uva 714 Copying Books

原题：

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

1,2,…,m) that may have different 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,… < bk−1≤ bk= m such that i-th scriber

gets a sequence of books with numbers between bi−1+ 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 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,

1 ≤ k ≤ m ≤ 500. At the second line, there are integers p1,p2,…,pmseparated 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,…pmdivided

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.

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

大意：

给你n本书，和k个手抄员，让你把这些书分给这些人。最后计算的时间是最后一个人抄完的用时。现在让你求最后一个人抄完的最小时间。（按给定的顺序来，不能排序！！！）

#include <bits/stdc++.h>
using namespace std;

long long m,k,tot;
int a[510],mark[510];
bool solve(long long index,int kk)//在限定值index下是否满足条件
{
long long tmp=0;
for(int i=0;i<m;i++)
{
if(a[i]>index)
return false;
if(kk==1)
{
tmp=0;
for(int j=i;j<m;j++)
tmp+=a[j];
if(tmp<=index)
return true;
else
return false;
}
if(tmp+a[i]<=index)
tmp+=a[i];
else
{
kk--;
tmp=0;
i--;
}
}
return true;
}
int main()
{
ios::sync_with_stdio(false);
long long t,tmp,seg;
long long l,r,mid;
cin>>t;
while(t--)
{
tmp=tot=0;
memset(mark,0,sizeof(mark));
cin>>m>>k;
for(int i=0;i<m;i++)
{
cin>>a[i];
tot+=a[i];
}
if(k==1)
{
for(int i=0;i<m;i++)
if(i!=m-1)
cout<<a[i]<<" ";
else
cout<<a[i]<<endl;
continue;
}
l=0;r=tot;
while(r-l>1)
{
mid = (l+r)/2;
if(solve(mid,k))
r=mid;
else
l=mid;
}
seg=l+1;
//      cout<<seg<<endl;
int q=1;
tmp=0;
for(int i=m-1;i>=0;i--)
{
if(tmp+a[i]>seg)
{
mark[i]=1;
tmp = a[i];
q++;
}
else
tmp+=a[i];
if(k-q==i+1)
{
for(int j=0;j<=i;j++)
mark[j]=1;
break;
}
}
for(int i=0;i<m-1;i++)
{
cout<<a[i]<<" ";
if(mark[i])
cout<<'/'<<" ";
}
cout<<a[m-1]<<endl;
}
return 0;
}

解答：

最大化最小值，最小化最大值，或者是最大，最小化平均值一般的解法都是用二分法考虑。此题也是刘汝佳那本小白书上的一个例题吧。这里考虑一个index值，如果这个index能满足把这些书分成k份，那么就缩小index的范围，二分的思想就由此产生。首先设置一个上界r，就是总的数页数tot，下界l就设置成0，如果当前index=（l+r)/2能满足正好分成k份，r=index否则l=index。

最后结果返回一个seg值，根据题目要求从后往前贪心选择，模拟就可以了~