Knowledge required: complete search or disjoint set

When one has Ebola, everyone in his class can potentially get Ebola. With that observation, we can express each class as a set. When person $i$ and $j$ are in the same class, we can simply merge the 2 sets that $i$ and $j$ belongs to. Hence, the intended solution to this problem is to use disjoint set to manage the contents of the sets.

The disjoint set data structure has 2 operations $\mathrm{Find}(a)$ and $\mathrm{Merge}(a,b)$. When person $i$ and $j$ are in the same class, we can simply do $\mathrm{Merge}(i,j)$. After processing all the classes, we can simply compare the value of $\mathrm{Find}(1)$ and $\mathrm{Find}(N)$ to determine if person $N$ is infected.

A properly implemented disjoint set has a time complexity of $\mathcal{O}(\alpha (N))$ for each operation, where $\alpha (N)$ is the Inverse Ackermann function, which is smaller than $5$ for all reasonable values of $N$.

Time Complexity: $\mathcal{O}(N+\sum K_i\alpha (N))$

Alternate solution

Create a vertex for each student and each class. If a student is in a class, then create an edge between that student and that class. In total, there are $N+M$ vertices and $\sum K_i$ edges.

Next, do a graph search starting at student $1$. Make sure to output the visited students. The visited classes can be ignored.

Time Complexity: $\mathcal{O}(N+\sum K_i)$