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.
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.
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[100001],b[100001]; vector<ll> edgea,edgeb; const ll mod=1e9+7; ll n,m; void solve(int Case) { ll visa[100001],visb[100001]; memset(visb,0,sizeof(visb)); memset(visa,0,sizeof(visa)); for(int i=0;i<m;i++) { if(visb[i]) continue; ll ret=1; ll next_pos=b[i]; visb[next_pos]=1; while(next_pos!=i) { ret++; next_pos=b[next_pos]; visb[next_pos]=1; } // cout<<ret<<endl; edgeb.push_back(ret); } ll f=1; for(int i=0;i<n;i++) { if(visa[i]) continue; ll ret=1; ll next_pos=a[i]; visa[next_pos]=1; while(next_pos!=i) { ret++; next_pos=a[next_pos]; visa[next_pos]=1; } ll temp=0; //cout<<ret<<endl; for(int j=0;j<edgeb.size();j++) { if(ret%edgeb[j]==0 || ret==edgeb[j]) temp=(temp+edgeb[j])%mod; } f=f*temp%mod; } printf("Case #%d: %d ",Case,f); } int main() { cin.sync_with_stdio(false); int Case=1; while(cin>>n>>m) { edgea.clear(); edgeb.clear(); for(int i=0;i<n;i++) cin>>a[i]; for(int j=0;j<m;j++) cin>>b[j]; solve(Case++); } return 0; }