Time Limit: 1000MSMemory Limit: 65536K
POJ 1015

Description

In Frobnia, a far-away country, the verdicts in court trials are determined by a jury consisting of members of the general public. Every time a trial is set to begin, a jury has to be selected, which is done as follows. First, several people are drawn randomly from the public. For each person in this pool, defence and prosecution assign a grade from 0 to 20 indicating their preference for this person. 0 means total dislike, 20 on the other hand means that this person is considered ideally suited for the jury. 
Based on the grades of the two parties, the judge selects the jury. In order to ensure a fair trial, the tendencies of the jury to favour either defence or prosecution should be as balanced as possible. The jury therefore has to be chosen in a way that is satisfactory to both parties. 
We will now make this more precise: given a pool of n potential jurors and two values di (the defence’s value) and pi (the prosecution’s value) for each potential juror i, you are to select a jury of m persons. If J is a subset of {1,…, n} with m elements, then D(J ) = sum(dk) k belong to J 
and P(J) = sum(pk) k belong to J are the total values of this jury for defence and prosecution. 
For an optimal jury J , the value |D(J) – P(J)| must be minimal. If there are several jurys with minimal |D(J) – P(J)|, one which maximizes D(J) + P(J) should be selected since the jury should be as ideal as possible for both parties. 
You are to write a program that implements this jury selection process and chooses an optimal jury given a set of candidates.

Input

The input file contains several jury selection rounds. Each round starts with a line containing two integers n and m. n is the number of candidates and m the number of jury members. 
These values will satisfy 1<=n<=200, 1<=m<=20 and of course m<=n. The following n lines contain the two integers pi and di for i = 1,…,n. A blank line separates each round from the next. 
The file ends with a round that has n = m = 0.

Output

For each round output a line containing the number of the jury selection round (‘Jury #1’, ‘Jury #2’, etc.). 
On the next line print the values D(J ) and P (J ) of your jury as shown below and on another line print the numbers of the m chosen candidates in ascending order. Output a blank before each individual candidate number. 
Output an empty line after each test case.

Sample Input

4 2 
1 2 
2 3 
4 1 
6 2 
0 0 

Sample Output

Jury #1  
Best jury has value 6 for prosecution and value 4 for defence:
2 3

HintIf

your solution is based on an inefficient algorithm, it may not execute in the allotted time.

Source

Southwestern European Regional Contest 1996

这是一道有多个“ 体积维度 ”的01背包问题。

物体:n名陪审团候选人。
背包容积:m名陪审团;
物品体积:
1.人数 j:即每个候选人的人数都为1。
2.辩方得分 d :即辩方给分,1~20的一个整数。
3.控方得分 p :即控方给分,1~20的一个整数。

因此,用bool数组dp[j][d][p]表示已有j个人被选入评审团,当前辩方总分为d,控方总分为p的状态是否可行。

那么很容易得到状态转移方程:dp[j][d][p]=dp[j][d][p]||dp[j-1][d-a[i]][p-b[i]];
初始化:dp[0][0][0]=true
目标:找到一个状态f[n][m][d][p],满足f[n][m][d][p]=1,并且|d-p|尽量小,如
果|d-p|相同,则d+p尽量大。

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