Codeforces Round #332 (Div. 2) B. Spongebob and Joke(水题,构造)
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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 sequencea1, a2, ..., am of lengthm, consisting of integers from 1 to n, not necessarily distinct. Then he picked some sequencef1, f2, ..., fn of lengthn and for each number ai got number bi = fai. To finish the prank he erased the initial sequenceai.
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 andm (1 ≤ n, m ≤ 100 000) — the lengths of sequencesfi andbi respectively.
The second line contains n integers, determining sequencef1, f2, ..., fn (1 ≤ fi ≤ n).
The last line contains m integers, determining sequenceb1, b2, ..., bm(1 ≤ bi ≤ n).
Print "Possible" if there is exactly one sequenceai, such thatbi = fai for alli 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 33 2 11 2 3
Possible3 2 1
3 31 1 11 1 1
Ambiguity
3 31 2 13 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 alli, so no sequence ai transforms into suchbi and we can say for sure that Spongebob has made a mistake.
【题意】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]].
【解题方法】简单构造。暴力存一下这个数在f[i]中出现了多少次,如果没出现就输出impossilbe,如果出现多次,就多解,如果出现一次就输出这个数。
【AC 代码】
#include <bits/stdc++.h>using namespace std;const int maxn = 100010;int f[maxn],b[maxn],a[maxn];vector<int>G[maxn];int main(){ int n,m; scanf("%d%d",&n,&m); for(int i=1; i<=n; i++) scanf("%d",&f[i]); for(int i=1; i<=m; i++) scanf("%d",&b[i]); for(int i=1; i<=n; i++){ G[f[i]].push_back(i); } bool ok=0; for(int i=1; i<=m; i++){ if(G[b[i]].size()==0){ puts("Impossible"); return 0; } if(G[b[i]].size()>1) ok=1; } if(ok==1){ puts("Ambiguity"); return 0; } puts("Possible"); for(int i=1; i<=m; i++){ printf("%d ",G[b[i]][0]); } return 0;}
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