动态规划: poj1505改编题

                                         
Copying Books
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

改变后:现在要计算出最小的时间内完成.

题意: n本书, m个人负责抄写. 问怎么划分可以使得时间花的最少. 每个人的抄写书本顺序是升序的.

解题思路:

1. 确定动态规划的状态: dp[i][j]: 表示前i个人完成前j本书的最小时间.

2. 假设: dp[i][j]. 前i-1个人完成了t本书的抄写.从t+1,t+2,...,j本书由第i个人

负责抄写. 显然, dp[i][j] = min(dp[i][j],max(dp[i-1][t],Pt+1+Pt+2+...+Pj));

(1 <= i <= m,i <= j <= m)

代码:

#include <cstdio>

#include <iostream>

#include <cstring>

using namespace std;

#define MAX 505

int n, m;

int p[MAX];

int dp[MAX][MAX]; //dp[i][j]: 前i个人完成前j本书的最小时间.

int add[MAX];

int num;

inline int min(int a,int b)

{
 return a < b ? a : b;

}

inline int max(int a,int b)

{
 return a > b ? a : b;

}

int main()

{
 int i, j, k;

// freopen("input.txt","r",stdin);
 while(scanf("%d %d",&n,&m) != EOF)  //n本书, m个人
 {
  memset(dp,0,sizeof(dp));
  memset(p,0,sizeof(p));
  memset(add,-1,sizeof(add));
  num = 0;
  for(i = 0; i < n; ++i)
  {
   scanf("%d",&p[i]);
   if(i == 0)
    dp[1][i] = p[i];
   else
    dp[1][i] = dp[1][i-1] + p[i];
  }

  int minsize;
  int sum;
  for(i = 2; i <= m; ++i)
  {
   for(j = 0; j < n; ++j)
   {
    minsize = dp[1][j];
    sum = 0;
    for(k = j; k >= 0; --k) //枚举状态转移时的j的划分
    {
     sum += p[k];
     minsize = min(minsize,max(dp[i-1][k-1],sum));
    }
    dp[i][j] = minsize;

   }
  }

  printf("%d\n",dp[m][n-1]);
 }
 return 0;

}