Your job is to prepare a lineup of n awardees at an award ceremony. You are given a list of m constraints of the form "i wants to receive an award before j." If you violate a constraint, it might upset the affected award recipient (i) and then i may leave the ceremony. Give an algorithm that prepares such a lineup, (or says that it is not possible) in O ( m + n ) time.

The idea is to use the concept of topological sorting.

Complexity of this algorithm is equal to complexity of DFS algorithm, that is, O(m+n).

Step-by-step explanation:

The idea is to use the concept of topological sorting.

A topological sort also called a topological ordering of a directed graph is a linear ordering of its vertices in a way that for every directed edge uv from vertex u to vertex v, u comes before v in the ordering. For instance, the vertices of the graph may represent tasks to be performed, and the edges may represent constraints that one task must be performed before another; in this application, a topological ordering is just a valid sequence for the tasks.

A topological sorting is possible if and only if the graph has no directed cycles, that is, if it is a directed acyclic graph (DAG).

For every edge (U - - >V), create a directed edge from U to V.

While creating an edge, if an edge creates a cycle, then it is not possible.

n= number of awardees

m= the number of constraint

Let n represent the vertices of a graph and constraints m is representing edges of a graph.

and graph is represented in adjacent of O(m+n)

Constraint i wants to receive award before j

Graphical representation - - - - - > (i) - - - - - - > (j) j depends on i

1. Presently run topological olding on graph G(n, m) where n is vertices and m is edges.

2. Run DFs on Graph, a temporary stack

3. Firstly call topological sorting for all its adjacent vertices and then push it to a stack. Finally, print content of stack.

4. Vertex is push to stack only when all adjacent vertices are already in stack.

Complexity of this algorithm is equal to complexity of DFS algorithm, that is, O(m+n).

i'm not sure i quite understand, could you provide more information?

answer: a video game has three difficulty levels: easy, medium, and expert. the game's scoring works in such a way that a player has to score a fixed number of points to advance to the next level, and then the scoring starts over. the total score needed to clear all three difficulty levels is 7,500. two times the number of points needed to clear the easy level is 500 more than the number of points needed to clear the medium level. twice the number of points needed to clear the medium difficulty level is 1,500 more than the number of points needed to clear the expert level. if x, y, and z represent the easy, medium, and expert levels, respectively, and a is the coefficient matrix of the system of equations modeling this situation, identify the inverse matrix, a-1, and the solution matrix, x, of the system of equations.

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