# CEMENT Editorial

CEMENT_FACTORY EDITORIAL.

SETTER : @jprm2
Editorialist : @jprm2

Difficulty : Easy-Medium

Pre-Requisites : observation , Dynamic programming ,Gravity , Bellman Ford Algorithm. Memoriazation Technique.

EXPLANATION

THe problem says you to transport the sacks between source station to destination steps with minimum cost first calculate cost between any two stations let’s say there are 2 stations 1 and 2 having heights as 3 and 5 respectively and distance between this two stations is lets say 4 so the cost of transporting from building 1 to building 2 is (5-3)4=24=8 against the gravity. And the cost to transport from 2 to 1 is different with (3-5)4=-24=-8.So both the paths are different with exactly opposite costs.

Now we understood that we needed to calculate the costs between all the T station queries given . Calculating this cost takes O(1) time if you already know the heights and difference between any two stations . You were given the distance between adjacent stations but you can find the distance between any 2 stations with the help of previously computed prefix sum and finding the difference between ith and jth value of the prefix sum array for ith and jth node in the asked query.

Now we knew all the nodes of a graph we knew cost of edges connecting 2 nodes of the asked query of this directed graph then we can easily calculate .The shortest
path using any of the known algorithm for single source shortest path but since negative edges and negative cycle both can be present in the test case then

you can use bellman ford algorithm to compute our minimum cost and if their is any negative cycle present over this directed graph then we will output -1
for all the queries because if any one negative cycle is present then we can always repeat along the cycle to make our value infinitely minimum then

if no path is possible for the given (2 nodes or stations) then we will directly output INFINITY THAT IS EQUAL TO 10000000000000000. in my solution.
By Dynamic programming you will store all the distances from given source in a single O(N*T) time by bellman ford algorithm and storing this information
answers all the given S queries in O(1) Time.

Links to study Bellman ford algorithm:-

Setter’s and Editorialist’s Solution:

Setters and Editorialists Solutions.