hdu 5399 Too Simple 2015多校联合训练赛#9 模拟

Too Simple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 797    Accepted Submission(s): 268

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. 
    
    
    
    
     
   
   
   
   

 

Author

xudyh

 

Source

2015 Multi-University Training Contest 9

如果有-1，且函数合法，那么就有（N！）^(x-1)，x是-1的个数

否则只需要判断函数映射是否会x= f(f(f....f(x))即可

函数合法的条件是：y = f(x) y包含1到n的每个数字

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
using namespace std;

int f[101][101];
#define ll long long
int jie[101];
int mod = 1000000007;
int ans[101];
int main(){
    int n,m;
    jie[0]  = 1;
    for(int i =1; i <= 100;i++)
        jie[i] =1ll*jie[i-1]*i%mod;
    while(scanf("%d%d",&n,&m)!=EOF){
        int res = 0;
        for(int i = 0;i < m; i++){
            scanf("%d",&f[i][1]);
            if(f[i][1] == -1) {
                res++;
                continue;
            }
            for(int j = 2;j <= n; j++)
                scanf("%d",&f[i][j]);
        }
        if(res == 0){
            for(int i = 1;i <= n; i++)
                ans[i] = i;
            for(int i = m-1; i >= 0; i--)
                for(int j = 1;j <= n; j++)
                    ans[j] = f[i][ans[j]];
            res = 1;
            for(int i = 1;i <= n; i++)
                if(ans[i] != i ) res = 0;
            if(res)puts("1");
            else puts("0");
        }
        else {
            int flag = 1;
            for(int i = 0;i < m && flag; i++){
                if(f[i][1] == -1) continue;
                memset(ans,0,sizeof(ans));
                for(int j = 1;j <= n;j++)
                    ans[f[i][j]] = 1;
                for(int j = 1;j <= n;j++)
                    if(ans[j] == 0)
                        flag = 0;
            }
            if(flag == 0) puts("0");
            else {
                ll t = 1;
                for(int i = 1;i < res; i++){
                    t = t*jie[n]%mod;
                }
                cout<<t<<endl;
            }
        }
    }
    return 0;
}