user: 这段汇编代码用C语言表示是什么样的：

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  lea    -0x9000(%rsp),%r11
L19:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L19
  sub    $0xdf8,%rsp
  lea    D41(%rip),%rdi
  xor    %r14d,%r14d
  xor    %r13d,%r13d
  lea    D29(%rip),%r12
  mov    %fs:0x28,%rax
  mov    %rax,0x9de8(%rsp)
  xor    %eax,%eax
  lea    0x1a0(%rsp),%r15
  call   <printf@plt>
  lea    0xc(%rsp),%rsi
  mov    %r12,%rdi
  call   <scanf@plt>
  lea    D42(%rip),%rdi
  xor    %eax,%eax
  call   <printf@plt>
  mov    0xc(%rsp),%edx
  test   %edx,%edx
  jle    L24
  nopl   (%rax)
L106:
  lea    (%r15,%r14,4),%rbp
  xor    %ebx,%ebx
  xchg   %ax,%ax
L71:
  mov    %rbp,%rsi
  mov    %r12,%rdi
  add    $0x1,%ebx
  add    $0x4,%rbp
  call   <scanf@plt>
  mov    0xc(%rsp),%eax
  cmp    %ebx,%eax
  jg     L71
  add    $0x1,%r13d
  add    $0x64,%r14
  cmp    %r13d,%eax
  jg     L106
L24:
  xor    %edi,%edi
  call   <insertq@plt>
  xor    %eax,%eax
  movl   $0x1,0x10(%rsp)
  call   <deleteq@plt>
  lea    D48(%rip),%rdi
  movslq %eax,%rbx
  xor    %eax,%eax
  lea    0x1(%rbx),%esi
  call   <printf@plt>
  cmp    $0xffffffff,%ebx
  je     L84
  lea    0x1a0(%rsp),%r13
  lea    0x10(%rsp),%r12
  lea    D13(%rip),%r14
L52:
  mov    0xc(%rsp),%eax
  test   %eax,%eax
  jle    L97
  lea    (%rbx,%rbx,4),%rax
  xor    %ebx,%ebx
  lea    (%rax,%rax,4),%rbp
  shl    $0x4,%rbp
  add    %r13,%rbp
  jmp    L83
  nopl   (%rax)
L133:
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jle    L97
L83:
  cmpl   $0x1,0x0(%rbp,%rbx,4)
  mov    %ebx,%edi
  jne    L133
  cmpl   $0x1,(%r12,%rbx,4)
  je     L133
  call   <insertq@plt>
  movl   $0x1,(%r12,%rbx,4)
  add    $0x1,%rbx
  cmp    %ebx,0xc(%rsp)
  jg     L83
  nopl   0x0(%rax)
L97:
  xor    %eax,%eax
  call   <deleteq@plt>
  movslq %eax,%rbx
  cmp    $0xffffffff,%ebx
  jne    L135
L84:
  mov    0x9de8(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L82
  add    $0x9df8,%rsp
  xor    %eax,%eax
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L135:
  lea    0x1(%rbx),%esi
  mov    %r14,%rdi
  xor    %eax,%eax
  call   <printf@plt>
  jmp    L52
L82:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: read_data
arguments: tag: D41
type_infered: string
name: read_data
arguments: tag: D29
type_infered: string
name: read_data
arguments: tag: D42
type_infered: string
name: read_data
arguments: tag: D48
type_infered: string
name: read_data
arguments: tag: D13
type_infered: string

tool: D41: 
Enter the number of vetices 

tool: D29: %d

tool: D42: 
Now enter the adjacency matrix 

tool: D48: The BFS travesal order is : %d

tool: D13:  %d

assistant: ```c
int main() {
  int A[100][100], B[100], n, i, j, x;
  printf("\nEnter the number of vetices ");
  scanf("%d", &n);
  printf("\nNow enter the adjacency matrix ");
  for (i = 0; i < n; i++)
    for (j = 0; j < n; j++) {
      scanf("%d", &A[i][j]);
    }
  insertq(0);
  B[0] = 1;
  x = deleteq();
  printf("The BFS travesal order is : %d", x + 1);
  while (x != -1) {
    for (j = 0; j < n; j++) {
      if (A[x][j] == 1 && B[j] != 1) {
        insertq(j);
        B[j] = 1;
      }
    }
    x = deleteq();
    if (x != -1) {
      printf(" %d", x + 1);
    }
  }
  return 0;
}
```