Facebook yesterday announced the winner of its 2012 Hacker Cup. Over the weekend, 25 of the world's best programmers came to Facebook to compete for the title of World Champion in Facebook's second-annual Hacker Cup finals. Roman Andreev (pictured above) received the title of World Champion for Facebook's 2012 Hacker Cup.
Facebook's Hacker Cup is an annual algorithmic programming contest open to engineers from around the world. Programmers are judged on accuracy and speed as they race to solve algorithmic problems to advance through up to five rounds of programming challenges.
This year, 8,000 people from 150 countries entered the qualification round in January. Over the course of three more rounds, the contestant pool was narrowed down to 25 finalists from Russia, Germany, Poland, Ukraine, China, South Korea, Japan, Taiwan, and the U.S. Two months later, the finalists were flown out to Facebook to compete for the World Champion title.
The finalists were judged on accuracy and speed as they raced to solve three programming challenges in three hours. The competition kicked off at 10:00 AM PST Saturday morning. For the next three hours, the finalists tackled the problems, in order of easiest to most difficult. The contestants were able to specify their programming language and developer environment preferences beforehand, and about 70 percent chose C++.
At 1:00 PM PST, an alarm sounded and the competition was over. First place and $5,000 went to Roman Andreev of Russia, who completed one problem correctly in one hour and four minutes. Tomek Czajka from the U.S. got second place and $2,000 by a margin of one minute, finishing one problem correctly in one hour and five minutes. Third place and $1,000 went to Tiancheng Lou from China, who completed one problem correctly in one hour and 44 minutes.
Do you think you could have done better? Below is a practice problem:
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 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 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.
I'm not surprised contestants only solved one out of three problems this year.