$Sol$
This is one having a plurality of "volume dimensions" $ 0 / $ 1 knapsack problem.
The $ N $ N $ $ considered as candidates number items, each item has the following three volumes:
1. "number" of each candidate "number" is $ 1 $, and ultimately to fill the volume of $ M $ backpack
2. "defense score", $ A [I] $
3. "anti-party score", $ B [I] $
The defense is required out of $ D $ and $ P $ prosecution out the absolute value of the difference between the $ | the DP | $ minimum, if the selection method is not unique, then select $ D + P $ maximum program
Therefore, the $ a [i] -b [i ] $ as one volume, the $ a [i] + b [ i] $ as the value of the article
Note here $ a [i] -b [i ] $ may be negative, so that uniform plus $ 4000 $
In consideration of the $ I $ individuals stage, $ f [j] [k ] $ $ I $ denotes have been previously selected for Custom $ J $ individuals, this time difference is out of both of $ k $ time , both of the maximum score.
$f[j][k]=max(f[j][k],f[j-1][k-(a[i]-b[i])]+a[i]+b[i])$
Since the required output scheme, the array plus a $ d [i] [j] [k] $ denotes $ f [i] [j] [k] $ $ I $ of candidates is not selected from the group selected from the group, so that the recording program.
$Code$
1 #include<iostream> 2 #include<cstdio> 3 #include<cstring> 4 #define Rg register 5 #define il inline 6 #define mem(a,b) memset(a,b,sizeof(a)); 7 #define go(i,a,b) for(Rg int i=a;i<=b;i++) 8 #define yes(i,a,b) for(Rg int i=a;i>=b;i--) 9 using namespace std; 10 il int read() 11 { 12 int x=0,y=1;char c=getchar(); 13 while(c<'0'||c>'9'){if(c=='-')y=-1;c=getchar();} 14 while(c>='0'&&c<='9'){x=(x<<3)+(x<<1)+c-'0';c=getchar();} 15 return x*y; 16 } 17 int T,n,m,hhh=400,t,ct,a[201],b[201],f[21][801],as[21]; 18 bool d[201][21][801]; 19 int main() 20 { 21 while(1) 22 { 23 n=read(),m=read();if(!n)break; 24 printf("Jury #%d\n",++T); 25 go(i,1,n)a[i]=read(),b[i]=read(); 26 mem(f,-777);mem(d,0); 27 f[0][hhh]=0; 28 go(i,1,n) 29 yes(j,m,1) 30 go(k,-400,400) 31 { 32 if(k-(a[i]-b[i])>=-400 && k-(a[i]-b[i])<=400 && f[j-1][k-(a[i]-b[i])+hhh]+a[i]+b[i]>=0 && f[j][k+hhh]<f[j-1][k-(a[i]-b[i])+hhh]+a[i]+b[i]) 33 f[j][k+hhh]=f[j-1][k-(a[i]-b[i])+hhh]+a[i]+b[i],d[i][j][k+hhh]=1; 34 //if(f[j][k+hhh]>=0)cout<<"i:"<<i<<" j:"<<j<<" k:"<<k<<" f:"<<f[j][k+hhh]<<" d:"<<d[i][j][k+hhh]<<endl; 35 } 36 go(i,0,400) 37 { 38 if(f[m][hhh+i]>=0){if(f[m][hhh+i]>f[m][hhh-i])t=i;else t=-i;break;} 39 else if(f[m][hhh-i]>=0){t=-i;break;} 40 } 41 printf("Best jury has value %d for prosecution and value %d for defence:\n",(f[m][hhh+t]+t)/2,(f[m][hhh+t]-t)/2); 42 int nw=n;ct=m; 43 while(nw>0) 44 { 45 int dd=d[nw][ct][t+hhh]; 46 if(dd)as[ct]=nw,ct--,t-=(a[nw]-b[nw]); 47 nw--; 48 } 49 go(i,1,m)printf(" %d",as[i]);printf("\n\n"); 50 } 51 return 0; 52 }