POJ1505 Copying Books（dp）

Copying Books

Time Limit: 3000MS Memory Limit: 10000KTotal Submissions: 8818 Accepted: 2743

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, ... < b_k-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

29 3100 200 300 400 500 600 700 800 9005 4100 100 100 100 100

Sample Output

100 200 300 400 500 / 600 700 / 800 900100 / 100 / 100 / 100 100

Source

Central Europe 1998

思路：最小化最大值问题，两种解法：1.二分搜索  2.dp解决；
二分搜索：二分查找最小化最大值即可，在查找过程中，每次枚举这个加和，对数列从左到右看一遍，看在每段加和不超过这个枚举值的前提下最少可以将这个数列分成几段。如果分段数小于等于k则这个枚举值偏大或者刚刚好，如果大于k则说明枚举值偏小。
dp解决：d[i][j]:=前i个抄写员复制前j本书的最小完成时间数；
状态转移方程：d[i][j]=min{max{d[i-1][h],p[h+1]+...+p[j]}}；
初始化：d[1][i]=d[1][i-1]+p[i]；
时间复杂度：O（k*m*m）

代码为了方便没输出划分，输出最小化最大值的值；

代码：（dp）

#include <bits/stdc++.h>using namespace std;const int maxn=501;int p[maxn],d[maxn][maxn];int m,k,sum,Min,temp; int main(){    while (scanf("%d%d",&m,&k)) {for (int i=1; i<=m; i++) scanf("%d",&p[i]);for(int i=1;i<=m;i++)    d[1][i]=d[1][i-1]+p[i];for(int i=2;i<=k;i++)    for(int j=1;j<=m;j++)    {    sum=0;    Min=0x3f3f3f3;        for(int h=j;h>=0;h--)//倒着相对比较方便         {        sum+=p[h];        temp=max(d[i-1][h-1],sum);Min=min(Min,temp);         }        d[i][j]=Min;    }printf("%d\n",d[k][m]);}return 0;}