Codeforces Round #332 (Div. 二) B. Spongebob and Joke
2016-01-08 20:14
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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
Input
Output
Input
Output
Input
Output
Hint
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:给你数组f[]和数组b[]问你是否存在数组a[]使得bi==fai;
题解:三种情况1、数组b中的数字数组f中没有 则输出Impossible 2、数组b中的数字在数组f中重复出现且数组b中的数字都在数组f中输出Ambiguity 3、数组b中的数字都在数组f中且无重复输出Possible以及数组a[]
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
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
Hint
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:给你数组f[]和数组b[]问你是否存在数组a[]使得bi==fai;
题解:三种情况1、数组b中的数字数组f中没有 则输出Impossible 2、数组b中的数字在数组f中重复出现且数组b中的数字都在数组f中输出Ambiguity 3、数组b中的数字都在数组f中且无重复输出Possible以及数组a[]
#include<stdio.h> #include<string.h> #include<stdlib.h> #include<algorithm> #define LL long long #define MAX 100100 using namespace std; int b[MAX],f[MAX]; int a[MAX],vis[MAX]; int main() { int n,m,i,j,ans; while(scanf("%d%d",&n,&m)!=EOF) { memset(vis,0,sizeof(vis)); for(i=1;i<=n;i++) { scanf("%d",&ans); f[ans]=i; vis[ans]++; } for(i=1;i<=m;i++) scanf("%d",&b[i]); int flag=0; int Flag=0; for(i=1;i<=m;i++) { if(vis[b[i]]==0) { flag=1; break; } if(vis[b[i]]>1) Flag=1; } if(flag) { printf("Impossible\n"); continue; } if(Flag) { printf("Ambiguity\n"); continue; } for(i=1;i<=m;i++) a[i]=f[b[i]]; printf("Possible\n"); for(i=1;i<m;i++) printf("%d ",a[i]); printf("%d\n",a[m]); } return 0; }
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