Codeforces 599B Spongebob and Joke 【水题】
2015-11-23 22:05
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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
output
input
output
input
output
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:给定一个由n个元素组成的序列f,一个由m个元素组成的序列b(n>=m),问是否唯一存在一个序列a[]使得
对任意的i,都有b[i] = f[a[i]]。
用两个map就可以轻松KO了。
AC代码:
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:给定一个由n个元素组成的序列f,一个由m个元素组成的序列b(n>=m),问是否唯一存在一个序列a[]使得
对任意的i,都有b[i] = f[a[i]]。
用两个map就可以轻松KO了。
AC代码:
#include <cstdio> #include <cstring> #include <cmath> #include <algorithm> #include <cstdlib> #include <queue> #include <stack> #include <map> #include <vector> #define INF 0x3f3f3f3f #define MAXN 1000000+10 #define MAXM 1010 #define eps 1e-8 #define LL long long using namespace std; int f[MAXN], b[MAXN]; map<int, int> fp; map<int, int> id; int main() { int n, m; scanf("%d%d", &n, &m); fp.clear(); id.clear(); for(int i = 1; i <= n; i++) scanf("%d", &f[i]), fp[f[i]]++, id[f[i]] = i; int sum = 0, tsum = 0; for(int i = 1; i <= m; i++) { scanf("%d", &b[i]); if(fp[b[i]] == 1) sum++; else if(fp[b[i]] > 1) tsum++; } if(sum + tsum < m) printf("Impossible\n"); else if(tsum > 0) printf("Ambiguity\n"); else { printf("Possible\n"); int used = 0; for(int i = 1; i <= m; i++) { if(fp[b[i]] == 1) { if(used > 0) printf(" "); printf("%d", id[b[i]]); used++; } } printf("\n"); } return 0; }
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