poj 1015 Jury Compromise

链接：http://poj.org/problem?id=1015

题目大意：求两组中m个的和之后的差最小，

题目思路：动态规划，可行方案dp(j-1, x)能演化成方案dp(j, k)的必要条件是：存在某个候选人i，i 在方案dp(j-1, x)中没有被选上，且x+V(i) = k。在所有满足该必要条件的dp(j-1, x)中，选出 dp(j-1, x) + S(i) 的值最大的那个，那么方案dp(j-1, x)再加上候选人i，就演变成了方案 dp(j, k)。

题目：

Jury Compromise

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 22306		Accepted: 5751		Special Judge

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

代码：

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
int n,m;
int dp[25][805];
int path[25][805];
int fix=400;
int p[205],d[205];
int dis[205],sum[205];
bool select(int i,int k,int j)
{
while(path[i][k]!=-1)
{
if(path[i][k]==j)
return true;
k=k-dis[path[i][k]];
i--;
}
return false;
}
int main()
{
int cas;
cas=1;
while(scanf("%d%d",&n,&m))
{
if(n+m==0)
break;
for(int i=0;i<n;i++)
{
scanf("%d%d",&p[i],&d[i]);
dis[i]=p[i]-d[i];
sum[i]=p[i]+d[i];
}
memset(dp,-1,sizeof(dp));
memset(path,-1,sizeof(path));
fix=m*20;
dp[0][fix]=0;
for(int i=1;i<=m;i++)
{
for(int k=0;k<=2*fix;k++)
{
if(dp[i-1][k]>=0)
{
for(int j=0;j<n;j++)
{
if(!select(i-1,k,j))
{
if(dp[i][k+dis[j]]<dp[i-1][k]+sum[j])
{
dp[i][k+dis[j]]=dp[i-1][k]+sum[j];
path[i][k+dis[j]]=j;
}
}
}
}
}
}
int k=0;
for(k=0;k<=fix;k++)
{
if(dp[m][fix-k]>=0||dp[m][fix+k]>=0)
break;
}
int dd=dp[m][fix-k]>dp[m][fix+k] ? fix-k : fix+k;
int pp[205];
int tmp=dd;
for(int j=m,k=0;path[j][tmp]!=-1;k++)
{
pp[k]=path[j][tmp];
tmp=tmp-dis[pp[k]];
j--;
}
sort(pp,pp+m);
printf("Jury #%d \n",cas++);
printf("Best jury has value %d for prosecution and value %d for defence: \n",(dp[m][dd]+dd-fix)/2,(dp[m][dd]-dd+fix)/2);
for(int i=0;i<m;i++)
{
printf(" %d",pp[i]+1);
}
printf("\n");
}
return 0;
}