user: 请将这段汇编代码片段转换成等效的C代码:

```asm
<main>:
  endbr64
  push   %rbp
  mov    %rsp,%rbp
  lea    -0x9000(%rsp),%r11
L77:
  sub    $0x1000,%rsp
  orq    $0x0,(%rsp)
  cmp    %r11,%rsp
  jne    L77
  sub    $0xdf0,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,-0x8(%rbp)
  xor    %eax,%eax
  lea    D53(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    -0x9df0(%rbp),%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  lea    D26(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  movl   $0x0,-0x9dec(%rbp)
  jmp    L127
L97:
  movl   $0x0,-0x9de8(%rbp)
  jmp    L80
L121:
  lea    -0x9c50(%rbp),%rcx
  mov    -0x9de8(%rbp),%eax
  movslq %eax,%rsi
  mov    -0x9dec(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rsi,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    %rax,%rsi
  lea    D34(%rip),%rax
  mov    %rax,%rdi
  call   <scanf@plt>
  addl   $0x1,-0x9de8(%rbp)
L80:
  mov    -0x9df0(%rbp),%eax
  cmp    %eax,-0x9de8(%rbp)
  jl     L121
  addl   $0x1,-0x9dec(%rbp)
L127:
  mov    -0x9df0(%rbp),%eax
  cmp    %eax,-0x9dec(%rbp)
  jl     L97
  mov    $0x0,%edi
  call   <insertq@plt>
  movl   $0x1,-0x9de0(%rbp)
  mov    $0x0,%eax
  call   <deleteq@plt>
  mov    %eax,-0x9de4(%rbp)
  mov    -0x9de4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D44(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  jmp    L38
L47:
  movl   $0x0,-0x9de8(%rbp)
  jmp    L111
L68:
  mov    -0x9de8(%rbp),%eax
  movslq %eax,%rcx
  mov    -0x9de4(%rbp),%eax
  movslq %eax,%rdx
  mov    %rdx,%rax
  shl    $0x2,%rax
  add    %rdx,%rax
  lea    0x0(,%rax,4),%rdx
  add    %rdx,%rax
  shl    $0x2,%rax
  add    %rcx,%rax
  mov    -0x9c50(%rbp,%rax,4),%eax
  cmp    $0x1,%eax
  jne    L101
  mov    -0x9de8(%rbp),%eax
  cltq
  mov    -0x9de0(%rbp,%rax,4),%eax
  cmp    $0x1,%eax
  je     L101
  mov    -0x9de8(%rbp),%eax
  mov    %eax,%edi
  call   <insertq@plt>
  mov    -0x9de8(%rbp),%eax
  cltq
  movl   $0x1,-0x9de0(%rbp,%rax,4)
L101:
  addl   $0x1,-0x9de8(%rbp)
L111:
  mov    -0x9df0(%rbp),%eax
  cmp    %eax,-0x9de8(%rbp)
  jl     L68
  mov    $0x0,%eax
  call   <deleteq@plt>
  mov    %eax,-0x9de4(%rbp)
  cmpl   $0xffffffff,-0x9de4(%rbp)
  je     L38
  mov    -0x9de4(%rbp),%eax
  add    $0x1,%eax
  mov    %eax,%esi
  lea    D23(%rip),%rax
  mov    %rax,%rdi
  mov    $0x0,%eax
  call   <printf@plt>
L38:
  cmpl   $0xffffffff,-0x9de4(%rbp)
  jne    L47
  mov    $0x0,%eax
  mov    -0x8(%rbp),%rdx
  sub    %fs:0x28,%rdx
  je     L46
  call   <__stack_chk_fail@plt>
L46:
  leave
  ret
```

tool_calls: name: disassemble_data
arguments: tag: D53
datatype: string
name: disassemble_data
arguments: tag: D34
datatype: string
name: disassemble_data
arguments: tag: D26
datatype: string
name: disassemble_data
arguments: tag: D44
datatype: string
name: disassemble_data
arguments: tag: D23
datatype: string

tool: D53: 
Enter the number of vetices 

tool: D34: %d

tool: D26: 
Now enter the adjacency matrix 

tool: D44: The BFS travesal order is : %d

tool: D23:  %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;
}
```