National Olympiad in Informatics, China, 2008
Day 2, Problem 3 - Tournament Matching
With the Olympics right around the corner, students' interest in physical education is skyrocketing. As NOI 2008 approaches, the school is currently planning to host a ping-pong tournament. Beings a very avid ping-pong enthusiast, this is the perfect opportunity for little Z to show off his skills. Thus, he quickly registers himself and begins his intensive preparation.
The tournament will follow a single-elimination format where only the winners of a round advances. There just happens to be n (n is a power of 2, so we may as well let n = 2k) participating students. After the first round, 2k−1 students will be eliminated, and the other 2k−1 will advance to the next round. After the second round, only 2k−2 students will advance, and so on until after round k, where the champion and runner-up will be determined. Every contestant is assigned a number from 1 to n, where little Z is contestant number 1. The IDs of all students are unique, and all students will be assigned to n slots. Then, the tournament process will follow a structure like the one below.
|Slot 1||Slot 2||Slot 3||Slot 4||Slot 5||Slot 6||Slot 7||Slot 8|
|Winner of Slots 1 and 2||Winner of Slots 3 and 4||Winner of Slots 5 and 6||Winner of Slots 7 and 8|
|Winner of Slots 1, 2, 3, and 4||Winner of Slots 5, 6, 7, and 8|
The tournament bracket for when n = 8.
To attract more students, the prizes of the tournament are very generous. The competitors being eliminated in round i are awarded ai dollars, and the champion will receive the largest prize of ak+1 dollars. Needless to say, the prize amounts will satisfy a1 < a2 < … < ak+1.
In the warm-up rounds prior to the real tournament, little Z was repeatedly defeated. After some careful analysis, he concluded that the key reason behind his failure was not an issue of his skill, but was because the students who defeated him just happened clash with his playing stylistically. Little Z thought to himself, how nice would it be if these students could be avoided in the actual tournament!
Let's assume that we already know the probabilities of winning for all contestants in pairwise matches. The probability of contestant A winning against contestant B is PA,B (where it is guaranteed that PA,B + PB,A = 1). Little Z wishes to know for a certain tournament structure (by reassigning the slots for each contestant), what is the largest possible prize that he can obtain. Can you help little Z design a matching scheme for this tournament so that the expectations for his prize money is as large as possible?
There are 10 test cases
match10.in that will be given to your program (through standard input). They can be downloaded here for you to study: match.zip
The first line of each test case contains an integer from 1 to 10, specifying the test case number (
matchx.in will have x on the first line).
The second line contains a single integer n, representing the total number of contestants. The data guarantees that k will be a nonnegative integer for 2k = n.
For the following n lines, each line contains n real numbers Pi,j between 0 and 1 inclusive, representing the probability of contestant i defeating contestant j. Each real value will be given to two decimal places. Especially take note of when Pi,j = 0.00.
For the following k + 1 lines, each line contains an integer describing the prize money for a particular round. The i-th of these lines contain ai.
Output n lines, where line i represents the number of the contestant that is matched to slot i. Little Z's number must be matched to the first slot.
0 4 0.00 0.70 0.60 0.80 0.30 0.00 0.60 0.40 0.40 0.40 0.00 0.70 0.20 0.60 0.30 0.00 1 2 3
1 4 2 3
After the first round is over, the probability for contestant 1 (little Z) to advance is 80%, the probability for contestant 2 to advance is 60%, the probability for contestant 3 to advance is 40%, and the probability for contestant 4 to advance is 20%.
In the second round (the championship match), the probability of contestant 1 (little Z) winning the first two matches is 80%*(60%*70% + 40%*60%) = 52.8%. Thus, the probability of little Z losing his first match is P1 = 1 − 0.8 = 0.2, the probability of him winning the first match but losing the second match is P2 = 0.8 − 0.528 = 0.272, and the probability of him being the champion is P3 = 0.528. From this, the expected value for his prize money is 0.2*1 + (0.8 − 0.528)*2 + 0.528*3 = 2.328.
We supply a tool match_check.py for you to test whether your solutions are accepted. The usage for this program is
match_check.py <input-file> <output-file>
When you execute
match_check.py, the program will process your input and output files and print the result to standard output. Possible results of the execution include:
- Abnormal termination: An unknown error occured.
Format error:The output file is incorrectly formatted.
Not a permutation:The output file is not a permutation of the integers from 1 to n.
OK. Your answer is xx:The output will be accepted. xx will be the expected prize money.
Each test case will be graded independently.
For each test case, if your output is illegal (i.e. has invalid formatting, doesn't fit with the requirements, etc.) then you will receive a score of 0.
Otherwise if the expected prize money for your output is your_ans, the prize money determined by the contest organizers' solution is our_ans, and a parameter for grading is d, then your score out of 10 for the test case will be:
- 12, if your_ans > our_ans.
- 1, if your_ans < our_ans*d.
- Otherwise, your score will be:
The mathematical expected value is the most fundamental property of a random variable. It reflects the size of the average value of a random variable, also known as the expectation or mean. It is an extension of the simple arithmetic mean, or a weighted average of all the possible values. Ex. If a city has 100,000 families, 1,000 of which do not have children, 90,000 of which has one child, 6,000 of which have two children, and 3,000 of which have three children, then the number of children in a family within this city is a random variable that can take on the values of 0, 1, 2, or 3. The probability for 0 children is 0.01, of 1 child is 0.9, of 2 children is 0.06, and of 3 children is 0.03. Its expected value is 0×0.01 + 1×0.9 + 2×0.06 + 3×0.03 = 1.11, and therefore the average family has 1.11 children.
To calculate the random variable for this problem, assume that little Z's chances of losing the first round is P1, chances of winning the first round but losing the second is P2, the chances of winning the first two rounds but losing the third is P3, and so on. Then the expected value for little Z's prize money is:
P1•a1 + P2•a2 + … + Pk+1•ak+1
Point Value: 40 (partial)
Time Limit: 10.00s
Memory Limit: 512M
Added: Aug 01, 2014
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