【汇编语言实战】八皇后问题

#include<stdio.h>
int place[8]={0};//皇后位置
bool flag[8]={1,1,1,1,1,1,1,1};//定义列
bool d1[15]={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1};
/*定义上对角线（共有15个对角线，因此定义一个长度为15的bool型数组，初值为1代表该对角线没有被皇后占领，若被皇后占领则赋值为0*/
bool d2[15]={1,1,1,1,1,1,1,1,1,1,1,1,1,1,1} ;//定义下对角线
int number=0;//记录输出次数
void print()//定义输出函数
{
  int col,i,j;
number++;//每调用一次输出函数number自加一次，记录摆放方法个数
  printf("No.%2d\n",number);
  int table[8][8]={0};//设置一个8*8的棋盘
  for (i=0;i<8;i++)
  {
  table[i][place[i]]=1;//将每一行皇后所在位置赋值为1
  }
  for (i=0;i<8;i++)
  {
  for (j=0;j<8;j++)
  {
  printf("%d|",table[i][j]);
  }printf("\n");
  }
}
int queen(int n )//定义递归回溯函数
{
  int col;
  for (col=0;col<8;col++)
  {
  if (flag[col]&&d1[n-col+7]&&d2[n+col])//判断皇后是否冲突
  {
    place[n]=col;//放置皇后
    flag[col]=false;
    d1[n-col+7]=false;
  d2[n+col]=false;//将该皇后所在的行、列、对角线设置为被占领
    if(n<7) {queen(n+1);}//当行数小于7时；递归调用下一行
  else{print();}//调用输出函数
    flag[col]=true;//回溯
    d1[n-col+7]=true;
    d2[n+col]=true;
  }
  }
  return number;
}
int main()
{
number=queen(0);//从第0行开始执行
printf("共有%d种摆放

INCLUDE Irvine32.inc
.data
    Queens dd 8 dup(0)
    Count dd 0
    msg db 'The number of the way setting is:',0
    sep db '===============',0
    eight dd 8
    seven dd 7
.code
main PROC
    mov eax,0
    push eax
    call eight_queen
    lea edx,msg
    call writestring
    mov eax,Count
    call writedec
    call crlf
    ; call print
    exit
main ENDP
eight_queen PROC
    push ebp
    mov ebp,esp
    pushad
    mov esi,[ebp+8]     ;esi:line
    mov edi,0           ;edi:list
    ; mov eax,esi
    ; call writedec
    ; call crlf
for_0:
    cmp edi,eight
    jge final
    push edi
    push esi
    call Check
    cmp eax,1
    jne for_0_end
    mov edx,offset Queens   ;edx:Queens
    mov [edx+esi*4],edi
    cmp esi,seven
    ; mov eax,edi
    ; call writedec
    ; call crlf
    jne next
    inc Count
    call print
    mov eax,0
    mov edx,offset Queens 
    mov [edx+esi*4],eax
    jmp final
next:
    mov eax,esi
    add eax,1
    push eax
    call eight_queen
    mov edx,offset Queens
    mov eax,0
    mov [edx+esi*4],eax
for_0_end:
    inc edi
    jmp for_0
final:
    popad
    pop ebp
    ret 4
eight_queen ENDP
Check PROC
    push ebp
    mov ebp,esp
    sub esp,4
    pushad
    mov ecx,[ebp+8]     ;ecx:line
    mov edx,[ebp+12]    ;edx:list
    mov esi,0           ;esi:index
for_0:
    cmp esi,ecx
    jge ret1
    mov ebx,offset Queens   ;ebx:Queens
    mov edi,[ebx+esi*4]     ;edi:data
    cmp edx,edi
    je ret0
    ; mov ebx,offset Queens 
    mov edi,esi
    add edi,[ebx+esi*4]
    sub edi,ecx
    sub edi,edx
    cmp edi,0
    je ret0
    ; mov ebx,offset Queens 
    mov edi,esi
    sub edi,[ebx+esi*4]
    sub edi,ecx
    add edi,edx
    cmp edi,0
    je ret0
for_0_end:
    inc esi
    jmp for_0
ret1:
    mov eax,1
    mov [ebp-4],eax
    jmp final
ret0:
    mov eax,0
    mov [ebp-4],eax
    jmp final
final:     
    popad
    mov eax,[ebp-4]
    add esp,4
    pop ebp
    ret 8
Check ENDP
print PROC
    push ebp
    mov ebp,esp
    pushad
    mov esi,0       ;esi:line
for_0:
    cmp esi,eight
    jge final
    mov edi,0       ;edi:list
for_1:
    mov edx,offset Queens   ;edx:Queens
    cmp edi,[edx+esi*4]
    jge next
    mov eax,0
    call writedec
for_1_end:
    inc edi
    jmp for_1
next:
    mov al,'#'
    call writechar
    mov edx,offset Queens   ;edx:Queens
    mov edi,[edx+esi*4]
    inc edi
for_2:
    cmp edi,eight
    jge for_0_end
    mov eax,0
    call writedec
for_2_end:
    inc edi
    jmp for_2
for_0_end:
    call crlf
    inc esi
    jmp for_0
final:
    lea edx,sep
    call writestring
    call crlf
    popad
    pop ebp
    ret
print ENDP
END main