hdu5399 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.

思路：若不存在-1，则要判断是否满足函数的对应关系，如果满足输出1，不满足输出0. 若存在-1 ，个数为 ans 分析得结果为 （n！）^（ans-1）。

代码

#include<iostream>
#include<cstdio>
#define ll __int64
#define mod 1000000007
using namespace std;
ll ff[105];

int main()
{
ff[1]=1;
for(int i=2;i<=101;i++)
ff[i]=ff[i-1]*i%mod;// 计算n！
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
int f[105][105];
int ans=0;ll res=1;
for(int i=1;i<=m;i++)
{
scanf("%d",&f[i][1]);
if(f[i][1]==-1) ans++;
else{
for(int j=2;j<=n;j++)
{
scanf("%d",&f[i][j]);
for (int k = j - 1; k; k--) if (f[i][j] == f[i][k]) res = 0;
//判断有没有函数对应关系是否为一一对应，不是则结果为0
}
}
}
// 没有-1时，判断i(1≤i≤n),f1(f2(⋯fm(i)))=i是否成立
if(ans==0)
{
int flag=1;
for(int j=1;j<=n;j++)
{
int a=f[m][j];
for(int i=m-1;i>=1;i--)
a=f[i][a];
if(a!=j)flag=0;
if(!flag) break;
}
if(!flag) {printf("0\n");continue;}
if(flag) {printf("1\n") ;continue;}
}
else{
for(int i=1;i<=ans-1;i++)
res=(res*ff
)%mod;
}
printf("%I64d\n",res);
}
return 0;
}