HDU5399-多校-模拟

Too Simple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1214    Accepted Submission(s): 406

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.

 

 

一开始看起来很复杂，想了想发现只要保证最后一个函数是任意的，中间的其他函数都没问题。

但是有一个坑的情况是，可能一个任意函数也没有，全都是固定的函数，这时要验证这组函数是否可行。我们当时验证的顺序弄反，卡了一场，真是too simple。

最近做的题都是模拟题啊，，，太没技术含量了，，，，

 

#include <cstdio>
#include <algorithm>
#include <cstring>
#include <ctype.h>
#include <cstdlib>
#include <stack>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <cmath>

using namespace std;
const long long MOD = 1e9+7;

int N,T,M;
int func[200][200],vis[200];
long long nn;

long long qpow(long long a,long long i,long long n)
{
    if(i == 0) return 1 % n;
    long long temp = qpow(a,i>>1,n);
        temp = temp * temp % n;
    if( i&1 ) temp = temp * a % n;
    return temp;
}

long long mi(long long a,int t)
{
    long long ans = 1;
    for(int i=0;i<t;i++) {ans *= a;ans %= MOD;}
    return ans;
}

long long solve()
{
    for(int i=1;i<=N;i++)
    {
        int ans = i;
        for(int j=M-1;j>=0;j--)
        {
            ans = func[j][ans-1];
        }
        if(ans != i) return 0LL;
    }
    return 1LL;
}

int main()
{
    while(~scanf("%d%d",&N,&M))
    {
        long long cnt = 0LL,ans = 0LL;
        nn = 1LL;
        int flag = 0;
        for(int i=1; i <= N;i++) {nn *= i; nn %= MOD;}

        for(int i=0;i<M;i++)
        {
            memset(vis,0,sizeof vis);
            if(scanf("%d",&func[i][0]) && (func[i][0] == -1))
            {
                cnt++;
            }
            else 
            {
                vis[func[i][0]]++;
                for(int j=1;j<N;j++) 
                {
                    scanf("%d",&func[i][j]);
                    if( vis[func[i][j]] ) flag = 1;
                    else vis[func[i][j]]++;
                }
            }
        }

        if(flag) ans = 0LL;
        else if(cnt > 0) { ans = mi(nn,cnt-1); ans %= MOD;}
        else
        {
            ans = solve();
        }

        printf("%I64d\n",ans%MOD);
    }
}