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 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.

解答：

这道题目比赛那会已经想出来规律，队友去敲的代码，只是注意几点小地方就行。判断有没有一行中有重复的数字。如果不存在-1，判断是不是符合那个条件。

#include <iostream>#include <cstdio>#include <cmath>#include <algorithm>#define MOD 1000000007using namespace std;long long int f[105];long long int a[105][105];int main(){ int n,m; f[0]=1; for(int i=1;i<=100;i++) f[i]=(f[i-1]*i)%MOD; while(scanf("%d%d",&n,&m)!=EOF) { long long ans=1,tot=0; for(int i=0;i<m;i++) { scanf("%lld",&a[i][1]); if(a[i][1]==-1) tot++; else for(int j=2;j<=n;j++) { scanf("%lld",&a[i][j]); for(int k=j-1;k>=1;k--) if(a[i][k]==a[i][j]) ans=0; } } for(int i=1;i<tot;i++) ans=(ans*f[n])%MOD; if(!tot) { int x[105]; for(int i=1;i<=n;i++) x[i]=i; for(int i=m-1;i>=0;i--) for(int j=1;j<=n;j++) { x[j]=a[i][x[j]]; } for(int i=1;i<=n;i++) if(x[i]!=i)ans=0; //不是自身映射，无解 } printf("%lld\n",ans); }}</algorithm></cmath></cstdio></iostream>