poj1015 Jury Compromise[背包]

每一件物品有两个属性。朴素思想是把这两种属性都设计到状态里，但空间爆炸。又因为这两个属性相互间存在制约关系（差的绝对值最小），不妨把答案设计入状态中，设$f[i][j]$选$i$个人，两者之差$j$。这样dp。于是$O(NM^3)$暴力背包即可，记录转移，也就是$g[i][j][k]$来表示当前到这个人$i$的时候选了$j$人，是怎么转移的，如果由$j-1$转移，说明这个人$i$选了。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 #include<queue>
 #define dbg(x) cerr<<#x<<" = "<<x<<endl
 #define _dbg(x,y) cerr<<#x<<" = "<<x<<"   "<<#y<<" = "<<y<<endl
 using namespace std;
 typedef long long ll;
 template<typename T>inline char MIN(T&A,T B){return A>B?A=B,:;}
 template<typename T>inline char MAX(T&A,T B){return A<B?A=B,:;}
 template<typename T>inline T _min(T A,T B){return A<B?A:B;}
 template<typename T>inline T _max(T A,T B){return A>B?A:B;}
 template<typename T>inline T read(T&x){
     x=;int f=;char c;while(!isdigit(c=getchar()))if(c=='-')f=;
     while(isdigit(c))x=x*+(c&),c=getchar();return f?x=-x:x;
 }
 const int N=+,M=+,INF=0x3f3f3f3f;
 int f[][M],g[N][][M],a[N],b[N],sen[M];
 int n,m,ans,thx,orz,T;

 int main(){//freopen("test.in","r",stdin);//freopen("test.out","w",stdout);
     while(read(n),read(m),n||m){
         thx=orz=ans=;
         for(register int i=;i<=m;++i)for(register int j=;j<=;++j)f[i][j]=-INF;
         memset(g,,sizeof g);
         for(register int i=;i<=n;++i)read(a[i]),read(b[i]);
         f[][]=;
         for(register int i=;i<=n;++i)
             for(register int j=_min(i,m);j;--j)
                 for(register int k=a[i]-b[i];k<=;++k)
                     if(MAX(f[j][k],f[j-][k-(a[i]-b[i])]+a[i]+b[i]))g[i][j][k]=j-;
                     else g[i][j][k]=j;
         for(register int i=;i<=;++i)if(f[m][+i]>=||f[m][-i]>=){
             ans=f[m][+i]>f[m][-i]?+i:-i;break;
         }
         int i=n,j=m;
         while(i){
             if(g[i][j][ans]==j-)sen[j]=i,thx+=a[i],orz+=b[i],ans-=(a[i]-b[i]),--i,--j;
             else --i;
         }
         printf("Jury #%d\n",++T);
         printf("Best jury has value %d for prosecution and value %d for defence: \n",thx,orz);
         for(register int i=;i<=m;++i)printf(" %d",sen[i]);
         puts("");
     }
     return ;
 }