HUD-5399 Too Simple（数学）

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 10^9+7.

Sample Input

3 3
1 2 3
-1
3 2 1

Sample Output

1
HintThe order in the function series is determined. What she can do is to assign the values to the unknown functions.

思路与官方的一样：

首先要求每个f_if​i​​是个排列，否则如果某个f_if​i​​将两个数映射向同一个数，那么最后这两个数得到的值一定相同。

如果还剩一个位置为-1−1，那么这个排列是唯一确定的，假设X*f_i*Y=IX∗f​i​​∗Y=I，那么f_i=X^{-1}*Y^{-1}f​i​​=X​−1​​∗Y​−1​​.

所以假设有c(c\geq
1)c(c≥1)个-1−1，那么答案为(n!)^{c-1}(n!)​c−1​​个可行的方案。

注意特判所有函数都已知的情况。

#include <cstdio>
#include <cstring>

using namespace std;

const long long MOD=1000000007;

long long f[101];
int a[101][101];
bool vis[101];

void Init() {
f[0]=f[1]=1;
for(long long i=2;i<=100;++i)
f[i]=(f[i-1]*i)%MOD;
}

long long quickpow(long long m,long long n) {
long long b=1;
while(n>0){
if (n&1)
b=(b*m)%MOD;
n=n>>1;
m=(m*m)%MOD;
}
return b;
}

int n,m;

bool Judge() {//特判所有函数已知的情形
int i,j,x;
for(i=1;i<=n;++i) {
x=i;
for(j=m-1;j>=0;--j)
x=a[j][x];
if(x!=i)
return false;
}
return true;
}

int main() {
int i,j,cnt;
bool ok;
Init();
while(2==scanf("%d%d",&n,&m)) {
cnt=0,ok=true;
for(i=0;i<m;++i) {
scanf("%d",&a[i][1]);
if(a[i][1]==-1)
++cnt;
else {
memset(vis,false,sizeof(vis));
vis[a[i][1]]=true;
for(j=2;j<=n;++j) {
scanf("%d",&a[i][j]);
if(vis[a[i][j]])//刚开始以为是单射函数，结果WA了好久。如果存在非单射函数，则结果为0
ok=false;
else
vis[a[i][j]]=true;
}
}
}
if(cnt==0)
printf("%d\n",Judge()?1:0);
else
printf("%I64d\n",ok?quickpow(f
,cnt-1):0);
}
return 0;
}