user: Translate the below assembly code into C programming language:

```asm
<main>:
  endbr64
  push   %r15
  push   %r14
  push   %r13
  push   %r12
  push   %rbp
  push   %rbx
  sub    $0x688,%rsp
  mov    %fs:0x28,%rax
  mov    %rax,0x678(%rsp)
  xor    %eax,%eax
  movl   $0x1,0xc(%rsp)
  lea    D11(%rip),%r13
  jmp    L52
L30:
  add    $0x1,%r15d
  cmp    $0x13,%r15d
  je     L50
L57:
  lea    0x1c(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  test   %rax,%rax
  jle    L95
  cmpl   $0x0,0x1c(%rsp)
  jle    L30
  mov    $0x0,%ebx
  lea    0x20(%rsp),%r14
  movslq %r15d,%r12
  lea    (%r12,%r12,4),%rbp
  shl    $0x2,%rbp
L17:
  mov    %r14,%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x20(%rsp),%eax
  sub    $0x1,%eax
  mov    %eax,0x20(%rsp)
  cltq
  lea    (%rax,%rax,4),%rdx
  lea    (%r12,%rdx,4),%rdx
  movl   $0x1,0x30(%rsp,%rdx,4)
  add    %rbp,%rax
  movl   $0x1,0x30(%rsp,%rax,4)
  add    $0x1,%ebx
  cmp    %ebx,0x1c(%rsp)
  jg     L17
  jmp    L30
L50:
  lea    0x30(%rsp),%rdi
  mov    $0x14,%esi
  call   <computeAllPairShortestPath@plt>
  mov    0xc(%rsp),%esi
  lea    D38(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  lea    0x24(%rsp),%rsi
  mov    %r13,%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  cmpl   $0x0,0x24(%rsp)
  jle    L76
  mov    $0x0,%ebx
  lea    0x2c(%rsp),%r12
  lea    0x28(%rsp),%rbp
L8:
  mov    %r12,%rdx
  mov    %rbp,%rsi
  lea    D25(%rip),%rdi
  mov    $0x0,%eax
  call   <scanf@plt>
  mov    0x28(%rsp),%esi
  mov    0x2c(%rsp),%edi
  lea    -0x1(%rdi),%edx
  movslq %edx,%rdx
  lea    -0x1(%rsi),%eax
  cltq
  lea    (%rax,%rax,4),%rax
  lea    (%rdx,%rax,4),%rax
  mov    0x30(%rsp,%rax,4),%ecx
  mov    %edi,%edx
  lea    D19(%rip),%rdi
  mov    $0x0,%eax
  call   <printf@plt>
  add    $0x1,%ebx
  cmp    %ebx,0x24(%rsp)
  jg     L8
L76:
  mov    $0xa,%edi
  call   <putchar@plt>
  addl   $0x1,0xc(%rsp)
L52:
  lea    0x30(%rsp),%rdi
  mov    $0x640,%edx
  mov    $0x0,%esi
  call   <memset@plt>
  mov    $0x0,%r15d
  jmp    L57
L95:
  mov    0x678(%rsp),%rax
  sub    %fs:0x28,%rax
  jne    L26
  mov    $0x0,%eax
  add    $0x688,%rsp
  pop    %rbx
  pop    %rbp
  pop    %r12
  pop    %r13
  pop    %r14
  pop    %r15
  ret
L26:
  call   <__stack_chk_fail@plt>
```

tool_calls: name: parse_data
arguments: tag: D11
type: string
name: parse_data
arguments: tag: D38
type: string
name: parse_data
arguments: tag: D25
type: string
name: parse_data
arguments: tag: D19
type: string

tool: D11: %d

tool: D38: Test Set #%d


tool: D25: %d %d

tool: D19: %2d to %2d: %d


assistant: ```c
int main(void) {
  int caseId, adjMat[20][20], numNeighbor, neighbor, stopFlg, city, time, numQuery, cityOne, cityTwo, query;
  caseId = 1;
  stopFlg = 0;
  while (1) {
    memset(adjMat, 0, 20 * 20 * sizeof(int));
    for (city = 0; city < 20 - 1; city++)
      if (scanf("%d", &numNeighbor) > 0) {
        for (time = 0; time < numNeighbor; time++) {
          scanf("%d", &neighbor);
          neighbor--;
          adjMat[neighbor][city] = 1;
          adjMat[city][neighbor] = 1;
        }
      } else {
        stopFlg = 1;
        break;
      }
    if (stopFlg)
      break;
    computeAllPairShortestPath(adjMat, 20);
    printf("Test Set #%d\n", caseId);
    scanf("%d", &numQuery);
    for (query = 0; query < numQuery; query++) {
      scanf("%d %d", &cityOne, &cityTwo);
      printf("%2d to %2d: %d\n", cityOne, cityTwo, adjMat[cityOne - 1][cityTwo - 1]);
    }
    printf("\n");
    caseId++;
  }
  return 0;
}
```