HDU 5399 Too Simple

最新推荐文章于 2018-12-30 11:22:22 发布

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本文详细探讨了在已知部分函数定义的情况下，如何通过逻辑推理确定剩余未知函数，以确保整个函数序列满足特定条件。通过实例分析，展示了计算不同函数序列的数量的方法，以及在面对多个未知函数时的策略。

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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

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.
 

这题其实很简单，对于多于两个-1的情况不管之前那个怎么选，只要最后一个固定就可以保证，所以答案就是(n!)^(m-1)   m是-1的个数
不过此题坑比较多，要注意判断
#pragma comment(linker, "/STACK:1024000000,1024000000") 
#include<set>
#include<map>
#include<queue>
#include<stack>
#include<cmath>
#include<cstdio>
#include<vector>
#include<string>
#include<cstring>
#include<algorithm>
#include<functional>
using namespace std;
typedef long long LL;
const LL base = 1e9 + 7;
const int maxn = 105;
LL T, n, m, f[maxn], a[maxn][maxn];

inline void read(int &x)
{
    char ch;
    while ((ch = getchar())<'0' || ch>'9');
    x = ch - '0';
    while ((ch = getchar()) >= '0' && ch <= '9') x = x * 10 + ch - '0';
}

int main()
{
    //read(T);
    for (int i = f[0] = 1; i <= 100; i++) f[i] = f[i - 1] * i % base;
    while (scanf("%I64d%I64d", &n, &m) != EOF)
    {
        LL tot = 0, ans = 1;
        for (int i = 1; i <= m; i++)
        {
            scanf("%I64d", &a[i][1]);
            if (a[i][1] == -1) tot++;
            else for (int j = 2; j <= n; j++)
            {
                scanf("%I64d", &a[i][j]);
                for (int k = j - 1; k; k--) if (a[i][j] == a[i][k]) ans = 0;
            }
        }
        for (int i = 1; i < tot; i++) ans = ans * f[n] % base;
        if (tot == 0)
        {
            for (int i = 1; i <= n; i++) a[0][i] = i;
            for (int i = m; i; i--)
                for (int j = 1; j <= n; j++) a[0][j] = a[i][a[0][j]];
            for (int i = 1; i <= n; i++) if (a[0][i] != i) ans = 0;
        }
        printf("%I64d\n", ans);
    }
    return 0;
}