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# LEMPLOY - Editorial

Contest

Author: Vaibhav Gosain

Editorialist: Vaibhav Gosain

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[i][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[i][m].
4. Edge from all nodes in layer (j-1) to corresponding nodes in layer j with capacity P[i][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[i][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[i]. We claim N*MAX - (cut of above graph) is the value of total productivity of the firm if person i is given role R[i] for all i.

Reason:

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

# AUTHOR'S SOLUTION:

Author's solution can be found here.

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question asked: 03 Oct '16, 22:11

question was seen: 948 times

last updated: 03 Oct '16, 22:21