2012 Facebook Hacker Cup World Champion Crowned

By David Cohen Comment

Roman Andreev of Russia was crowned the 2012 Facebook Hacker Cup World Champion, taking home $5,000 for completing one of the final problems of the five-round programming competition in one hour and four minutes.

More than 8,000 programmers from 150 countries entered the qualification round for the Hacker Cup in January, and the 25 finalists spent the weekend at Facebook headquarters in Menlo Park, California competing for the title.

The finalists hailed from Russia, Germany, Poland, Ukraine, China, South Korea, Japan, Taiwan, and the U.S.

The final competition took place this past Saturday from 10 a.m. to 1 p.m. Pacific Time, as the finalists were given three programming challenges, with obvious solutions that would take too long to implement, forcing them to be creative and “hack out” new answers.

Joining Andreev (pictured below) on the winners list were Tomek Czajka of the U.S., who finished second, completing one problem in one hour and five minutes and receiving $2,000; and China’s Tiancheng Lou, who took one hour and 44 minutes to complete one problem and won $1,000.

Do you think you have what it takes to compete for the Hacker Cup? Check out the sample problem below, from Facebook’s engineering group:

Party Time

You’re throwing a party for your friends, but since your friends may not all know each other, you’re afraid a few of them may not enjoy your party. So to avoid this situation, you decide that you’ll also invite some friends of your friends. But who should you invite to throw a great party?

Luckily, you are in possession of data about all of the friendships of your friends and their friends. In graph theory terminology, you have a subset G of the social graph, whose vertices correspond to your friends and their friends (excluding yourself), and edges in this graph denote mutual friendships. Furthermore, you have managed to obtain exact estimates of how much food each person in G will consume during the party if he were to be invited.

You want to choose a set of guests from G. This set of guests should include all of your friends, and the subgraph of G formed by the guests must be connected. You believe that this will ensure that all of your friends will enjoy your party since any two of them will have something to talk about.

In order to save money, you want to pick the set of guests so that the total amount of food needed is as small as possible. If there are several ways of doing this, you prefer one with the fewest number of guests.

The people/vertices in your subset G of the social graph are numbered from 0 to N – 1. Also, for convenience, your friends are numbered from 0 to F – 1, where F is the number of your friends that you want to invite. You may also assume that G is connected. Note again that you are not yourself represented in G.


The first line of the input consists of a single number T, the number of test cases. Each test case starts with a line containing three integers N, the number of nodes in G, F, the number of friends, and M, the number of edges in G. This is followed by M lines each containing two integers. The ith of these lines will contain two distinct integers u and v, which indicates a mutual friendship between person u and person v. After this follows a single line containing N space-separated integers with the ith representing the amount of food consumed by person i.


Output T lines, with the answer to each test case on a single line by itself. Each line should contain two numbers, the first being the minimum total quantity of food consumed at a party satisfying the given criteria and the second the minimum number of people you can have at such a party.


T = 50

1 ? F ? 11

F ? N-1

2 ? N ? 250

N-1 ? M ? N * (N – 1) / 2

G is connected, and contains no self-loops or duplicate edges.

For each person, the amount of food consumed is an integer between 0 and 1000, both inclusive.