Codeforces 299 B Spongebob and Joke

B. Spongebob and Joketime limit per test2 secondsmemory limit per test256 megabytesinputstandard inputoutputstandard outputWhile 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.InputThe 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).OutputPrint "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

NoteIn 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.题目大意：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]]，然后输出a[i]，如果有多种可能的话输出

Ambiguity

，没有的话输出

Impossible

解体思路：

#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <vector>#include <queue>#include <algorithm>#include <set>using namespace std;#define MM(a) memset(a,0,sizeof(a))typedef long long LL;typedef unsigned long long ULL;const int maxn = 1e5+5;const int mod = 1e9+7;const double eps = 1e-8;const int INF = 0x3f3f3f3f;LL gcd(LL a, LL b){    if(b == 0)        return a;    return gcd(b, a%b);}int b[maxn],f[maxn];int vis[maxn];int data[maxn];int cnt[maxn];int main(){    int n, m;    while(~scanf("%d%d",&n,&m))    {        MM(vis);        MM(cnt);        for(int i=1; i<=n; i++)        {            scanf("%d",&b[i]);            vis[b[i]] = i;            cnt[b[i]]++;        }        for(int i=1; i<=m; i++)            scanf("%d",&f[i]);        bool ok = false, yes = false, flag = false;        for(int i=1,j=1; i<=m; i++)        {            if(vis[f[i]] == 0)            {                ok = true;                break;            }            if(cnt[f[i]] > 1)                yes = true;            else                data[j++] = vis[f[i]];        }        if(ok)            puts("Impossible");        else        {            if(yes)                puts("Ambiguity");            else            {                puts("Possible");                for(int i=1; i<m; i++)                    cout<<data[i]<<" ";                cout<<data[m]<<endl;            }        }    }    return 0;}