D. Artsem and Saunders
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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 , and h(g(x)) = f(x) for all , or determine that finding these is impossible.
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).
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.
31 2 3
31 2 31 2 3
32 2 2
11 1 12
22 1
-1
题意:给你一个序列f[x]。x和fx都属于1到n。然后让你找出上面两个约束符合条件的gx(x和gx都属于1到n)和hx(x和hx)都属于1到m。
思路:首先我们看一个约束条件:h(g(x)) = f(x)
可以联想到gx充当的是一个下标的角色。也就说实际上hx的值应当是和fx的值是相等的。我们把hx中和fx中的值相等的做成一个映射关系。也就说hx是fx去重之后的一个结果。
而此时的gx就应该是fx中的元素的一个标号。
也就说我们把fx的不同元素进行标号,然后hx按照标号来放置这些不同的元素。
那么g1,g2,g4就应该是a的标号1.
然后我们就能初步求出gx和hx了。
然后我们在查看一下求出的序列满不满足第二个约束条件。
满足就是符合条件的,不满足就输出-1.
#include <bits/stdc++.h>using namespace std;const int MAXN=1e5+7;int n,m;int f[MAXN];int h[MAXN],g[MAXN];map<int,int>ha;int main(){ int i; int n; scanf("%d",&n); int cnt=0; for(i=1; i<=n; ++i) { scanf("%d",&f[i]); if(!ha[f[i]]) { ha[f[i]]=++cnt; h[cnt]=f[i]; } } for(i=1; i<=n; ++i)g[i]=ha[f[i]]; for(i=1; i<=cnt; ++i)if(g[h[i]]!=i)break; if(i<=cnt)puts("-1"); else { printf("%d\n",cnt); for(i=1; i<=n; ++i)printf(i==n?"%d\n":"%d ",g[i]); for(i=1; i<=cnt; ++i)printf(i==m?"%d\n":"%d ",h[i]); } return 0;}
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