0/1 knapsack problem

Code Listing 1:

1#include <iostream>2#include <vector>3 using namespacestd;4 Const intmin=0x80000000;5 Const intn=3;//Number of items6 Const intv=5;//Backpack Capacity7 intf[n+1][v+1];8 9 intPackage (int*w,int*c,intNintV);Ten voidMainintargcChar*argv[]) One { A  intw[4]={0,7,5,8};//Item Weights -  intc[4]={0,2,3,4};//Item Size -  intresult=Package (w,c,n,v); the  if(result>0) -  { -cout<<Endl; -cout<<"The opt value:"<<result<<Endl; +   inti=n,j=V; -    while(i) +   { A    if(f[i][j]== (f[i-1][j-c[i]]+W[i])) at    { -cout<<i<<":"<<"w="<<W[i]<<", c="<<C[i]<<Endl; -j-=C[i]; -    } -i--; -   } in  } -  Else tocout<<"can not find the OPT value"<<Endl; +  return; - } the  * intPackage (int*w,int*c,intNintV) $ {Panax Notoginseng  inti,j; -Memset (F,0,sizeof(f));//initialized to 0 the  +   for(i=0; i<=n;i++) A   for(j=1; j<=v;j++)//This step is to resolve whether the pack capacity is exactly met, theF[i][j]=min;//if "just" satisfies the backpack capacity, that is, full pack, then add this step, if you do not need "exactly", then initialized to 0 +      -   for(i=1; i<=n;i++) $    for(j=c[i];j<=v;j++) $   { -F[i][j]= (f[i-1][j]>f[i-1][j-c[i]]+w[i])? f[i-1][J]:(f[i-1][j-c[i]]+w[i]); -cout<<"f["<<i<<"]["<<j<<"]="<<f[i][j]<<Endl; the   } -  returnF[n][v];Wuyi}

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Code Listing 2:

1 Code 22 3#include <iostream>4#include <vector>5 using namespacestd;6 Const intmin=0x80000000;7 Const intn=3;//Number of items8 Const intv=5;//Backpack Capacity9 intf[v+1];Ten  One intPackage (int*w,int*c,intNintV); A voidMainintargcChar*argv[]) - { -  intw[4]={0,7,5,8};//Item Weights the  intc[4]={0,2,3,4};//Item Size -  intresult=Package (w,c,n,v); -  if(result>0) -  { +cout<<Endl; -cout<<"The opt value:"<<result<<Endl; +  } A  Else atcout<<"can not find the OPT value"<<Endl; -  return; - } -  - intPackage (int*w,int*c,intNintV) - { in  inti,j; -Memset (F,0,sizeof(f));//initialized to 0 to  +   for(i=1; i<=v;i++)//This step is to resolve whether the pack capacity is exactly met, -F[i]=min;//if "just" satisfies the backpack capacity, that is, full pack, then add this step, if you do not need "exactly", then initialized to 0 the      *   for(i=1; i<=n;i++) $    for(j=v;j>=c[i];j--)//Note that there is a difference in order between this and the solution, to clarify the reasonPanax Notoginseng   { -F[j]= (F[j]>f[j-c[i]]+w[i])? F[J]:(f[j-c[i]]+w[i]); thecout<<"f["<<i<<"]["<<j<<"]="<<f[j]<<Endl; +   } A  returnF[v]; the}

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0/1 knapsack problem