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Combinatorics Difficulty 6.5 National Olympiad Prove it Soviet Union

Problem:

A country contains nn cities and some towns. There is at most one road between each pair of towns and at most one road between each town and each city, but all the towns and cities are connected, directly or indirectly. We call a route between a city and a town a gold route if there is no other route between them which passes through fewer towns. Show that we can divide the towns and cities between nn republics, so that each belongs to just one republic, each republic has just one city, and each republic contains all the towns on at least one of the gold routes between each of its towns and its city.

Solution

Solution:

Let the cities be C1C_1, C2C_2, ..., CnC_n. For each town TT take the shortest path from TT to a city. If there are shortest paths to more than one city, then take one to the city with the smallest index. We assign TT to that city.

Now suppose TT is assigned to CiC_i. Let GG be a gold route from TT to CiC_i of length nn. Suppose TT' is another town on GG. Note that TT' cannot be a city, or TT would be assigned to TT', not CiC_i. Suppose the part of GG between TT' and CiC_i has length kk. There cannot be a path from TT' to any city of length <k< k, otherwise we would have a path from TT to a city shorter than nn. Nor can there be a path of length kk from TT' to CjC_j with j<ij < i, otherwise there would be a path of length nn from TT to CjC_j and TT should be assigned to CjC_j. Hence TT' is also assigned to CiC_i, as required.

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Source: MathNet, licensed CC-BY-4.0. Statement reproduced verbatim; metadata (topic, difficulty) added by this project.