HDU 6038 (2017 多校训练赛1 1006） Function(图论)

2017 Multi-University Training Contest - Team 1 1006

Function

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 0    Accepted Submission(s): 0


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

 
题意：

给出两个数列 a 和b   满足 f(i)=b[f(a[i])] 问有多少种组合满足这种情况

分析：

看起来像是数论组合数学之类的 但其实是图论

分析一下样例：

第一个样例  a={1,0,2}   b={0,1}

那么f(0)=b[f(1)]    f(1)=b[f(0)]    f(2)=b[f(2)]

这里有两个环分别为 f(0)->f(1)   和f（2）

那么要使得满足条件带入的b必须成环而且是 a环长度的因子

按照这个思路就可以得到答案了~

AC代码：

#include<stdio.h>
#include<string.h>
#include<vector>
using namespace std;
#define mod 1000000007
vector<int>v;
vector<int>G[100010];
void init()
{

for(int i=1;i<=100000;i++)
{
for(int j=i;j<=100000;j+=i)
G[j].push_back(i);
}
}
int a[100010],b[100010],vis[100010],blen[100010];
int dfs(int pr,int *p)
{
if(vis[pr])
return 0;
vis[pr]=1;
return dfs(p[pr],p)+1;
}
int main()
{
int n,m,cas=0;
init();   //预处理因子
while(scanf("%d%d",&n,&m)==2)
{
v.clear();
for(int i=0;i<n;i++)
scanf("%d",&a[i]);
for(int i=0;i<m;i++)
scanf("%d",&b[i]);
memset(vis,0,sizeof(vis));
for(int i=0;i<n;i++)
{
if(!vis[i])
v.push_back(dfs(i,a));    //找a的环 和记录环的长度
}
memset(blen,0,sizeof(blen));
memset(vis,0,sizeof(vis));
for(int i=0;i<m;i++)
{
if(!vis[i])
blen[dfs(i,b)]++;        //找b 的环，和记录长度数量用于和a匹配
}
long long ans=1;
for(int i=0;i<v.size();i++)
{
long long t=0;
for(int j=0;j<G[v[i]].size();j++)
{
t+=G[v[i]][j]*blen[G[v[i]][j]];
t%=mod;
}
ans*=t;
ans%=mod;
}
printf("Case #%d: %lld\n",++cas,ans);
}
}