【Codeforces Round 332 (Div 2)B】【扭转题意 位置映射】Spongebob and Joke 给b[]中的每个数找f[]中的位置

B. Spongebob and Joke

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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, ..., a_m of length m, consisting of integers from 1 to n, not necessarily distinct. Then he picked some sequence f₁, f₂, ..., f_n of length n and for each number a_i got number b_i = f_ai. To finish the prank he erased the initial sequence a_i.

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 f_i and b_i respectively.

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

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

Output

Print "Possible" if there is exactly one sequence a_i, such that b_i = f_ai for all i from 1 to m. Then print m integers a₁, a₂, ..., a_m.

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

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

Sample test(s)

input

3 33 2 11 2 3

output

Possible3 2 1 

input

3 31 1 11 1 1

output

Ambiguity

input

3 31 2 13 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 f_i ≠ 3 for all i, so no sequence a_i transforms into such b_i and we can say for sure that Spongebob has made a mistake.

#include<stdio.h>#include<iostream>#include<string.h>#include<string>#include<ctype.h>#include<math.h>#include<set>#include<map>#include<vector>#include<queue>#include<bitset>#include<algorithm>#include<time.h>using namespace std;void fre(){freopen("c://test//input.in","r",stdin);freopen("c://test//output.out","w",stdout);}#define MS(x,y) memset(x,y,sizeof(x))#define MC(x,y) memcpy(x,y,sizeof(x))#define MP(x,y) make_pair(x,y)#define ls o<<1#define rs o<<1|1typedef long long LL;typedef unsigned long long UL;typedef unsigned int UI;template <class T1,class T2>inline void gmax(T1 &a,T2 b){if(b>a)a=b;}template <class T1,class T2>inline void gmin(T1 &a,T2 b){if(b<a)a=b;}const int N=1e5+10,M=0,Z=1e9+7,ms63=1061109567;int casenum,casei;int n,m;int f[N];int e[N];int p[N];int main(){while(~scanf("%d%d",&n,&m)){MS(e,0);for(int i=1;i<=n;++i){int x;scanf("%d",&x);if(e[x]==0)e[x]=i;else if(e[x]>0)e[x]=-1;}int ans=1;for(int i=1;i<=m;++i){int x;scanf("%d",&x);if(ans==0)continue;if(e[x]==0){ans=0;continue;}//Impossibleif(ans==-1)continue;if(e[x]==-1){ans=-1;continue;}//Ambiguityp[i]=e[x];}if(ans==0)puts("Impossible");else if(ans==-1)puts("Ambiguity");else {puts("Possible");for(int i=1;i<=m;++i)printf("%d ",p[i]);puts("");}}return 0;}/*【trick&&吐槽】有些题，就是给你绕啊绕，我们要好好读题目，再理清思路，做起来就很简单了！【题意】这是理清后题意的版本——发现理清之后就是傻叉题啊！给你一个表示映射关系的数组f[]，元素个数为n（1e5），每个数的范围都在[1,n]。给你一个映射之后的数列b[]，元素个数为m（1e5），每个数的范围都在[1,n]。对于b的每个位置的数b[i]，都是由f[]某个位置的数生成的。是哪个位置呢？记录为a[i]。如果存在至少一个b[i]，不能够生成，答案就是Impossible否则，如果至少存在一个b[i]，所对应的a[i]不唯一，那么答案就是Ambiguity否则输出Possible和解。【类型】模拟【分析】按照理清之后的题意。我们发现，对于b[]中的每个数，f[]中都要恰好只出现一次。于是我们对于f[]中的所有数x，记录位置。x没出现过，e[]便为0x只出现过一次，e[]便为位置x出现过多次，e[]便为-1然后扫描一遍b[]，这道题就做完AC啦！*/