D. Artsem and Saunders----数学思维

来源：互联网发布：淘宝怎么定时上架软件编辑：程序博客网时间：2024/05/21 19:45

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 $\text{[math]}$ , and h(g(x)) = f(x) for all $\text{[math]}$ , 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 printm numbers 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
31 2 3
output
31 2 31 2 3
input
32 2 2
output
11 1 12
input
22 1
output
-1

题目链接：http://codeforces.com/contest/765/problem/D

前天的cf了，结果昨天vj上有场比赛，也没有补这个题，今天补上；

看的卿学姐的博客，然后又从网上看了篇博客才懂。。。我差不多已经是个废C~K了

主要还是数学思想吧，算是一个置换的水题吧。

题目中给出你两个公式，g(h(x))==x,h(g(x))==f(x);现给你f（x），让你求符合条件的g序列和h序列。

现在我们来把这两个式子变一下形。

我们知h(g(x))==f(x);

又知g(h(x))==x；

所以h(g(h(x)))==f(h(x)) ——>h(x)=f(h(x));

所以X==f(X)//整体代换

所以f(X)==f(f(X))；

这个是用来判断是否能够构成序列的条件。

下面这一段话来自于大牛：http://www.cnblogs.com/zzuli2sjy/p/6400877.html

由h(g(x)) = f(x),假设f(x1) = f(x2);那么有h(g(x1)) = h(g(x2)),那么我们假设g(x1)!=g(x2),那么就有h中存在两个相等的数，那么g(h(g(x1))) = g(x1),g(h(g(x2))) = g(x2)就有个g(x1) = g(x2)所以假设不成立，那么h中的数都是不一样的，那么就为f中的不同种类数，m就确定了。于是确定h(x),那么h(x)有许多的排列，假设x1，x2，h(x1) = a,h(x2) = b,f(y1) = a,f(y2) = b;通过条件1求出g,h(g(a)) = f(a),h(g(b)) = f(b);-> g(a) = x1,g(b) = x2,那么有h(x1) = f(a),h(x2) = f(b);那么假设h(x1) = b,h(x2) = a,可以推得g(a) = x2,g(b) = x1,有h(x2) = f(a),h(x1) = f(b),总是可以的f（a）= a，f（b） = b;

大牛的证法很详细，由此我们可以得出h序列其实就是f序列的去重序列，然后我们由h(g(x))==f(x)这个公式就可以推出来g序列了。

附上卿学姐的链接：http://www.cnblogs.com/qscqesze/p/6401529.html

代码：

#include <cstdio>#include <cstring>#include <iostream>#include <map>using namespace std;int a[200000];int b[200000];int c[200000];map<int ,int>H;int main(){    int n;    scanf("%d",&n);    for(int i=1;i<=n;i++){        scanf("%d",&a[i]);    }    for(int i=1;i<=n;i++){        if(a[i]!=a[a[i]]){            return 0*printf("-1\n");        }    }    int m=0;    for(int i=1;i<=n;i++){        if(!H[a[i]]){            m++;            H[a[i]]=m;            c[m]=a[i];        }        b[i]=H[a[i]];    }    cout<<m<<endl;    for(int i=1;i<=n;i++){        printf(i==1?"%d":" %d",b[i]);    }    cout<<endl;    for(int i=1;i<=m;i++){        printf(i==1?"%d":" %d",c[i]);    }    cout<<endl;    return 0;}

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