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📁 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>#define N 8 <BR><BR>int column[N+1]; // =
=A6P=C4=E6=ACO=A7_=A6=B3=AC=D3=A6Z=A1A1=AA=ED=A5=DC=A6=B3 <BR>int =
rup[2*N+1]; // =A5k=A4W=A6=DC=A5=AA=A4U=ACO=A7_=A6=B3=AC=D3=A6Z <BR>int =
lup[2*N+1]; // =A5=AA=A4W=A6=DC=A5k=A4U=ACO=A7_=A6=B3=AC=D3=A6Z <BR>int =
queen[N+1] =3D {0}; <BR>int num; // =B8=D1=B5=AA=BDs=B8=B9 <BR><BR>void =
backtrack(int); // =BB=BC=B0j=A8D=B8=D1 <BR><BR>int main(void) { <BR>    =
int i; <BR>    num =3D 0; <BR><BR>    for(i =3D 1; i &lt;=3D N; i++) =
<BR>        column[i] =3D 1; <BR><BR>    for(i =3D 1; i &lt;=3D 2*N; =
i++) <BR>        rup[i] =3D lup[i] =3D 1; <BR><BR>    backtrack(1); =
<BR><BR>    return 0; <BR>} <BR><BR>void showAnswer() {<BR>    int x, =
y;<BR>    printf("\n=B8=D1=B5=AA %d\n", ++num);<BR>    for(y =3D 1; y =
&lt;=3D N; y++) {<BR>        for(x =3D 1; x &lt;=3D N; x++) {<BR>        =
    if(queen[y] =3D=3D x) {<BR>                printf(" Q");<BR>         =
   }<BR>            else {<BR>                printf(" .");<BR>          =
  }<BR>        }<BR>        printf("\n");<BR>    }<BR>}<BR><BR>void =
backtrack(int i) { <BR>    int j;<BR><BR>    if(i &gt; N) { <BR>        =
showAnswer();<BR>    } <BR>    else { <BR>        for(j =3D 1; j &lt;=3D =
N; j++) { <BR>            if(column[j] =3D=3D 1 &amp;&amp; <BR>          =
       rup[i+j] =3D=3D 1 &amp;&amp; lup[i-j+N] =3D=3D 1) { <BR>          =
      queen[i] =3D j; <BR>                // =
=B3]=A9w=AC=B0=A6=FB=A5=CE<BR>                column[j] =3D rup[i+j] =3D =
lup[i-j+N] =3D 0; <BR>                backtrack(i+1); <BR>               =
 column[j] =3D rup[i+j] =3D lup[i-j+N] =3D 1; <BR>            } <BR>     =
   } <BR>    } <BR>} <BR></PRE><BR>
<UL>
  <LI>Java </LI></UL><PRE>public class Queen {<BR>    // =
=A6P=C4=E6=ACO=A7_=A6=B3=AC=D3=A6Z=A1A1=AA=ED=A5=DC=A6=B3 <BR>    =
private int[] column;<BR>    // =
=A5k=A4W=A6=DC=A5=AA=A4U=ACO=A7_=A6=B3=AC=D3=A6Z<BR>    private int[] =
rup; <BR>    // =A5=AA=A4W=A6=DC=A5k=A4U=ACO=A7_=A6=B3=AC=D3=A6Z<BR>    =
private int[] lup;<BR>    // =B8=D1=B5=AA<BR>    private int[] =
queen;<BR>    <BR>    // =B8=D1=B5=AA=BDs=B8=B9<BR>    private int =
num;<BR>    <BR>    public Queen() {<BR>        column =3D new =
int[8+1];<BR>        rup =3D new int[2*8+1];<BR>        lup =3D new =
int[2*8+1];<BR>        <BR>        for(int i =3D 1; i &lt;=3D 8; i++) =
<BR>            column[i] =3D 1; <BR><BR>        for(int i =3D 1; i =
&lt;=3D 2*8; i++) <BR>            rup[i] =3D lup[i] =3D 1; <BR>        =
<BR>        queen =3D new int[8+1];<BR>    }<BR>    <BR>    public void =
backtrack(int i) {<BR>        if(i &gt; 8) { <BR>            =
showAnswer();<BR>        } <BR>        else { <BR>            for(int j =
=3D 1; j &lt;=3D 8; j++) { <BR>                if(column[j] =3D=3D 1 =
&amp;&amp; <BR>                   rup[i+j] =3D=3D 1 &amp;&amp; <BR>      =
             lup[i-j+8] =3D=3D 1) { <BR>                    queen[i] =3D =
j; <BR>                    // =B3]=A9w=AC=B0=A6=FB=A5=CE<BR>             =
       column[j] =3D rup[i+j] =3D lup[i-j+8] =3D 0; <BR>                 =
   backtrack(i+1); <BR>                    column[j] =3D rup[i+j] =3D =
lup[i-j+8] =3D 1; <BR>                } <BR>            } <BR>        } =
<BR>    }<BR>    <BR>    protected void showAnswer() {<BR>        =
num++;<BR>        System.out.println("\n=B8=D1=B5=AA " + num);<BR>       =
 for(int y =3D 1; y &lt;=3D 8; y++) {<BR>            for(int x =3D 1; x =
&lt;=3D 8; x++) {<BR>                if(queen[y] =3D=3D x) {<BR>         =
           System.out.print(" Q");<BR>                }<BR>              =
  else {<BR>                    System.out.print(" .");<BR>              =
  }<BR>            }<BR>            System.out.println();<BR>        =
}<BR>    }<BR>    <BR>    public static void main(String[] args) {<BR>   =
     Queen queen =3D new Queen();<BR>        queen.backtrack(1);<BR>    =
}<BR>} </PRE><BR><BR><BR></BODY></HTML>

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