背包問題（knapsack problem）.mht

23种算法C与JAVA实现 23种算法C与JAVA实现

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<UL>
  <LI>C </LI></UL><PRE>#include &lt;stdio.h&gt; <BR>#include =
&lt;stdlib.h&gt; <BR><BR>#define LIMIT 8   // =AD=AB=B6q=AD=AD=A8=EE =
<BR>#define N 5       // =AA=AB=AB~=BA=D8=C3=FE <BR>#define MIN 1     // =
=B3=CC=A4p=AD=AB=B6q <BR><BR>struct body { <BR>    char name[20]; <BR>   =
 int size; <BR>    int price; <BR>}; <BR><BR>typedef struct body object; =
<BR><BR>int main(void) { <BR>    int item[LIMIT+1] =3D {0}; <BR>    int =
value[LIMIT+1] =3D {0}; <BR>    int newvalue, i, s, p; <BR><BR>    =
object a[] =3D {{"=A7=F5=A4l", 4, 4500}, <BR>                  =
{"=C4=AB=AAG", 5, 5700}, <BR>                  {"=BE=EF=A4l", 2, 2250}, =
<BR>                  {"=AF=F3=B2=F9", 1, 1100}, <BR>                  =
{"=B2=A2=A5=CA", 6, 6700}}; <BR><BR>    for(i =3D 0; i &lt; N; i++) { =
<BR>        for(s =3D a[i].size; s &lt;=3D LIMIT; s++) { <BR>            =
p =3D s - a[i].size; <BR>            newvalue =3D value[p] + a[i].price; =
<BR>            if(newvalue &gt; value[s]) {// =
=A7=E4=A8=EC=B6=A5=ACq=B3=CC=A8=CE=B8=D1 <BR>                value[s] =
=3D newvalue; <BR>                item[s] =3D i; <BR>            } <BR>  =
      } <BR>    } <BR><BR>    printf("=AA=AB=AB~\t=BB=F9=AE=E6\n"); <BR> =
   for(i =3D LIMIT; i &gt;=3D MIN; i =3D i - a[item[i]].size) { <BR>     =
   printf("%s\t%d\n", <BR>                  a[item[i]].name, =
a[item[i]].price); <BR>    } <BR><BR>    printf("=A6X=ADp\t%d\n", =
value[LIMIT]); <BR><BR>    return 0; <BR>}  <BR></PRE><BR>
<UL>
  <LI>Java </LI></UL><PRE>class Fruit {<BR>    private String name;<BR>  =
  private int size;<BR>    private int price;<BR>    <BR>    public =
Fruit(String name, int size, int price) {<BR>        this.name =3D =
name;<BR>        this.size =3D size;<BR>        this.price =3D =
price;<BR>    }<BR>    <BR>    public String getName() {<BR>        =
return name;<BR>    }<BR><BR>    public int getPrice() {<BR>        =
return price;<BR>    }<BR><BR>    public int getSize() {<BR>        =
return size;<BR>    }<BR>}<BR><BR>public class Knapsack {<BR>    public =
static void main(String[] args) {<BR>        final int MAX =3D 8;<BR>    =
    final int MIN =3D 1;<BR>        int[] item =3D new int[MAX+1]; <BR>  =
      int[] value =3D new int[MAX+1];  <BR><BR>        Fruit fruits[] =
=3D {<BR>                new Fruit("=A7=F5=A4l", 4, 4500), <BR>          =
      new Fruit("=C4=AB=AAG", 5, 5700), <BR>                new =
Fruit("=BE=EF=A4l", 2, 2250), <BR>                new =
Fruit("=AF=F3=B2=F9", 1, 1100), <BR>                new =
Fruit("=B2=A2=A5=CA", 6, 6700)}; <BR><BR>        for(int i =3D 0; i &lt; =
fruits.length; i++) { <BR>            for(int s =3D fruits[i].getSize(); =
s &lt;=3D MAX; s++) { <BR>                int p =3D s - =
fruits[i].getSize(); <BR>                int newvalue =3D value[p] + =
<BR>                                   fruits[i].getPrice(); <BR>        =
        if(newvalue &gt; value[s]) {// =
=A7=E4=A8=EC=B6=A5=ACq=B3=CC=A8=CE=B8=D1 <BR>                    =
value[s] =3D newvalue; <BR>                    item[s] =3D i; <BR>       =
         } <BR>            } <BR>        } <BR><BR>        =
System.out.println("=AA=AB=AB~\t=BB=F9=AE=E6"); <BR>        for(int i =
=3D MAX; <BR>            i &gt;=3D MIN; <BR>            i =3D i - =
fruits[item[i]].getSize()) { <BR>            =
System.out.println(fruits[item[i]].getName()+ <BR>                    =
"\t" + fruits[item[i]].getPrice()); <BR>        } <BR><BR>        =
System.out.println("=A6X=ADp\t" + value[MAX]);  <BR>    =
}<BR>}</PRE><BR><BR><BR></BODY></HTML>

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