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.
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).
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".
3 3 3 2 1 1 2 3
Possible 3 2 1
3 3 1 1 1 1 1 1
Ambiguity
3 3 1 2 1 3 3 3
Impossible
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.
用哈希过的
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cstdlib>
#include <algorithm>
using namespace std;
const int max1 = 101000;
struct node {
int x, y;
}a[max1];
int b[max1];
int f[max1];
int a1[max1];
int main() {
memset(f, 0, sizeof(f));
memset(a, 0, sizeof(a));
memset(a1, 0, sizeof(a1));
memset(b, 0, sizeof(b));
for (int i = 0; i<max1; i++)
a[i].x = 0;
int n, m;
cin >> n>> m;
for (int i = 1; i<=n; i++) {
cin >> f[i];
a[f[i]].x ++ ;
a[f[i]].y = i;
}
for (int i = 1; i<=m; i++)
cin >> b[i];
int flag1 = 0;
int flag = 0;
int num = 1;
for (int i = 1; i<=m; i++) {
if (a[b[i]].x == 1) {
a1[num++] = a[b[i]].y;
}
else if (a[b[i]].x >= 2) {
a1[num++] = a[b[i]].y;
flag = 1;
}
else if (a[b[i]].x == 0) {
flag = 2;
break;
}
}
if (flag == 2)
cout << "Impossible" << endl;
else if (flag == 1)
cout << "Ambiguity" << endl;
else {
cout << "Possible" << endl;
for (int i = 1; i<num-1; i++)
cout << a1[i] << " " ;
cout << a1[num-1] << endl;
}
return 0;
}

海绵宝宝在帕特里克购物时开了一场玩笑,将帕特里克的一系列整数通过另一组映射序列进行了转换并删除了原始序列。本篇探讨了如何根据剩余信息还原原始序列的可能性及其唯一性。
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