Codeforces Round #332 (Div. 2)B. Spongebob and Joke

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, ..., 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.

打了好几8次cf了 一直掉到灰名 昨天终于上分了 绿了 关键的一题（纪念FQ半夜三更打cf的日子）

题意： 输入 n,m

第二行 n个整数 f1, f2, ..., fn (1 ≤ fi ≤ n).

第三行 m个整数 b1, b2, ..., bm (1 ≤ bi ≤ n).

判断 有且只有一个bi = fai 输出 fai的位置；

当b 没有在 f出现时输出 Impossible

当b 在f中出现 但f的位置不唯一时输出Ambiguity

除此之外 输出 Possible 及b1...bm 对应的位置 看懂题意 代码很简单

比赛的时候 被hack了一次

早点睡了 c题要补

#include<bits/stdc++.h>
using namespace std;
map<int,int>mp1;
map<int,int>mp2;
map<int,int>mp3;
int n,m;
int a,b;
int s[100005];
int re[100005];
int main()
{
mp1.clear();
mp2.clear();
mp3.clear();
memset(s,0,sizeof(s));
memset(re,0,sizeof(re));
scanf("%d%d",&n,&m);
for(int i=0;i<n;i++)
{
scanf("%d",&a);
mp1[a]++;
mp2[a]=i+1;
}
for(int i=0;i<m;i++)
{
scanf("%d",&s[i]);
mp3[s[i]]=1;
}
for(int i=0;i<m;i++)
{
if(mp2[s[i]]==0)
{
printf("Impossible\n");
return 0;
}
re[i]=mp2[s[i]];
}
for(int i=1;i<=n;i++)
{
if(mp1[i]>=2&&mp3[i])
{
printf("Ambiguity\n");
return 0;
}
}
printf("Possible\n");
printf("%d",re[0]);
for(int i=1;i<m;i++)
printf(" %d",re[i]);
printf("\n");
return 0;
}