HDU 5399 Too Simple(过程中略微用了一下dfs)——多校练习9

Posted 2020-10-02 yxysuanfa

Too Simple

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

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.

 

/*********************************************************************/

题意：给你m个函数f1,f2,?,fm:{1,2,?,n}→{1,2,?,n}(即全部的x∈{1,2,?,n},相应的f(x)∈{1,2,?,n})。已知当中一部分函数的函数值，问你有多少种不同的组合使得全部的i(1≤i≤n),满足f1(f2(?fm(i)))=i

对于函数集f1,f2,?,fm and g1,g2,?,gm。当且仅当存在一个i(1≤i≤m),j(1≤j≤n),fi(j)≠gi(j)，这种组合才视为不同。

假设还是不理解的话，我们来解释一下例子，

3 3

1 2 3

-1

3 2 1

例子写成函数的形式就是

n=3,m=3

f1(1)=1,f1(2)=2,f1(3)=3

f2(1)、f2(2)、f2(3)的值均未知

f3(1)=3,f3(2)=2,f3(3)=1

所以要使全部的i(1≤i≤n),满足f1(f2(?fm(i)))=i。仅仅有一种组合情况，即f2(1)=3,f2(2)=2,f2(3)=1这么一种情况

解题思路：事实上。细致想想。你就会发现，此题的解跟-1的个数有关，当仅仅有一个-1的时候，由于相应关系都已经决定了。所以仅仅有1种可行解，而当你有两个-1时，当中一个函数的值能够依据还有一个函数的改变而确定下来，故有n!种解。

依此类推，当有k个-1时，解为

所以，我们仅仅须要提前将n！

计算好取模存下来。剩下的就是套公式了

有一点须要提醒的是，当-1的个数为0时。即不存在-1的情况，解并不一定为0，若不存在-1的情况下仍满足对于全部的i(1≤i≤n),满足f1(f2(?fm(i)))=i，输出1。否则输出0。所以加个dfs推断一下方案是否可行。多校的时候就被这一点坑了。看来还是考虑得不够多。

若对上述有什么不理解的地方，欢迎提出。

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<stdio.h>
#include<string.h>
#include<stdlib.h>
#include<queue>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<stdlib.h>
#include<cmath>
#include<string>
#include<algorithm>
#include<iostream>
#define exp 1e-10
using namespace std;
const int N = 105;
const int inf = 1000000000;
const int mod = 1000000007;
__int64 s[N];
bool v[N];
int w[N][N];
int dfs(int t,int x)
{
    if(t==1)
        return w[t][x];
    return dfs(t-1,w[t][x]);
}
int main()
{
    int n,m,i,j,k,x;
    bool flag;
    __int64 ans;
    s[0]=1;
    for(i=1;i<N;i++)
        s[i]=(s[i-1]*i)%mod;
    while(~scanf("%d%d",&n,&m))
    {
        flag=true;
        for(k=0,i=1;i<=m;i++)
        {
            scanf("%d",&x);
            if(x==-1)
                k++;
            else
            {
                memset(v,false,sizeof(v));
                v[x]=true;w[i][1]=x;
                for(j=2;j<=n;j++)
                {
                    scanf("%d",&x);
                    v[x]=true;
                    w[i][j]=x;
                }
                for(j=1;j<=n;j++)
                    if(!v[j])
                        break;
                if(j<=n)
                    flag=false;
            }
        }
        /*for(i=1;i<=m;i++)
        {
            for(j=1;j<=n;j++)
                printf("%d ",w[i][j]);
            printf("\n");
        }*/
        if(!flag)
            puts("0");
        else if(!k)
        {
            for(i=1;i<=n;i++)
                if(dfs(m,i)!=i)
                    break;
            //printf("%d*\n",i);
            if(i>n)
                puts("1");
            else
                puts("0");
        }
        else
        {
            for(ans=1,i=1;i<k;i++)
                ans=ans*s[n]%mod;
            printf("%I64d\n",ans);
        }
    }
    return 0;
}

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