2017多校赛 Function

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 838    Accepted Submission(s): 369

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 from 0 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 2
1 0 2
0 1
3 4
2 0 1
0 2 3 1

 

Sample Output

Case #1: 4

Case #2: 4

 

比赛的时候想到了环这个东西，怀疑自己的思路，没有继续向下推了。果然专注度和练习的强度不够。继续加油。

 AC代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <stack>
#include <vector>
using namespace std;
typedef long long ll;
ll a[],b[];
vector<ll> edgea,edgeb;
const ll mod=1e9+;
ll n,m;
void solve(int Case)
{
    ll visa[],visb[];
    memset(visb,,sizeof(visb));
    memset(visa,,sizeof(visa));
    for(int i=;i<m;i++)
    {
        if(visb[i]) continue;
        ll ret=;
        ll next_pos=b[i];
        visb[next_pos]=;
        while(next_pos!=i)
        {

            ret++;
            next_pos=b[next_pos];
            visb[next_pos]=;
        }
       // cout<<ret<<endl;
        edgeb.push_back(ret);
    }
    ll f=;
    for(int i=;i<n;i++)
    {
        if(visa[i]) continue;
        ll ret=;
        ll next_pos=a[i];
        visa[next_pos]=;
        while(next_pos!=i)
        {
            ret++;
            next_pos=a[next_pos];
            visa[next_pos]=;
        }
        ll temp=;
        //cout<<ret<<endl;
        for(int j=;j<edgeb.size();j++)
        {
            if(ret%edgeb[j]== || ret==edgeb[j]) temp=(temp+edgeb[j])%mod;
        }
        f=f*temp%mod;
    }
    printf("Case #%d: %d\n",Case,f);
}
int main()
{
    cin.sync_with_stdio(false);
    int Case=;
    while(cin>>n>>m)
    {
        edgea.clear();
        edgeb.clear();
        for(int i=;i<n;i++) cin>>a[i];
        for(int j=;j<m;j++) cin>>b[j];
        solve(Case++);
    }
    return ;
}