UVa714 Copying Books

714 Copying Books

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 nished 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
1; 2; : : : ; m) that may have dierent number of pages (p1; p2; : : : ; pm) and you want to make one copy of
each of them. Your task is to divide these books among k scribes, k  m. 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 0 = b0 < b1 < b2; : : : < bk1
 bk = m such that i-th scriber
gets a sequence of books with numbers between bi1
+ 1 and bi. 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 nd the optimal
assignment.

Input

The input consists of N cases. The rst line of the input contains only positive integer N. Then follow
the cases. Each case consists of exactly two lines. At the rst line, there are two integers m and k,
1  k  m  500. At the second line, there are integers p1; p2; : : : pm separated by spaces. All these
values are positive and less than 10000000.

Output

For each case, print exactly one line. The line must contain the input succession p1; p2; : : : pm 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 rst scriber,
then to the second scriber etc. But each scriber must be assigned at least one book.

Sample Input

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

Sample Output

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

Solution

“最大值最小”，很容易让我们想到二分答案。

• 检查时，每次尽量从左向右划分即可

• 输出时，因为要求字典序最小,所以有一个贪心过程，每次尽量从右往左划分即可

• 代码

#include
#include
#include
using namespace std;
const int maxm = 500 + 5;
int m, k, p[maxm];

int solve(long long maxp) { //二分答案中的检查步骤
  long long done = 0;
  int ans = 1;
  for(int i = 0; i < m; i++) {
    if(done + p[i] <= maxp) done += p[i];
    else { ans++; done = p[i]; }
  }
  return ans;
}

int last[maxm]; 
void print(long long ans) { //含贪心过程的输出
  long long done = 0;
  memset(last, 0, sizeof(last));
  int remain = k;
  for(int i = m-1; i >= 0; i--) {
    if(done + p[i] > ans || i+1 < remain) {
      last[i] = 1; remain--; done = p[i];
    }
    else {
      done += p[i];
    }
  }
  for(int i = 0; i < m-1; i++) {
    printf("%d ", p[i]);
    if(last[i]) printf("/ ");
  }
  printf("%d\n", p[m-1]);
}

int main() {
    freopen("input.in", "r", stdin);
    freopen("output.out", "w", stdout);
    int T;
    scanf("%d", &T);
    while(T--) {
      scanf("%d%d", &m, &k);
      long long tot = 0;
      int maxp = -1;
      for(int i = 0; i < m; i++) {
        scanf("%d", &p[i]);
        tot += p[i];
        maxp = max(maxp, p[i]);
      }
      long long L = maxp, R = tot;
      while(L < R) {
        long long M = L + (R-L)/2;
        if(solve(M) <= k) R = M; else L = M+1;
      }
      print(L);
  }
  return 0;
}