codeforces 599B Spongebob and Joke

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 test(s)

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

Note

In the first sample 3 is replaced by 1 and vice versa, while 2 never changes. The answer exists and is unique.

In the second sample all numbers are replaced by 1, so it is impossible to unambiguously restore the original sequence.

In the third sample fi ≠ 3 for all i, so no sequence ai transforms into such bi and we can say for sure that Spongebob has made a mistake.

题目传送门

题意：给n个数字f序列，给m个数字b序列，是否存在连续的m个数字a[i]使f[a[i]] == b[i]，有且只有一种情况就输出possible然后输出这个，多种情况输出Ambiguity，没有就输出Impossible。

可转换a[i] = f[b[i]]
利用vector，把第f[i]个数字存为序号i 。然后看f[b[i]].size()是否存在数字，如果不存在就输出Impossible，如果存在f[b[i]].size()>1，则有多种方案，输出Ambiguity,只有一种情况就输出possible，然后输出这个数组a。这个a[i]就是b[i]在f[]里面存的序号。

CODE

#include"stdio.h"
#include"iostream"
#include"algorithm"
#include"string.h"
#include"math.h"
#include"vector"

using namespace std;

int n,m;
vector<int>f[100010];
int b[100010];
int f1[100010];

int main(void)
{
    scanf("%d%d",&n,&m);
    for(int i = 1;i <= n;i++)
    {
        int t;
        scanf("%d",&t);
        f[t].push_back(i);
    }
    for(int i = 1;i <= m;i++)
    {
        scanf("%d",&b[i]);
    }
    int flag = 0;
    for(int i = 1;i <= m;i++)
    {
        if(f[b[i]].size() == 0)
            return puts("Impossible ");
        if(f[b[i]].size() > 1)
            flag = 1;
    }
    if(flag == 1)
        return puts("Ambiguity");
    puts("Possible");
    for(int i = 1;i <= m;i++)
    {
        printf("%d ",f[b[i]][0]);
    }
    puts("");
    return 0;
}