LEMPLOY - Editorial

editorial
graph
hard
maxflow
mincut-maxflow

#1

PROBLEM LINK:

Practice

Contest

Author: Vaibhav Gosain

Editorialist: Vaibhav Gosain

DIFFICULTY:

HARD

PREREQUISITES:

maxflow, min cut: Topcoder Tutorial

EXPLANATION:

This problem can be solved via the concept of minimum cut.

We model the following graph, containing M+2 layers:

0th and (M+1)th layer for source and sink, M other layers for each role in firm, each containing N nodes.

Edges:

  1. Among all pairs of nodes i,j within the same layer, an edge of capacity D*[j] from j to i.
  2. Edge from source to all nodes of layer 1, with capacity INF.
  3. Edge from all nodes in layer M to sink , with capacity MAX-P*[m].
  4. Edge from all nodes in layer (j-1) to corresponding nodes in layer j with capacity P*[j-1] for 2<=j<=M
  5. Edge from all nodes in layer j to corresponding nodes in layer (j-1) with capacity INF for 2<=j<=M

where MAX = maximum possible value of P*[j]

The required maximum productivity of the firm = N*MAX - mincut

Why does this work?

Let S be the source and T be the sink.

The infinite edge from layer j to layer j-1 ensures that if ith node in layer j is on the S side in the cut, corresponding node in layer j-1 will also be on the S side in the cut.

Now, say the last layer whose ith node lies in S side of the cut is R*. We claim NMAX - (cut of above graph) is the value of total productivity of the firm if person i is given role R for all i.

Reason:

  1. Productivity contributed by P*[R*] is due to the edge between layer R* and layer R*+1.
  2. For any 2 people i and j such that R[j] > R*, there will be R[j]-R* layers for which node j is on S side of cut and node i is on the T side. Hence the value D*[j] will be added R[j]-R* times to the mincut.

AUTHOR’S SOLUTION:

Author’s solution can be found here.