HDU 5399 Too Simple

Problem Description

Rhason Cheung had a simple problem, and asked Teacher Mai for help. But Teacher Mai thought this problem was too simple, sometimes naive. So she ask you for help.

Teacher Mai has  m  functions  f1,f2,⋯,fm:{1,2,⋯,n}→{1,2,⋯,n} (that means for all  x∈{1,2,⋯,n},f(x)∈{1,2,⋯,n} ). But Rhason only knows some of these functions, and others are unknown.

She wants to know how many different function series  f1,f2,⋯,fm  there are that for every  i(1≤i≤n) , f1(f2(⋯fm(i)))=i . Two function series  f1,f2,⋯,fm  and  g1,g2,⋯,gm  are considered different if and only if there exist  i(1≤i≤m),j(1≤j≤n) , fi(j)≠gi(j) .

 

Input

For each test case, the first lines contains two numbers  n,m(1≤n,m≤100) .

The following are  m  lines. In  i -th line, there is one number  −1  or  n  space-separated numbers.

If there is only one number  −1 , the function  fi  is unknown. Otherwise the  j -th number in the  i -th line means  fi(j) .

 

Output

For each test case print the answer modulo  109+7 .

 

Sample Input

   
   
   
   

    
    
    
    3 3
1 2 3
-1
3 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1 
    
    
    
    
 
     
 
      Hint 
     
The order in the function series is determined. What she can do is to assign the values to the unknown functions. 
    
    
    
    
    
 
    
    
    
    
 
    
    
    
    
这题其实很简单，对于多于两个-1的情况不管之前那个怎么选，只要最后一个固定就可以保证，所以答案就是(n!)^(m-1) m是-1的个数
    
    
    
    
不过此题坑比较多，要注意判断
    
    
    
    

    
    
    
    
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const LL base = 1e9 + 7;
const int maxn = 105;
LL T, n, m, f[maxn], a[maxn][maxn];

inline void read(int &x)
{
    char ch;
    while ((ch = getchar())<'0' || ch>'9');
    x = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';
}

int main()
{
    //read(T);
    for (int i = f[0] = 1; i <= 100; i++) f[i] = f[i - 1] * i % base;
    while (scanf("%I64d%I64d", &n, &m) != EOF)
    {
        LL tot = 0, ans = 1;
        for (int i = 1; i <= m; i++)
        {
            scanf("%I64d", &a[i][1]);
            if (a[i][1] == -1) tot++;
            else for (int j = 2; j <= n; j++)
            {
                scanf("%I64d", &a[i][j]);
                for (int k = j - 1; k; k--) if (a[i][j] == a[i][k]) ans = 0;
            }
        }
        for (int i = 1; i < tot; i++) ans = ans * f[n] % base;
        if (tot == 0)
        {
            for (int i = 1; i <= n; i++) a[0][i] = i;
            for (int i = m; i; i--)
                for (int j = 1; j <= n; j++) a[0][j] = a[i][a[0][j]];
            for (int i = 1; i <= n; i++) if (a[0][i] != i) ans = 0;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}