Long Year

Store the number of days in each month in an array. Update february based on whether given year was leap or not. Check $$$1 \leq month \leq 12$$$ and date is valid for that month. Then it's just taking sum of number of days in months before current month, and the number of days in current month.

Virus transmission

Each person represent a circular region. Two regions overlap if distance between centres $$$\leq$$$ sum of radius of both circles. For each infected person, mark all persons as infected (in a boolean array) whose region intersects with any initially infected person.

Clever Criminals

We need number of houses visited odd number of times. House $$$x$$$ is visited only by $$$f(i)-1$$$ criminals (one less, since there's no thief stealing from consecutive houses, representing factor $$$1$$$), where $$$f(i)$$$ is the number of divisors. $$$M \geq N$$$ is important, since it means all divisors upto $$$N$$$ are tried.

Hence, only houses visited even number of times are houses numbered with value $$$x$$$ where $$$x$$$ is a perfect square. This count can be counted naively in $$$O(\sqrt{N})$$$ and subtracted from $$$N-1$$$ House numbered $$$0$$$ is handled separately.

Vaccine-Distribution

Edges of graph forms connected components. All edges in a component get vaccination from the vertex providing vaccine at minimum cost within current component. It can be easily simulated using Disjoint Set Union.

Vaccine Time

Only when there are cyclic dependencies, that we need to merge them. Hence, for the given Directed dependency graph, build Strongly connected components, merge them. Resulting graph would be a DAG, run DP on this DAG to find minimum time at which ingredient at node x would be ready for all nodes $$$1 \leq x \leq N$$$ of original graph.

Now we can sort these times and run binary search. It's easy to prove no dependency is ignored here.

Let's Go Pawri !!

This problem is classical TwoSAT where each type of condition represent one (or more) implications, and we need to check if TwoSAT can be satisfied.

• condition 1 u v: !u => v, !v => u
• condition 2 u v: u => !v, v => !u
• condition 3 u v: u <=> v, !u <=> !v
• condition 4 u v: !v => !u

Read more on 2-SAT here

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