每一个函数都必须是一个排列,经过连续的一段确定函数后数字不能少.

满足上面的条件的话,只要有一个-1函数特别的排列一下就可以满足要求,剩下的可以随意填

没有-1的话特判

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.

Author

xudyh

Source

2015 Multi-University Training Contest 9

/* ***********************************************
Author        :CKboss
Created Time  :2015年08月19日 星期三 10时25分44秒
File Name     :HDOJ5399.cpp
************************************************ */

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <string>
#include <cmath>
#include <cstdlib>
#include <vector>
#include <queue>
#include <set>
#include <map>

using namespace std;

typedef long long int LL;
const LL mod=1e9+7LL;
const int INF=50000000;

int n,m;
int f[110][110];

int cn;
int color[110];

bool COLOR(int x,int c)
{
	if(color[x]==c) return false;
	color[x]=c; cn++;
	return true;
}

LL QuickPow(LL a,LL n)
{
	LL e=1;
	while(n)
	{
		if(n&1) e=(a*e)%mod;
		a=(a*a)%mod;
		n/=2;
	}
	return e%mod;
}

bool used[110];

LL jc(LL n)
{
	LL ret=1;
	for(int i=2;i<=n;i++)
		ret=(ret*i)%mod;
	return ret%mod;
}

bool check_P(int x,int L=1,int R=m)
{
	if(used[x]==true) return false;
	int nx=x;
	for(int i=R;i>=L;i--)
	{
		nx=f[i][nx];
	}
	if(nx==x)
	{
		if(used[x]==false)
		{
			used[x]=true;
			return true;
		}
		else return false;
	}
	return false;
}

bool check_Range(int L,int R)
{
	memset(used,false,sizeof(used));
	bool flag=true;
	for(int i=1;i<=n&&flag;i++)
	{
		int nx=i;
		for(int j=R;j>=L;j--)
		{
			nx=f[j][nx];
		}
		if(used[nx]==true) return false;
		else used[nx]=true;
	}
	return true;
}

int main()
{
	//freopen("in.txt","r",stdin);
	//freopen("out.txt","w",stdout);

	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int nig=0;
		bool check=true;
		memset(color,0,sizeof(color));

		for(int i=1;i<=m;i++)
		{
			scanf("%d",&f[i][1]);
			if(f[i][1]==-1) nig++;
			else
			{
				cn=0;
				check=COLOR(f[i][1],i);
				for(int j=2;j<=n;j++)
				{
					scanf("%d",&f[i][j]);
					check=COLOR(f[i][j],i);
				}
				if(cn!=n) check=false;
			}
		}

		if(check==false)
		{
			puts("0");
		}
		else if(check&&nig==0)
		{
			/// tePan
			memset(used,false,sizeof(used));
			bool flag=true;
			for(int i=1;i<=n&&flag;i++)
			{
				if(check_P(i)==false) flag=false;
			}
			if(flag==true) puts("1");
			else puts("0");
		}
		else
		{
			//// (n!^(nig-1))
			bool flag=true;
			int Left=INF,Right=-INF;
			f[m+1][1]=-1;
			for(int i=1;i<=m+1&&flag;i++)
			{
				if(f[i][1]==-1)
				{
					if(Left!=INF&&Right!=-INF)
					{
						/// check;
						flag=check_Range(Left,Right);
					}
					Left=INF; Right=-INF;
				}
				else
				{
					Left=min(Left,i); Right=max(Right,i);
				}
			}
			if(flag==false) puts("0");
			else printf("%lld\n",QuickPow(jc(n),nig-1)%mod);
		}
	}

    return 0;
}

版权声明：来自: 码代码的猿猿的AC之路 http://blog.csdn.net/ck_boss