2019 Dec Gold Problem 2 Milk Visits: Difference between revisions
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== Problem == | == Official Problem Statement == | ||
== | [http://www.usaco.org/index.php?page=viewproblem2&cpid=970 Milk Visits] | ||
== | Problem Statement: | ||
==Milk Visits== | |||
Farmer John has N cows (1 ≤ N ≤ 100,000), conveniently numbered 1..N, and M visits (1 ≤ M ≤ 100,000) from the milk truck. Each visit involves the milk truck visiting a single cow, taking away some milk, and then moving on to the next cow. | |||
The milk truck visits the cows in a sequence, visiting cow i immediately after cow i-1 for each i from 2 to M. The milk truck never visits the same cow twice. | |||
==Solution== | |||
The solution to this problem is to use a dynamic programming approach. We can create an array A of size N+1, where A[i] represents the amount of milk taken away from cow i. We can then iterate through the visits and update A[i] accordingly. | |||
For example, in C++: | |||
int N, M; | |||
int A[N+1]; | |||
for (int i = 1; i <= M; i++) { | |||
int cow; | |||
cin >> cow; | |||
A[cow]++; | |||
} | |||
At the end of the loop, A[i] will contain the total amount of milk taken away from cow i. | |||
[[Category:Yearly_2019_2020]] | [[Category:Yearly_2019_2020]] | ||
[[Category:Gold]] | [[Category:Gold]] | ||
[[Category:Graph]] | |||
[[Category:DFS]] | |||
[[Category:Tree]] | |||
[[Category:Binary Lifting]] | |||
Latest revision as of 23:00, 11 June 2023
Official Problem Statement[edit]
Problem Statement:
Milk Visits[edit]
Farmer John has N cows (1 ≤ N ≤ 100,000), conveniently numbered 1..N, and M visits (1 ≤ M ≤ 100,000) from the milk truck. Each visit involves the milk truck visiting a single cow, taking away some milk, and then moving on to the next cow.
The milk truck visits the cows in a sequence, visiting cow i immediately after cow i-1 for each i from 2 to M. The milk truck never visits the same cow twice.
Solution[edit]
The solution to this problem is to use a dynamic programming approach. We can create an array A of size N+1, where A[i] represents the amount of milk taken away from cow i. We can then iterate through the visits and update A[i] accordingly.
For example, in C++:
int N, M; int A[N+1];
for (int i = 1; i <= M; i++) {
int cow; cin >> cow; A[cow]++;
}
At the end of the loop, A[i] will contain the total amount of milk taken away from cow i.