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 1000000007using 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[n])%mod;        }        printf("%I64d\n",res);    }    return 0;}