[hdu5399 Too Simple]YY

题意：m个{1,2...n}→{1,2...,n}的函数，有些已知有些未知，求对任意i∈{1,2,...,n}，f₁(f₂(...(f_m(i)))=i的方案总数，为了方便简记为F(i)

思路：如果存在一个f，当i!=j时，有f(i)=f(j)，那么方案数为0，因为由里到外进行f运算，两个不同的数到这里来了变成了i和j，然后变成了同一个数，最终还是等于同一个数，所以在最外面至少有一个不会满足F(x)=x。如果f全部确定了，那么只需对每个i计算一下F(i)即可确定答案。如果f没确定的个数为cnt，则答案就是n!^cnt-1，因为对后(cnt-1)个未确定的f，对于它们的每种合法情况，第一个f有且仅有唯一一种情况使得F(i)=i成立。

#pragma comment(linker, "/STACK:10240000")
#include <map>
#include <set>
#include <cmath>
#include <ctime>
#include <deque>
#include <queue>
#include <stack>
#include <vector>
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>

using namespace std;

#define X                   first
#define Y                   second
#define pb                  push_back
#define mp                  make_pair
#define all(a)              (a).begin(), (a).end()
#define fillchar(a, x)      memset(a, x, sizeof(a))
#define copy(a, b)          memcpy(a, b, sizeof(a))

typedef long long ll;
typedef pair<int, int> pii;
typedef unsigned long long ull;

//#ifndef ONLINE_JUDGE
void RI(vector<int>&a,int n){a.resize(n);for(int i=;i<n;i++)scanf("%d",&a[i]);}
void RI(){}void RI(int&X){scanf("%d",&X);}template<typename...R>
void RI(int&f,R&...r){RI(f);RI(r...);}void RI(int*p,int*q){int d=p<q?:-;
while(p!=q){scanf("%d",p);p+=d;}}void print(){cout<<endl;}template<typename T>
void print(const T t){cout<<t<<endl;}template<typename F,typename...R>
void print(const F f,const R...r){cout<<f<<", ";print(r...);}template<typename T>
void print(T*p, T*q){int d=p<q?:-;while(p!=q){cout<<*p<<", ";p+=d;}cout<<endl;}
//#endif
template<typename T>bool umax(T&a, const T&b){return b<=a?false:(a=b,true);}
template<typename T>bool umin(T&a, const T&b){return b>=a?false:(a=b,true);}

const double PI = acos(-1.0);
const int INF = 1e9 + ;
const double EPS = 1e-12;

/* -------------------------------------------------------------------------------- */

const int md = 1e9 + ;

int n, m, f[][], fac[];

int powermod(int a, int n, int md) {
    int ans = , tmp = a;
    while (n) {
        if (n & ) ans = (ll)ans * tmp % md;
        tmp = (ll)tmp * tmp % md;
        n >>= ;
    }
    return ans;
}

bool chk() {
    for (int i = ; i <= n; i ++) {
        int p = i;
        for (int j = m - ; j >= ; j --) {
            p = f[j][p];
        }
        if (p != i) return false;
    }
    return true;
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
#endif // ONLINE_JUDGE
    fac[] = ;
    for (int i = ; i <= ; i ++) fac[i] = (ll)fac[i - ] * i % md;
    int x;
    while (cin >> n >> m) {
        int cnt = ;
        bool ok = true;
        for (int i = ; i < m; i ++) {
            scanf("%d", &x);
            if (x == - ) cnt ++;
            else {
                bool vis[] = {};
                vis[x] = true;
                f[i][] = x;
                for (int j = ; j < n; j ++) {
                    scanf("%d", &x);
                    vis[x] = true;
                    f[i][j + ] = x;
                }
                for (int i = ; i <= n; i ++) {
                    if (!vis[i]) ok = false;
                }
            }
        }
        if (!ok) puts("0");
        else {
            if (cnt) printf("%d\n", powermod(fac[n], cnt - , md));
            else printf("%d\n", chk());
        }
    }
    return ;
}