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Today Tomya buys N retort pouches of food in Chef Ciel's restaurant. The i-th retort pouch contains Vi units of food, and its use-by date is the Ui-th day (1-origin). Moreover, food from the i-th retort pouch can be eaten only within Li days after it was opened. Namely, if Tomya will open the i-th retort pouch on the xi-th day, she can eat food units from it on the k-th day if and only if xi ≤ k ≤ min(Ui, xi+Li−1).
The retort pouches in Ciel's restaurant are designed in such a way that if Vi < Vj, then Ui ≤ Uj.
Tomya will eat at most two units of food everyday. However after the min(Ui, xi+Li−1)-th day, uneaten food units in the i-th retort pouch must be discarded. Your task is to decide when to open each of the retort pouches and which food unit(s) to eat each day in such a way that the number of discarded units will be minimum.
There is also a restriction that once you open some pouch, food units from previously opened pouches should be discarded.
The above restriction is an addition to the problem statement that was made after the contest. We are extremely sorry about that. We all had missed the possibility to achieve better score without this restriction in some situations. We are grateful to Mikhail Kever for pointing out our mistake.
The first line of the input contains an integer T denoting the number of test cases. The description of T test cases follows.
The first line of each test case contains an integer N denoting the number of retort pouches.
Each of the following N lines contains 3 space separated integers Vi, Ui, Li denoting the parameters of the i-th pouch.
For each test case, output a single line containing the minimum number of food units which Tomya must discard.
- 1 ≤ T ≤ 2013
- 1 ≤ N ≤ 2013
- 1 ≤ Vi, Ui, Li ≤ 20000000000000 (2 * 1013)
- If Vi < Vj, then Ui ≤ Uj
- The sum of N in one test file does not exceed 20130.
Input: 3 3 8 9 5 7 5 3 10 100 1 3 5 4 3 7 8 5 8 10 4 1 10000 1 10000 Output: 9 0 9998
Example case 1. One of the optimal ways is the following:
The 1st pouch is opened on the day x1 = 4,
the 2nd pouch is opened on the day x2 = 1,
the 3rd pouch is opened on the day x3 = 8,
the 1st pouch is available at the interval [4, 8],
the 2nd pouch is available at the interval [1, 3],
the 3rd pouch is available at the interval [8, 8].
On days 1, 2, 3, Tomya eats 2 units of the 2nd pouch each day.
On days 4, 5, 6, 7 Tomya eats 2 units of the 1st pouch each day.
On day 8, Tomya eats 2 units of the 3rd pouch.
In this case, Tomya discard the 1 unit of the 1st pouch, and 8 units of the 3rd pouch.
Example case 2. There is only one optimal way as follows:
The 1st pouch is opened on the day x1 = 1,
the 2nd pouch is opened on the day x2 = 3,
the 3rd pouch is opened on the day x3 = 7,
the 1st pouch is available at the interval [1, 3],
the 2nd pouch is available at the interval [3, 7],
the 3rd pouch is available at the interval [7, 10].
On days 1, 2, Tomya eats 2 units of the 1st pouch each day.
On day 3, Tomya eats 1 unit of the 1st pouch and 1 unit of the 2nd pouch.
On days 4, 5, 6, Tomya eats 2 units of the 2nd pouch each day.
On days 7, 8, 9, 10, Tomya eats 2 units of the 3rd pouch each day.
In this case, Tomya can eat all units of food.
Example case 3. Please note that here L1 is larger than U1.
|Tags||cook30, greedy, laycurse, medium-hard|
|Time Limit:||2.013 sec|
|Source Limit:||50000 Bytes|
|Languages:||C, JAVA, PYTH, PYTH 3.6, CS2, PAS fpc, PAS gpc, RUBY, PHP, GO, NODEJS, HASK, SCALA, D, PERL, FORT, WSPC, ADA, CAML, ICK, BF, ASM, CLPS, PRLG, ICON, SCM qobi, PIKE, ST, NICE, LUA, BASH, NEM, LISP sbcl, LISP clisp, SCM guile, JS, ERL, TCL, PERL6, TEXT, CLOJ, FS|
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