HDU 5399(Too Simple-判定映射)

Too Simple

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

Total Submission(s): 1035 Accepted Submission(s): 353



[align=left]Problem Description[/align]
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).

[align=left]Input[/align]
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).

[align=left]Output[/align]
For each test case print the answer modulo
109+7.

[align=left]Sample Input[/align]

3 3
1 2 3
-1
3 2 1

[align=left]Sample Output[/align]

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

[align=left]Author[/align]
xudyh

[align=left]Source[/align]
2015 Multi-University Training Contest 9

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已知函数有一个不是双射，必定无解

若所有函数均已知，直接判断

否则，只要有一个 -1, 那么必能构造出解

有k个-1，前k-1个随便排 ，最后那个就能推出 ans=(n!)^(k-1)

#include<bits/stdc++.h>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (1000000007)
#define MAXN (100+10)

typedef long long ll;
ll mul(ll a,ll b){return (a*b)%F;}
ll add(ll a,ll b){return (a+b)%F;}
ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}
void upd(ll &a,ll b){a=(a%F+b%F)%F;}
int n,m;
bool b[MAXN];
int f[MAXN][MAXN];
int g[MAXN];
int main()
{
//	freopen("D.in","r",stdin);

while(cin>>n>>m) {
int cnt=0; bool flag=0;
For(i,m)
{
MEM(b)
int p;
scanf("%d",&p);
if (p==-1) { ++cnt;continue;}
else f[i][1]=p;
b[p]=1;
Fork(j,2,n) {
scanf("%d",&p);f[i][j]=p;
if (!b[p]) b[p]=1; else flag=1;
}
}

if (flag) {
puts("0");continue;
}

if (!cnt) {
For(i,n) g[i]=i;
ForD(i,m)
For(j,n) g[j]=f[i][g[j]];
bool flag=0;
For(i,n) if (g[i]^i) flag=1;
if (flag) puts("0");else puts("1");
continue;
}

if (cnt==1)
{
puts("1");
continue;
}

ll ans=1,p2=1;
For(i,n) p2=mul(p2,i);
For(i,cnt-1) ans=mul(p2,ans);
cout<<ans<<endl;

}

return 0;
}