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

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.

题意：给你n个f，m个b，按照公式让你判断是否能决定a

思路：结果可以有三种：1.不能，此时，f中存在两个或两个以上的a相同。2.不确定，此时一个b对应多个f，3.确定

f[num]存储a应有的值，然后用b查询，注意不存在比不确定优先判断。。

总结：水题，不难想，读懂就能写了

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 1000100
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
using namespace std;
int f[MAXN];
int a[MAXN],b[MAXN];
int main()
{
int n,m;
int i;
while(scanf("%d%d",&n,&m)!=EOF)
{
mem(f);
int num;
for(i=1;i<=n;i++)
{
scanf("%d",&num);
if(f[num])
f[num]=-1;
else
f[num]=i;
}
for(i=1;i<=m;i++)
scanf("%d",&b[i]);
int bz1=0,bz2=0,bz3=0;
for(i=1;i<=m;i++)
{
if(f[b[i]]==0)
bz1=1;
else if(f[b[i]]==-1)
bz2=1;
else
a[i]=f[b[i]];

}
if(bz1)
printf("Impossible\n");
else if(bz2)
printf("Ambiguity\n");
else
{
printf("Possible\n");
for(i=1;i<m;i++)
printf("%d ",a[i]);
printf("%d\n",a[i]);
}
}
return 0;
}