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

B. Spongebob and Joke

Time Limit: 20 Sec

Memory Limit: 256 MB

题目连接

http://codeforces.com/contest/599/problem/B

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

3 3
3 2 1
1 2 3

Sample Output

Possible
3 2 1 

HINT

题意

给你n个f[i],m个b[i]，然后问你能不能找到m个a[i]，使得b[i]=f[a[i]]

题解：

暴力存一下这个数在f[i]中出现了多少次，如果没出现就输出impossilbe，如果出现多次，就多解，如果出现一次就输出这个数。（注意impossible的优先级大于多解，这是hack点

注意ai可以等于aj

代码

#include<iostream>
#include<math.h>
#include<vector>
#include<stdio.h>
using namespace std;
#define maxn 100005
int a[maxn];
int f[maxn];
int b[maxn];
vector<int> Q[maxn];
int main()
{
    int n,m;scanf("%d%d",&n,&m);
    for(int i=;i<n;i++)
        scanf("%d",&f[i]);
    for(int i=;i<m;i++)
        scanf("%d",&b[i]);
    for(int i=;i<n;i++)
        Q[f[i]].push_back(i);
    int flag = ;
    for(int i=;i<m;i++)
    {
        if(Q[b[i]].size()==)
            return puts("Impossible");
        if(Q[b[i]].size()>)
            flag = ;
    }
    if(flag == )
        return puts("Ambiguity");
    puts("Possible");
    for(int i=;i<m;i++)
        printf("%d ",Q[b[i]][]+);
    printf("\n");
}