[POJ1015] Jury COMPROMISE[DP]

Test instructions: $n $ items Selected $m$, each item has two weights $a[i]$ and $b[i]$, required to select $m$ items $\left|\sum a[i]-\sum b[i]\right|$ Minimum, when this difference is the same, select $\sum a[i]+\sum B [i]$ maximum, output $\sum a[i]$,$\sum b[i]$, scheme.

Exercises

Make $v=\left|\sum a[i]-\sum b[i]\right|$

$S =\sum a[i]+\sum b[i]$

$s [i]=a[i]+b[i]$

$v [i]=a[i]-b[i]$

Set $f[j][k]$ indicates the maximum value of b[i]$ when the $j$ is selected $k =\sum a[i]-\sum $s$ (Note that there is no absolute value), if $f[j][k]==-1$ indicates that no legal scheme exists.

$path [j][k]$ means that $f[j][k]$ is the last item selected by $f[j-1][k]$ after $path[j][k]$ is selected, that is, state $f[j][k]$.

The output of the final scheme is traced forward with the $path$ array, and the previous selection of $path[j][k]$ is $path[j-1][k-v[path[j][k]]]$.

The transfer is as follows:

If(f[j-1][k]+s[I]> F[j][k+v[I]]&&! selected (J-1 , K,i) )//i not selected by previous state f[j][k+v< Span class= "Sh_symbol" >[i]]=f[j-1][k]+s[i], Path[j][k+v[i]]=i;  
   /span>                

So the final answer is $\left|. K \right|$ the smallest legal $f[m][k] (f[m][k]!=-1) $.

Because $f[m][k]=\sum a[i]+\sum b[i]$

$k =\sum a[i]-\sum b[i]$

So the answer to the $\sum a[i]= (f[m][k]+k)/2$

$\sum b[i]= (f[m][k]-k)/2$

Finally, the output of the scheme is traced forward with the $path$ array, and the previous selection of $path[j][k]$ is $path[j-1][k-v[path[j][k]]]$.

The resulting scenarios are sorted out. To avoid negative numbers the following table, to move the second dimensional plane, pay attention to the details.

1#include <iostream>2#include <algorithm>3#include <cstdio>4#include <cstdlib>5#include <cstring>6#include <ctime>7#include <cmath>8#include <vector>9 Ten using namespacestd; One  A intn,m,a[ About],b[ About],v[ About],s[ About],fix,kase; - intf[1100][1100],path[1100][1100]; -  the BOOLSelected (intJintKConst inti) - { -      while(j>0&& path[j][k]!=i) -{k-=v[path[j][k]]; j--; }  +     returnJ; - } +  A intMain () at { -      while(~SCANF ("%d%d", &n,&m) && n &&m) -     { -          for(intI=1; i<=n;++i) -scanf"%d%d",&a[i],&B[i]), -v[i]=a[i]-b[i],s[i]=a[i]+B[i]; in          -fix=m* -; Memset (F,0xFF,sizeof(f)); f[0][fix]=0; to  +          for(intj=1; j<=m;++j) for(intk=0;k<=fix<<1;++k) -         { the             if(f[j-1][k]<0)Continue; *              for(intI=1; i<=n;++i) $             {Panax Notoginseng                 if(f[j-1][k]+s[i] > F[j][k+v[i]] && -! Selected (J-1, k,i)) the                 { +f[j][k+v[i]]=f[j-1][k]+S[i]; Apath[j][k+v[i]]=i; the         }    }    } +  -         inttt for(intk=0; k<=fix;++k) $             if(f[m][fix-k]>=0|| f[m][fix+k]>=0) {tt=k; Break; } $  -         intTemp= (F[m][fix-tt]>f[m][fix+tt]? ( FIX-TT):(fix+tt)); -printf"Jury #%d\nbest jury have value%d for prosecution and" the             "value%d for defence: \ n",++Kase, -(F[m][temp]+temp-fix) >>1, (F[m][temp]-temp+fix) >>1);Wuyi  thevector<int> VEC; tt=temp; -          for(intj=m;j;--j) Wu{Vec.push_back (Path[j][tt]); tt-=V[vec.back ()];} -  About sort (Vec.begin (), Vec.end ()); $          for(intI=0;i< (int) vec.size (); ++i) printf ("%d", Vec[i]); -  -printf"\ n"); -     } A     return 0; +}

[POJ1015] Jury COMPROMISE[DP]