[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}，f(f(...(f(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!，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 ;

}