CodeForces - 757C - Felicity is Coming!（思维）

C. Felicity is Coming!

time limit per test
2 seconds

memory limit per test
256 megabytes

input
standard input

output
standard output

It's that time of the year, Felicity is around the corner and you can see people celebrating all around the Himalayan region. The Himalayan region has n gyms.
The i-th gym has gi Pokemon
in it. There are m distinct Pokemon types in the Himalayan region numbered from 1 to m.
There is a special evolution camp set up in the fest which claims to evolve any Pokemon. The type of a Pokemon could change after evolving, subject to the constraint that if two Pokemon have the same type before evolving, they will have the same type after
evolving. Also, if two Pokemon have different types before evolving, they will have different types after evolving. It is also possible that a Pokemon has the same type before and after evolving.
Formally, an evolution plan is a permutation f of {1, 2, ..., m},
such that f(x) = y means
that a Pokemon of type x evolves into a Pokemon of type y.
The gym leaders are intrigued by the special evolution camp and all of them plan to evolve their Pokemons. The protocol of the mountain states that in each gym, for every type of
Pokemon, the number of Pokemon of that type before evolving any Pokemon should be equal the number of Pokemon of that type after evolving all the Pokemons according to the evolution plan. They now want to find out how many distinct evolution
plans exist which satisfy the protocol.
Two evolution plans f1 and f2 are
distinct, if they have at least one Pokemon type evolving into a different Pokemon type in the two plans, i. e. there exists an i such
that f1(i) ≠ f2(i).
Your task is to find how many distinct evolution plans are possible such that if all Pokemon
in all the gyms are evolved, the number of Pokemon of each type in each of the gyms remains the same. As the answer can be large, output it modulo 109 + 7.

Input
The first line contains two integers n and m (1 ≤ n ≤ 105, 1 ≤ m ≤ 106) —
the number of gyms and the number of Pokemon types.
The next n lines
contain the description of Pokemons in the gyms. The i-th of these lines begins with the
integer gi (1 ≤ gi ≤ 105) —
the number of Pokemon in the i-th gym. After that gi integers
follow, denoting types of the Pokemons in the i-th gym. Each of these integers is between 1 and m.
The total number of Pokemons (the sum of all gi)
does not exceed 5·105.

Output
Output the number of valid evolution plans modulo 109 + 7.

Examples

input

2 3
2 1 2
2 2 3

output

1

input

1 3
3 1 2 3

output

6

input

2 4
2 1 2
3 2 3 4

output

2

input

2 23 2 2 12 1 2

output

1

input

3 7
2 1 22 3 4
3 5 6 7

output

24

Note
In the first case, the only possible evolution plan is:



In the second case, any permutation of (1,  2,  3) is
valid.
In the third case, there are two possible plans:





In the fourth case, the only possible evolution plan is:



题意：给你n个集合，每个集合里面有k个数，每个数的大小代表一个种类，让你f[i]的转换，使得每个种类的数量没变，求转换方法的个数（就题意略难懂）

思路：将每个种类所在的集合存起来，必须保证原来n个集合里面的x和y的个数一样才能转换

利用vector进行存储

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<vector>

using namespace std;

const long long MOD = 1e9+7;
const int N = 1e6+5;
vector<int> G
;
int main()
{
int n,m,k,x;
cin >> n >> m;
for(int i = 1;i <= n;i++)
{
scanf("%d",&k);
while(k--)
{
scanf("%d",&x);
G[x].push_back(i);
}
}

sort(G+1,G+1+m);
long long cnt = 1,res = 1;
for(int i = 1;i <= m-1;i++)
{
if(G[i] == G[i+1])
{
cnt++;
res = res * cnt % MOD;
}
else
cnt = 1;
}
printf("%lld\n",res);
return 0;
}