二分查詢+貪心 ||degree - 3

1 . 題目描述

Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so called scribers. The scriber had been given a book and after several months he finished its copy. One of the most famous scribers lived in the 15th century and his name was Xaverius Endricus Remius Ontius Xendrianus (Xerox

). Anyway, the work was very annoying and boring. And the only way to speed it up was to hire more scribers.

Once upon a time, there was a theater ensemble that wanted to play famous Antique Tragedies. The scripts of these plays were divided into many books and actors needed more copies of them, of course. So they hired many scribers to make copies of these books. Imagine you have m

 books (numbered ) that may have different number of pages ( ) and you want to make one copy of each of them. Your task is to divide these books among k scribes, . Each book can be assigned to a single scriber only, and every scriber must get a continuous sequence of books. That means, there exists an increasing succession of numbers 

 such that i-th scriber gets a sequence of books with numbers between b_i-1+1 and b_i. The time needed to make a copy of all the books is determined by the scriber who was assigned the most work. Therefore, our goal is to minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal assignment.

Input 

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly two lines. At the first line, there are two integers m and k,. At the second line, there are integers  separated by spaces. All these values are positive and less than 10000000.

For each case, print exactly one line. The line must contain the input succession  divided into exactly k parts such that the maximum sum of a single part should be as small as possible. Use the slash character (`/') to separate the parts. There must be exactly one space character between any two successive numbers and between the number and the slash.

If there is more than one solution, print the one that minimizes the work assigned to the first scriber, then to the second scriber etc. But each scriber must be assigned at least one book.

2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100

100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100

2 . 測試用例

input: 3
12 5
48 65 53 3 90 3 90 30 52 75 54 22
10 4
82 36 6 11 33 33 25 79 9 7
9 5
64 16 75 91 1 71 43 81 13


output: 48 65 / 53 3 90 / 3 90 30 / 52 75 / 54 22
82 / 36 6 11 / 33 33 25 / 79 9 7
64 / 16 75 / 91 1 / 71 43 / 81 13

3 .  演算法描述

    1. 主體思路：

             題目要求算出單個“抄寫者”能夠抄寫的最大值（在這裡記為max），問題1怎麼找出這個max？              問題1解答：具體實現為先計算出所有pages的合sum，然後在0-sum之間進行二分查詢，問題2就是怎樣判斷當前的二分查詢的值(mid)就是目標值（max）呢，或者當前的值是大於還是小於目標值？              問題2解答：這裡用到了貪心演算法，對於每一個scriber，按順序（從後往前，因為題目要求如果有相似解的情況，後面scriber抄寫的頁數應該大於前面的）依次累加page，直到超過max的值。迴圈計算直到第一個scriber，計算餘下pages的和，如果這個合的值大於當前的max，那麼證明當前max的值偏小，二分查詢應當向後搜尋了；如果這個合的值小於當前max，則證明當前max的偏大的，二分查詢向前搜尋；如果找到則返回查詢成功

    2. 演算法實現細節

             從上面的描述可以看到程式中最關鍵的實現在於怎樣確定二分查詢的mid值與我們想要找到的max值之間的關係. public static int judge(long mid){ boolean found = false; long sum = 0; int index = bookNum-1; for (int i=scriberNum-1;i>0;i--){ sum = 0; for (;index>=i&&(sum = sum + pages[index])<=mid;index--){ if (sum==mid){ found = true; } } seperators[i-1] = index; } sum = 0; for(;index>=0;index--){ sum = sum + pages[index]; if(sum > mid){ return BIG; } } if(sum==mid){ return FOUND; }else{ if (found){ return FOUND; }else{ return SMALL; } } } 

4 . 程式中遇到的問題

        實現這個程式中遇到的最大問題就是：計算部分合的過程中，出現有的抄寫者沒有分到書的情況（eg : input :5 4,100 100 100 100 100），這種情況在計算部分和的時候就有可能出現第四個與第三個抄寫者分配到了200頁但是第一個抄寫者沒有分配到書的情況。。         解決辦法：         在對“抄寫者”計算“頁數部分合”的時候，除了要滿足“部分合”少於“當前最大值”，還要滿足餘下的書必須能夠保證足夠分配給剩下的“抄寫者”。主要是下面程式碼的判斷部分......         for (int i=scriberNum-1;i>0;i--){ sum = 0; for (;index>=i&&(sum = sum + pages[index])<=mid;index--){ //這裡的"i"數值上等於剩下的抄寫者人數 if (sum==mid){ found = true; } } seperators[i-1] = index; } 

5 . 需要改進的地方

          無