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Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) D. Artsem and Saunders

2024-09-05 04:08:38   217人阅读

地址：http://codeforces.com/contest/765/problem/D

题目：

D. Artsem and Saunders
time limit per test
2 seconds
memory limit per test
512 megabytes
input
standard input
output
standard output

Artsem has a friend Saunders from University of Chicago. Saunders presented him with the following problem.

Let [n] denote the set {1,?...,?n}. We will also write f:?[x]?→?[y] when a function f is defined in integer points 1, ..., x, and all its values are integers from 1 to y.

Now then, you are given a function f:?[n]?→?[n]. Your task is to find a positive integer m, and two functions g:?[n]?→?[m], h:?[m]?→?[n], such that g(h(x))?=?x for all 技术分享, and h(g(x))?=?f(x) for all 技术分享, or determine that finding these is impossible.

Input

The first line contains an integer n (1?≤?n?≤?105).

The second line contains n space-separated integers — values f(1),?...,?f(n) (1?≤?f(i)?≤?n).

Output

If there is no answer, print one integer -1.

Otherwise, on the first line print the number m (1?≤?m?≤?106). On the second line print n numbers g(1),?...,?g(n). On the third line print mnumbers h(1),?...,?h(m).

If there are several correct answers, you may output any of them. It is guaranteed that if a valid answer exists, then there is an answer satisfying the above restrictions.

Examples
input
3
1 2 3
output
3
1 2 3
1 2 3
input
3
2 2 2
output
1
1 1 1
2
input
2
2 1
output
-1

思路：这题思路清晰的人应该可以很快做出来的。

　　g(h(x))?=?x ，可以推出m<=n;

　　h(g(x))=f(x),可以推出m>=f(x)中不同元素的个数。

　　先保证第二个条件h(g(x))=f(x)：

　　　　h[]=去重后的f[]。同时记录f[i]在h[]中的hash值：g[i]hash[f[i]]。

　　然后扫一遍g[h[i]]，判断值是否为x。

　　我不知道怎么证明可以这么做，感觉可行后莽一发就ac了。

　　有谁知道了可以交流下。

 1 #include <bits/stdc++.h>
 2 
 3 using namespace std;
 4 
 5 #define MP make_pair
 6 #define PB push_back
 7 typedef long long LL;
 8 typedef pair<int,int> PII;
 9 const double eps=1e-8;
10 const double pi=acos(-1.0);
11 const int K=1e5+7;
12 const int mod=1e9+7;
13 
14 int n,m,ans;
15 int h[K],g[K],f[K];
16 map<int,int>hs;
17 
18 int main(void)
19 {
20     cin>>n;
21     for(int i=1,x;i<=n;i++)
22     {
23         scanf("%d",&x),f[i]=x;
24         if(!hs[x])  h[++m]=x,hs[x]=m;
25     }
26     for(int i=1;i<=n;i++)
27         g[i]=hs[f[i]];
28     for(int i=1;i<=m;i++)
29     if(g[h[i]]!=i) ans=1;
30     if(ans)
31         printf("-1\n");
32     else
33     {
34         printf("%d\n",m);
35         for(int i=1;i<=n;i++)
36             printf("%d ",g[i]);
37         puts("");
38         for(int i=1;i<=m;i++)
39             printf("%d ",h[i]);
40         puts("");
41     }
42     return 0;
43 }

 

 

Codeforces Round #397 by Kaspersky Lab and Barcelona Bootcamp (Div. 1 + Div. 2 combined) D. Artsem and Saunders


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