Filling a bookshelf (2) P35814


Statement
 

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The statement of this problem is almost identical to the problem : “”, with two exceptions: Now, when filling the bookshelf, the relative order of the books in the input can be changed. And b can be as large as 105.

I.e., the problem is: Given b books, each one with width wi and height hi, use them to fill a bookshelf as much as possible. The second book (if any) must be shorter than the first book, the third book must be taller than the second book, …, and the last book must be taller than the penultimate book. Note that “short” and “tall” refer to the hi’s, and that the goal is to maximize the sum of the wi’s of the chosen books.

Input

Input consists of several cases. Each case begins with b, followed by b pairs with wi and hi. Assume 1 ≤ b ≤ 105 and 1 ≤ wi, hi ≤ 109. A special case with b = 0 marks the end of input.

Output

For every case, print the maximum possible sum of the widths of the chosen books.

Public test cases
  • Input

    3  900000000 8  700000000 4  800000000 6
    2  2 8  3 6
    4  8 2  9 3  6 1  7 4
    2  5 7  4 7
    4  4 20  6 10  3 20  8 10
    6  15 3  11 1  12 3  10 2  14 2  15 3
    6  15 3  11 1  12 3  10 2  14 3  15 3
    6  11 1  15 2  12 2  10 3  14 2  15 2
    0
    

    Output

    2400000000
    3
    24
    5
    15
    67
    65
    41
    
  • Information
    Author
    Salvador Roura
    Language
    English
    Official solutions
    C++
    User solutions
    C++
    Event
    Novè Concurs de Programació de la UPC - Semifinal
    Date
    2011-06-29