POJ 1015 Jury Compromise

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 

Hint

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

Source

Southwestern European Regional Contest 1996

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DP+神奇的思路~

有点类似于状压DP，只是不能记录状态，只能向前递推。

因为又要记录和差，又要记录顺序，所以我们开两个数组，f[i][j]表示已经选了i人，差为j的最大和，而pa[i][j]表示f[i][j]最后选上的人的编号，那么每次用f[i][k]更新f[i+1][k-d[j]+p[j]]，其中j是没有选过的人。j用到pa[i][j]来判断，把pa[i][j]向前递推，如果没有经过j，则j可以被更新。

这样最后输出路径也可以通过递推来实现。

（题目真难懂……读到吐血……最终还是看了翻译……2333）

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>using namespace std;int n,m,d[201],p[201],f[21][801],pa[21][801],zer,ans0,ans1,ans[21],totcas;bool flag;int main(){while(scanf("%d%d",&n,&m)==2 && n){for(int i=1;i<=n;i++) scanf("%d%d",&d[i],&p[i]);memset(f,-1,sizeof(f));memset(pa,0,sizeof(pa));zer=m*20;f[0][zer]=0;for(int i=0;i<m;i++)  for(int k=0;k<=zer*2;k++)    if(f[i][k]>=0)    {    for(int j=1;j<=n;j++)      if(f[i][k]+d[j]+p[j]>f[i+1][k-d[j]+p[j]])      {      int k1=i,k2=k;      while(pa[k1][k2]!=j && k1) k2=k2-p[pa[k1][k2]]+d[pa[k1][k2]],k1--;      if(!k1)      {      f[i+1][k-d[j]+p[j]]=f[i][k]+d[j]+p[j];      pa[i+1][k-d[j]+p[j]]=j;}  }}ans0=0;flag=0;for(int i=0;i<=zer;i++){if(f[m][zer+i]!=-1) ans0=f[m][zer+i],flag=1;if(f[m][zer-i]!=-1) ans0=max(ans0,f[m][zer-i]),flag=1;if(flag){if(f[m][zer+i]>f[m][zer-i]) ans1=i;else ans1=-i;break;}}printf("Jury #%d\nBest jury has value %d for prosecution and value %d for defence:\n",++totcas,(ans0-ans1)/2,(ans0+ans1)/2);int k1=m,k2=zer+ans1,now=0;while(k1){ans[++now]=pa[k1][k2];k1--;k2-=p[ans[now]]-d[ans[now]];}sort(ans+1,ans+m+1);for(int i=1;i<=m;i++) printf(" %d",ans[i]);printf("\n\n");}return 0;}