CodeForces599BSpongebob and Joke

Description
While Patrick was gone shopping, Spongebob decided to play a little trick on his friend. The naughty Sponge browsed through Patrick's personal stuff and found a sequence a1, a2, ..., am of length m, consisting of integers from 1 to n, not necessarily distinct. Then he picked some sequence f1, f2, ..., fn of length n and for each number ai got number bi = fai. To finish the prank he erased the initial sequence ai.

It's hard to express how sad Patrick was when he returned home from shopping! We will just say that Spongebob immediately got really sorry about what he has done and he is now trying to restore the original sequence. Help him do this or determine that this is impossible.

Input
The first line of the input contains two integers n and m (1 ≤ n, m ≤ 100 000) — the lengths of sequences fi and bi respectively.

The second line contains n integers, determining sequence f1, f2, ..., fn (1 ≤ fi ≤ n).

The last line contains m integers, determining sequence b1, b2, ..., bm(1 ≤ bi ≤ n).

Output
Print "Possible" if there is exactly one sequence ai, such that bi = fai for all i from 1 to m. Then print m integers a1, a2, ..., am.

If there are multiple suitable sequences ai, print "Ambiguity".

If Spongebob has made a mistake in his calculations and no suitable sequence ai exists, print "Impossible".

Sample Input
Input
3 3
3 2 1
1 2 3
Output
Possible
3 2 1
Input
3 3
1 1 1
1 1 1
Output
Ambiguity
Input
3 3
1 2 1
3 3 3
Output
Impossible

代码：（大犇的代码）

#include <stdio.h>using namespace std;int used[100005];int f[100005];int o[100005];int freq[100005], rev[100005];int main() {  int n, m, x;  scanf("%d %d", &n, &m);  for (int i = 1; i <= n; i++) {    scanf("%d", &f[i]);    freq[f[i]]++;    rev[f[i]] = i;  }  bool amb = false;  for (int i = 1; i <= m; i++) {    scanf("%d", &x);    if (freq[x] == 1) {      o[i] = rev[x];    } else if (freq[x] == 0) {      printf("Impossible\n");      return 0;    } else {      amb = true;    }  }  if (amb) {    printf("Ambiguity\n");    return 0;  }  printf("Possible\n");  for (int i = 1; i <= m; i++) {    printf("%d",o[i]);    if(i!=m)    printf(" ");    else    printf("\n");  }  return 0;}

思路：这题题目太绕了读晕了都，还是看别人的代码才看懂啥意思，简单说就是给你两个串f和b，如果总有f[a[i]]，满足b[i]==f[a[i]]输出a[n]这个数组。所以开始输入时对所有输入的值保存一个数组为1，记录当前地址。然后输入b时判断这个数是否存在于f，存在保存地址在另一个数组o中，此时因为i==a[i],所以保存地址的数组o即为所求的a[i]。如果有一个数不存在与f输出Impossible，如果有的数存在于b还有f而且不止1个输出Ambiguity。