week, will be happy to know if people find it helpful :)UPD : added first 5, do share suggestions if any.
•  » » Thank You so much, it will be very helpful.
•  » » » Do share suggestions/feedback if any :)
 » This is really helpful.Thank You arham_doshi !!
 » 13) flight routes give TLE as its O(nmk)
•  » » Its o(m*k), i was writing there max(n, m)*k
•  » » » 3 months ago, # ^ | ← Rev. 2 →   I calculated the complexity. Shouldn't it be O(k(m+nlogn))?. It is still giving me TLE for the above algorithm.Secondly, why did you run a classical djikstra before? you are not using the distance array in the algorithm.
•  » » » » 3 months ago, # ^ | ← Rev. 2 →   i am finding all the possible paths . you can think it as bfs . it i optimised brute force where you can only take take vertex at max k time. i was solving all the question in line so in 2-3 ques i needed dijkstra so i just kept it there ignore it. dm me your code.
•  » » » » here's my code https://pastebin.com/r8sMf9ea. hope it may help you
 » Auto comment: topic has been updated by arham_doshi (previous revision, new revision, compare).
 » CAN SOMEONE PLEASE HELP ME???for Investigation question above, i am doing exactly what the author does except for one thing:he uses a visited array and if already visited, continues..i do something like pair = pq.top(); if(pair.first > distance[pair.second]) continue; i want to know constraint / testcase wise why this will give TLE. it was giving tle for 3 large testcasesthe issue was resolved on introducing a boolean visited arraymycode = https://ideone.com/e6ZTWL (not needed tho)
 » Thank you so much! I got stuck at some and tried to find hints and solutions months ago, so it is relieving to find these.
 » 7 weeks ago, # | ← Rev. 3 →   https://cses.fi/paste/cfd81e7f5b4db001e2466/what is wrong in this solution? CycleFinding?????
•  » » Not sure but may be possible that the distance you are initating with LLONG_MAX may get over fload when you do distance[i] + someting
•  » » » I have a check distance[source]!=LLONG_MAX
 » can someone help me understanding this lines in Labyrinth problem? p = from[p.x][p.y]; what does from means? i was unable to googling it. is there any keyword? thank you
•  » » 7 weeks ago, # ^ | ← Rev. 2 →   It is just 2d vector, i have declared it on 3rd line, it helps us in recreating the path.
•  » » » what does FRM and DIR stand for? You mentioned it on the same Labyrinth problem. Thanks for this blog btw.
•  » » » » FRM stands for FROM means from which cell it has come to this cell, and DIR stands for directon means from previous cell in which directon do we move (right, left, up, down) to reach current cell. Both these auxiliary array help us in recreating path after wards.
•  » » » » » got it thanks, i did a similar thing but i thought FRM or DIR was some advanced thing that ive never heard of
•  » » » got it, thanks!
 » Thanks for the editorial!!Really great job!Just want to add that in the problem Monsters, after applying multi source bfs we can use dfs as well instead of bfs to find the path from 'A' to boundary, as we need not find the shortest path. My Implementation#include using namespace std; int n,m; vector > a(1000,vector (1000)); vector > d(1000,vector (1000,-1)); vector > vis(1000,vector (1000,false)); string ans; void dfs(int i,int j) { vis[i][j]=true; if(i==n-1 or i==0 or j==0 or j==m-1) { cout<< "YES\n"; cout<ans.size()+1) { ans+="D"; dfs(i+1,j); } // up if(i-1>=0 and not vis[i-1][j] and d[i-1][j]>ans.size()+1) { ans+="U"; dfs(i-1,j); } //right if(j+1ans.size()+1) { ans+="R"; dfs(i,j+1); } //left if(j-1>=0 and not vis[i][j-1] and d[i][j-1]>ans.size()+1) { ans+="L"; dfs(i,j-1); } // backtrack vis[i][j]=false; if(ans.size()>0) { ans.pop_back(); } } int main() { cin>>n>>m; vector > mons; pair start; for(int i=0;i>a[i][j]; if(a[i][j]=='M') { mons.push_back({i,j}); } else if(a[i][j]=='A') { start={i,j}; } else if(a[i][j]=='#') { vis[i][j]=true; } } } queue > q; for(pair p:mons) { q.push(p); vis[p.first][p.second]=true; d[p.first][p.second]=0; } while(not q.empty()) { pair v=q.front(); q.pop(); int i=v.first; int j=v.second; // down if(i+1=0 and not vis[i-1][j]) { vis[i-1][j]=true; d[i-1][j]=d[i][j]+1; q.push({i-1,j}); } //right if(j+1=0 and not vis[i][j-1]) { vis[i][j-1]=true; d[i][j-1]=d[i][j]+1; q.push({i,j-1}); } } // reset vis matrix for(int i=0;i
•  » » Thanks dhruv, one more thing i noticed is that if you use dfs you don't need the auxiliary array to recreate path, nice one.
•  » » » what is the time complexity if we run BFS on multiple monsters? For each monster we might take n.m time right?then overall complexity will be too high?
•  » » » » no the time complexity will still be n*m , we will just start bfs with all monsers simalteniosly.
 » IN THE FLIGHT ROUTES PROBLEM if you want to find k routes each vertex is visited atmax k times in k routes. Can somebody explain me why it is so?
 » How to solve Planet Queries II ?
 » Too much helpful. Thanks a lot.
 » In Highscores I dont understand how you detected positive cycle using dfs , can anyone please explain a bit more..
•  » » We first find positive cycles using Bellman-Ford, and then we check if these cycles can be reached on a path from 1->n. We can do this by reversing the edges and running a dfs from node n, and running a normal dfs from node 1.
 » Nice work arham_doshi
 » Can anyone help me out in the implementation of Dijkstra's Algo in python using heaps, I'm getting TLE in 2 test cases only My codefrom heapq import * n,m=map(int,input().split()) l=[] d={i:[] for i in range(1,n+1)} for i in range(m): a,b,c=map(int,input().split()) d[a].append([b,c]) dis=[float('inf') for i in range(n+1)] dis=0 h=[] heappush(h,[0,1]) #distance, node while h: z=heappop(h) distance,cur=z for i in d[cur]: node,di=i if distance+ di < dis[node]: dis[node]=distance + di heappush(h,[dis[node],node]) print(*dis[1:]) Thanks
•  » » I suggest you to use c++ or java as the test cases in cses are very tight.
 » It gives wrong answer for last 2 cases
 » arham_doshi I have a different approach for the problem "Flight Discount" But getting wrong answer on 3 TC can you help me please,I can't find any TC.My Solution