Codeforces Round #332 (Div. 2) 599B Spongebob and Joke(STL)

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.

题目链接:点击打开链接

给出n, m, 长度为n的序列f, 长度为m的序列b, 问是否存在长度为m的序列a, 使得f[a[i]] = b[i].

STL的使用, 读入序列f的时候进行标记, 如果已经存在则标记为-1, 未出现则记录下标. 读入序列b的时候进行判断, 如果未出现则标记

flag1, 出现多次则标记flag2, 否则就记录下标, 根据不同情况输出答案即可.

AC代码:

#include "iostream"
#include "cstdio"
#include "cstring"
#include "algorithm"
#include "cmath"
#include "utility"
#include "map"
#include "set"
#include "vector"
using namespace std;
typedef long long ll;
const int MOD = 1e9 + 7;
const int MAXN = 1e5 + 5;
int n, m;
int f[MAXN], b[MAXN], ans[MAXN];
map<int, int> mp;
bool flag1 = false, flag2 = false;
int main(int argc, char const *argv[])
{
scanf("%d%d", &n, &m);
for(int i = 1; i <= n; ++i) {
scanf("%d", &f[i]);
if(mp.count(f[i])) mp[f[i]] = -1;
else mp[f[i]] = i;
}
for(int i = 1; i <= m; ++i) {
scanf("%d", &b[i]);
if(!mp.count(b[i])) flag1 = true;
else if(mp[b[i]] < 0) flag2 = true;
else ans[i] = mp[b[i]];
}
if(flag1) printf("Impossible\n");
else if(flag2) printf("Ambiguity\n");
else {
printf("Possible\n");
for(int i = 1; i <= m; ++i)
printf("%d ", ans[i]);

printf("\n");
}
return 0;
}