Copying Books
Time Limit: 2 Seconds Memory Limit: 65536 KB
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, ... 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 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
【题意】
给出一个数列有n个数，要求用分割分把这个数列分成m段，不能改变原数列的顺序。每段至少一个数。求使得加和最大的那段的加和最小的划分方案。如果有多组解的话先要保证第一段和尽量小，若仍有多组解，要先保证第二段和尽量小，以此类推。
【解题思路】
因为这个数列是有序的，并且要求最大的最小可以考虑二分。将数列和作为二分上限，数列的最大值作为二分的下限，二分和，然后检验这个和能将数列分成几份，若比给定的m大，说明这个和不够大，则l=mid+1。当然为了保证如果有多解，需要保证第一段的和尽量小这一条件，需要从后面往前分，因为这样如果分成的段数比m小的话就在前面加就好啦。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int maxn=505;
int a[maxn],n,m,vis[maxn];
int check(int x)
{
int sum=0,cnt=0;
memset(vis,0,sizeof(vis));
for(int i=n-1;i>=0;i--)
{
sum+=a[i];
if(sum>x)
{
sum=a[i];
cnt++;
vis[i]=1;
}
}
return cnt+1;
}
int binarysearch(int l,int r)
{
while(l<r)
{
int mid=(l+r)/2;
int t=check(mid);
if(t>m)l=mid+1;
else r=mid;
}
return r;
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
int l=0,r=0;
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%d",&a[i]);
l=max(l,a[i]);
r+=a[i];
}
int t=binarysearch(l,r);
int num=check(t);
for(int i=0;i<n&& num<m;i++)
{
if(!vis[i])
{
num++;
vis[i]=1;
}
}
for(int i=0;i<n;i++)
{
printf("%d",a[i]);
if(vis[i])printf(" /");
if(i<n-1)printf(" ");
}
printf("\n");
}
return 0;
}