BAISED - Biased Standings
Usually, results of competitions are based on the scores of participants. However, we are planning a change for the next year of IPSC. During the registration each team will be able to enter a single positive integer : their preferred place in the ranklist. We would take all these preferences into account, and at the end of the competition we will simply announce a ranklist that would please all of you.
But wait... How would that ranklist look like if it won't be possible to satisfy all the requests?
Suppose that we already have a ranklist. For each team, compute the distance between their preferred place and their place in the ranklist. The sum of these distances will be called the badness of this ranklist.
Given team names and their preferred placements find one ranklist with the minimal possible badness.
The first line of the input file contains an integer T specifying the number of test cases. Each test case is preceded by a blank line.
Each test case looks as follows: The first line contains N : the number of teams participating in the competition. Each of the next N lines contains a team name (a string of letters and numbers) and its preferred place (an integer between 1 and N, inclusive). No two team names will be equal.
For each of the test cases output a single line with a single integer : the badness of the best ranklist for the given teams.
Input: 2 7 noobz 1 llamas 2 Winn3rz 2 5thwheel 1 NotoricCoders 5 StrangeCase 7 WhoKnows 7 3 ThreeHeadedMonkey 1 MoscowSUx13 1 NeedForSuccess 1 Output: 5 3Explanation:
In the first test case, one possible ranklist with the minimal badness is:
1. noobz 2. llamas 3. Winn3rz 4. 5thwheel 5. NotoricCoders 6. WhoKnows 7. StrangeCase
In the second test case all ranklists are equally good.Note: the input file will not exceed 5MB.
use long long int instead of int
*USE long long int instead of int*
In python, make sure to consider the empty line between inputs (before each test case). Cost me a runtime error!.
use vector of pairs;)
How can we prove that sorting of preferred ranks would give us optimal answer ?
This problem can be solved in O(n) time using counting sort but the rank is too high so it is preferable to use
Try to solve it in O(n) time.
Can anyone tell what is wring with this approach? I'm making a boolean array(size n) with all values set to 0(1 indexed) . In each input,I go to the kth (k==preferred place) index in the array and if it is 0 then i do nothing. But is it is 1 then I find the minimum distance among the left and right of the kth index where arr[index]=zero and count the distance and also set arr[index] to 1.
Why does SPOJ have this bad habit of not providing constraints? Took me a WA for not using long long :(
can anyone provide me the test cases with the answers