HDU 5399(Too Simple-判定映射)

Too Simple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1035    Accepted Submission(s): 353

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 numbersn,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 31 2 3-13 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.
 

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

 

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已知函数有一个不是双射，必定无解

若所有函数均已知，直接判断

否则，只要有一个 -1, 那么必能构造出解

有k个-1，前k-1个随便排 ，最后那个就能推出 ans=(n!)^(k-1)

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (1000000007)#define MAXN (100+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n,m;bool b[MAXN];int f[MAXN][MAXN];int g[MAXN];int main(){//freopen("D.in","r",stdin);while(cin>>n>>m) {int cnt=0; bool flag=0;For(i,m){MEM(b)int p;scanf("%d",&p);if (p==-1) { ++cnt;continue;}else f[i][1]=p;b[p]=1;Fork(j,2,n) {scanf("%d",&p);f[i][j]=p;if (!b[p]) b[p]=1; else flag=1;}}if (flag) {puts("0");continue;}if (!cnt) {For(i,n) g[i]=i;ForD(i,m) For(j,n) g[j]=f[i][g[j]];bool flag=0;For(i,n) if (g[i]^i) flag=1;if (flag) puts("0");else puts("1");continue;}if (cnt==1){puts("1");continue;}ll ans=1,p2=1;For(i,n) p2=mul(p2,i);For(i,cnt-1) ans=mul(p2,ans);cout<<ans<<endl;}return 0;}