hdu 6038 Function

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1621    Accepted Submission(s): 758

Problem Description

You are given a permutation a from 0 to n−1 and a permutation b from 0 to m−1.

Define that the domain of function f is the set of integers from 0 to n−1, and the range of it is the set of integers from 0 to m−1.

Please calculate the quantity of different functions f satisfying that f(i)=bf(ai) for each i from 0 to n−1.

Two functions are different if and only if there exists at least one integer from0 to n−1 mapped into different integers in these two functions.

The answer may be too large, so please output it in modulo 109+7.

 

Input

The input contains multiple test cases.

For each case:

The first line contains two numbers n,m.(1≤n≤100000,1≤m≤100000)

The second line contains n numbers, ranged from 0 to n−1, the i-th number of which represents ai−1.

The third line contains m numbers, ranged from 0 to m−1, the i-th number of which represents bi−1.

It is guaranteed that ∑n≤106,∑m≤106.

 

Output

For each test case, output "Case #x:y" in one line (without quotes), where x indicates the case number starting from 1 and y denotes the answer of corresponding case.

 

Sample Input

3 21 0 20 13 42 0 10 2 3 1

 

Sample Output

Case #1: 4Case #2: 4

 

Source

2017 Multi-University Training Contest - Team 1

因为a,b都是分别0~n-1,0~m-1内的数，所以a,b内一定存在环

例如

2 0 1
0 2 3 1

f(0)=b(f(2))
f(1)=b(f(0))
f(2)=f(f(1))

因此a中存在2->0->1->2的环

并且b中也存在环
0 1 2 3
0 2 3 1    --> 0->0,1->2->3->1的环

所以这里就有一个结论:
a中的环能映射到b中的条件:b环的长度必须是a环长度的因子

这样对a环枚举每个b环，若长度满足条件，则加上b环长度(==存在的函数数量)
最后乘起来就是答案了

这里还有就是找环的时候可以用类似深搜的思想沿着a数组一直搜下去知道遇到i==a[j]时停止

#include<cstdio>#include<cstring>#include<iostream>#include<algorithm>#include<vector>#define MAXN 1000010#define MAXM 100010using namespace std;typedef long long int ll;const int MOD = 1e9+7;ll n,m;ll a[MAXM],b[MAXM];ll vis[MAXM];vector<int> len1,len2;  //记录a,b内每个环的长度以及总共的个数void solve(ll a[],ll s,int flag)  //类似深搜，将一个环的归为一类{memset(vis,0,sizeof(vis));for(int i=0;i<s;i++){if(vis[i]) continue;ll k=a[i];ll slen=1;vis[i]=1;while(k!=i){slen++;vis[k]=1;k=a[k];}if(flag==1)len1.push_back(slen);elselen2.push_back(slen);}}int main(){ll ans;int t=0;while(scanf("%lld%lld",&n,&m)!=EOF){t++;ans=1;len1.clear();len2.clear();for(ll i=0;i<n;i++)scanf("%lld",&a[i]);for(ll i=0;i<m;i++)scanf("%lld",&b[i]);solve(a,n,1);solve(b,m,2);for(int i=0;i<len1.size();i++){ll res=0;for(int j=0;j<len2.size();j++){if(len1[i]%len2[j]==0)   //一个a环对应的B环是加的关系，表示一个a环的节点可以对应b环中这么多的数res=(res+len2[j])%MOD;}ans=(ans*res)%MOD;   //注意每一个a之间是乘的关系即使是a环内有相同的节点数，也要相乘，因为是不同的两个环}printf("Case #%d: %lld\n",t,ans%MOD);}return 0;}