Q482: Permutation Arrays

/*
    pos[2] = 0          pos[0] = 1            pos[1] = 2
         arr[0] = 32.0       arr[1] = 54.7         arr[2] = -2

 先印 1 所對到的 再印2 再印3

 => arr[pos[0]] -> arr[pos[1]] -> arr[pos[2]]

 Run time: 0.000
*/

#include <stdio.h>
#include <string.h>
#define N 1000
int pos[N];
char arr[N][N];
int main()
{
  int n;
  int i;
  int tmp;
  int cnt;
  bool state = true;
  scanf("%d\n", &n);

  while (n--)
  {
      if (!state)  printf("\n");
      state = false;

      cnt = 0;

      while (true)
      {
          scanf("%d", &tmp);
          pos[tmp-1] = cnt++;

          if (getchar() == '\n') // 當讀到尾巴
              break;
      }

      for ( i = 0; i < cnt; i++)
          scanf("%s", arr[i]);

      for ( i = 0; i < cnt; i++)
          printf("%s\n", arr[pos[i]]);


  }
}