2017 Multi-University Training Contest 1 && HDOJ 6038 Function 【强连通找环】

Function

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

Total Submission(s): 348    Accepted Submission(s): 136


Problem Description

You are given a permutation a from 0 ton−1 and apermutationb
from 0 tom−1.

Define that the domain of function f is the set of integers from 0 ton−1, and the range of it is
the set of integers from 0 tom−1.

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

Two functions are different if and only if there exists at least one integer from 0 ton−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 ton−1, thei
-th number of which representsai−1.

The third line contains m numbers,ranged from 0 tom−1, thei
-th number of which representsbi−1.

It is guaranteed that ∑n≤106,∑m≤106.

 
 
Output

For each test case, output "Case #x:y"in
one line (without quotes), wherex indicates the case number starting from 1 andy 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】给出两个数组a,b，a数组是[0~n-1] 的排列，b数组是[0~m-1]的排列，问有多少种赋值，使得F(i)=b[F(a[i])]。

【思路】举个例子，对于样例二，，我们可以得到以下等式：

(1) F(0)=b[F(2)]
(2) F(1)=b[F(0)]
(3) F(2)=b[f(1)]

可以发现，对于这个样例，F(0),F(1),F(2)形成了一个环，我们可以先设定F(2)的值，进而可以根据(1),（2）得到F(0),F(1)的值，然后利用(3)又回到了代回F(2)。要使这个环成立，必须在值域(即b)中也要有一个长度相等或为其因子的环.

那么这道题就转化为用强连通找环了，我们分别找出数组a和数组b中环的个数及每个环的长度，并求出数组b对应的长度为某个值的所有环包含元素的个数，然后枚举数组a中每个环的长度X，如果数组b中有一个环长度为d，且X%d==0，那么这个环的方案数加上该长度的所有环包含的元素num(因为第一个元素的取值有num种可能)，最后数组a的所有环的方案数累乘即可。

#include <cstdio>
#include <cmath>
#include <vector>
#include <iostream>
#include <map>
#include <cstring>
#include <algorithm>
using namespace std;
#define mst(a,b) memset((a),(b),sizeof(a))
#define rush() int T;scanf("%d",&T);while(T--)

typedef long long ll;
const int maxn = 100005;
const ll mod = 1e9+7;
const int INF = 0x3f3f3f;
const double eps = 1e-9;

int n,m;
int cnt,tot,id;
int num_huan;
int a[maxn];
int b[maxn];
int low[maxn],dfn[maxn];
int vis[maxn];
int stack[maxn];
int color[maxn];
int num1[maxn],num2[maxn];  //每个环的元素个数
ll ans[maxn];
vector<int>vec[maxn];
map<int,ll>mp;             //某一长度的环所有元素的个数

void init()
{
cnt=1,id=0,tot=-1;
mst(stack,0),mst(vis,0);
mst(low,0),mst(dfn,0);
mst(color,0);
}

void tarjan(int u)
{
vis[u]=1;
dfn[u]=low[u]=cnt++;
stack[++tot]=u;
for(int i=0;i<vec[u].size();i++)
{
int v=vec[u][i];
if(vis[v]==0) tarjan(v);
if(vis[v]==1) low[u]=min(low[u],low[v]);
}
if(dfn[u]==low[u])
{
id++;
do
4000
{
color[stack[tot]]=id;
vis[stack[tot]]=-1;
}
while(stack[tot--]!=u);
}
}

void solve1()
{
init();
mst(num1,0);
for(int i=1;i<=n;i++)
{
if(vis[i]==0)
tarjan(i);
}
for(int i=1;i<=n;i++)
{
num1[color[i]]++;
}
num_huan=id;
}

void solve2()
{
init();
mst(num2,0);
for(int i=1;i<=m;i++)
{
if(vis[i]==0)
tarjan(i);
}
for(int i=1;i<=m;i++)
{
num2[color[i]]++;
}
for(int i=1;i<=id;i++)
{
mp[num2[i]]+=num2[i];
}
}

int main()
{
int cas=1;
while(~scanf("%d%d",&n,&m))
{
mp.clear();
mst(ans,0);
for(int i=1;i<=n;i++)
{
scanf("%d",&a[i]);
a[i]++;
vec[i].clear();
}
for(int i=1;i<=n;i++)
{
vec[a[i]].push_back(i);
}
solve1();
//printf("***%d\n",num_huan);
for(int i=1;i<=m;i++)
{
scanf("%d",&b[i]);
b[i]++;
vec[i].clear();
}
for(int i=1;i<=m;i++)
{
vec[b[i]].push_back(i);
}
solve2();
for(int i=1;i<=num_huan;i++)
{
for(int j=1;j*j<=num1[i];j++)
{
if(num1[i]%j==0)
{
ans[i]+=mp[j];
if(j*j!=num1[i])
ans[i]+=mp[num1[i]/j];

}
}
}
ll output=1;
for(int i=1;i<=num_huan;i++)
{
output=(output*ans[i])%mod;
}
printf("Case #%d: %I64d\n",cas++,output);
}
return 0;
}