題目連結：

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

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

題目意思：

有n個人，每個人有個d值和p值，求選出m個人，使得|sum(d)-sum(p)|最小，如果最小值相同，則選擇|sum(d)+sum(p)|較大的。輸出選出的人。

n<=200 d,p<20 m<=20

解題思路：

dp,實際上就是一個揹包。

|sum(d)-sum(p)|最小，實際上就是找能湊成的最靠近0的.由於-20=<d-p<=20,所以把每個d-p都加上20，變成非負的整數。然後算出最靠近20*m的就是滿足要求的最小的。

dp[i][j]：表示當sum(d)-sum(p)為i，且有j個人時，能否湊成。儲存一個add[i][j]：表示這j個人的sum(d)+sum(p)

最後在20*m附近找.

程式碼：

//#include<CSpreadSheet.h>

#include<iostream>
#include<cmath>
#include<cstdio>
#include<sstream>
#include<cstdlib>
#include<string>
#include<string.h>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
#include<set>
#include<stack>
#include<list>
#include<queue>
#include<ctime>
#include<bitset>
#include<cmath>
#define eps 1e-6
#define INF 0x3f3f3f3f
#define PI acos(-1.0)
#define ll __int64
#define LL long long
#define lson l,m,(rt<<1)
#define rson m+1,r,(rt<<1)|1
#define M 1000000007
//#pragma comment(linker, "/STACK:1024000000,1024000000")
using namespace std;

#define Maxn 8100

int dp[Maxn][25],add[Maxn][25];
int savea[220],saveb[220];
vector<int>re[Maxn][25];
int n,m;


int main()
{
   //freopen("in.txt","r",stdin);
   //freopen("out.txt","w",stdout);
   int ca=0;

   while(scanf("%d%d",&n,&m)&&n+m)
   {
       for(int i=1;i<=n;i++)
           scanf("%d%d",&savea[i],&saveb[i]);
       memset(dp,0,sizeof(dp));
       memset(add,0,sizeof(add));
       dp[0][0]=1;

       int lim=20*n*2;
       for(int i=0;i<=lim;i++)
            for(int j=0;j<=m;j++)
                re[i][j].clear();

       for(int i=1;i<=n;i++)
       {
           int temp=savea[i]-saveb[i]+20;

           for(int j=lim;j>=temp;j--)
           {
               for(int k=m;k>=1;k--)
               {
                   if(dp[j-temp][k-1])
                   {
                       int te=add[j-temp][k-1]+savea[i]+saveb[i];
                       if(!dp[j][k])
                       {
                           add[j][k]=te;
                           re[j][k]=re[j-temp][k-1];
                           re[j][k].push_back(i);
                           dp[j][k]=1;
                       }
                       else if(te>add[j][k])
                       {
                           add[j][k]=te;
                           re[j][k]=re[j-temp][k-1];
                           re[j][k].push_back(i);
                       }
                   }
               }
           }
       }
       int ansa,ansb,a;

       int p=0,pp=20*m;
       int k=m;

       while(!dp[pp+p][k]&&!dp[pp-p][k])
           p++;
       if(!dp[pp+p][k])
       {
           ansa=abs((pp-p+add[pp-p][k]-20*m))/2;
           ansb=(add[pp-p][k]-pp+p+20*m)/2;
           a=pp-p;
           //printf("Best jury has value %d for prosecution and value %d for defence:\n",ansa,)
       }
       else if(!dp[pp-p][k])
       {
           ansa=abs((pp+p+add[pp+p][k]-20*m))/2;
           ansb=(add[pp+p][k]-pp-p+20*m)/2;
           a=pp+p;
       }
       else
       {
           if(add[pp+p][k]>add[pp-p][k])
           {
               ansa=abs((pp+p+add[pp+p][k]-20*m))/2;
               ansb=(add[pp+p][k]-pp-p+20*m)/2;
               a=pp+p;
           }
           else
           {
               ansa=abs((pp-p+add[pp-p][k]-20*m))/2;
               ansb=(add[pp-p][k]-pp+p+20*m)/2;
               a=pp-p;
           }
       }
       printf("Jury #%d\n",++ca);
       printf("Best jury has value %d for prosecution and value %d for defence:\n",ansa,ansb);
       for(int i=0;i<re[a][k].size();i++)
            printf(" %d",re[a][k][i]);
       printf("\n\n");
   }
   return 0;
}