Description
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
Source
假設(shè)有M本書(shū)（編號(hào)為1，2，…，M），想將每本復(fù)制一份，M本書(shū)的頁(yè)數(shù)可能不同（分別是P1，P2，…，PM）。任務(wù)時(shí)將這M本書(shū)分給K個(gè)抄寫(xiě)員（K≤M〉，每本書(shū)只能分配給一個(gè)抄寫(xiě)員進(jìn)行復(fù)制，而每個(gè)抄寫(xiě)員所分配到的書(shū)必須是連續(xù)順序的。
意思是說(shuō)，存在一個(gè)連續(xù)升序數(shù)列0=b0＜b1＜b2＜…＜bk-1＜bk=m，這樣，第i號(hào)抄寫(xiě)員得到的書(shū)稿是從bi-1+1到第bi本書(shū)。復(fù)制工作是同時(shí)開(kāi)始進(jìn)行的，并且每個(gè)抄寫(xiě)員復(fù)制的速度都是一樣的。所以，復(fù)制完所有書(shū)稿所需時(shí)間取決于分配得到最多工作的那個(gè)抄寫(xiě)員的復(fù)制時(shí)間。試找一個(gè)最優(yōu)分配方案，使分配給每一個(gè)抄寫(xiě)員的頁(yè)數(shù)的最大值盡可能小（如存在多個(gè)最優(yōu)方案，只輸出其中一種）。
設(shè)dp[i,j]表示前j個(gè)人復(fù)制前i本書(shū)所需要的最少時(shí)間，有狀態(tài)轉(zhuǎn)移方程dp[i,j]=min(dp[i,j],max(dp[v,j-1],sum[v+1,i]))，其中1<=i<=m，1<=j<=k，j-1<=v<=i-1，sum[v+1,j]表示第v+1本書(shū)到第i本書(shū)的頁(yè)數(shù)之和。
#include<iostream>
using namespace std;
const int MAXN = 510;
int sum[MAXN],path[MAXN],dp[MAXN][MAXN];
int main()
{
int m,k,i,j,v,ca,p,t;
scanf("%d",&ca);
while(ca--){
scanf("%d %d",&m,&k);
for(sum[0]=0,i=1;i<=m;i++){
scanf("%d",&p);
sum[i]=sum[i-1]+p;
}
memset(dp,-1,sizeof(dp));
for(dp[0][0]=0,i=1;i<=m;i++)
for(j=1;j<=i && j<=k;j++){
if(j==1) dp[i][j]=sum[i];
else
for(v=j-1;v<=i-1;v++){
t=max(dp[v][j-1],sum[i]-sum[v]);
if(dp[i][j]==-1 || t<=dp[i][j])
dp[i][j]=t;
}
}
for(i=m,j=k-1,p=0;i>=1;i--){
p+=sum[i]-sum[i-1];
if(p>dp[m][k] || i<=j){
path[j--]=i+1;
p=sum[i]-sum[i-1];
}
}
for(i=j=1;i<=m;i++){
if(i>1) printf(" ");
if(j<k && path[j]==i){
printf("/ ");
j++;
}
printf("%d",sum[i]-sum[i-1]);
}
printf("\n");
}
return 0;
}