hdu 5399 Too Simple(思路题，函数)

最新推荐文章于 2021-02-17 21:48:11 发布

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最新推荐文章于 2021-02-17 21:48:11 发布

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分类专栏：思路文章标签： ACM HDU

本文链接：https://blog.youkuaiyun.com/acm_cxq/article/details/51712460

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本文探讨了一个关于寻找特定函数序列数量的问题，通过分析已知部分函数，并计算未知函数的可能组合，最终求解出所有符合条件的函数序列总数。

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Too Simple

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1567 Accepted Submission(s): 510

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.

思路：无论给了多少个-1，我们只要保证最后一个对应了最开始给的i即可，所以一共有n!^(m-1)种可能性，注意输入有坑

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
using namespace std;
#define LL long long
#define N 110
LL mod=1000000007;
LL ma[N][N],vis[N];
LL get(LL n,LL m)
{
    LL ans=1;
    for(LL i=2; i<=n; i++)
        ans=ans*i%mod;
    LL l=1;
    for(LL i=1; i<m; i++)
        l=ans*l%mod;
    return l;
}
int main()
{
    LL n,m;
    int flag;
    while(~scanf("%lld %lld",&n,&m))
    {
        LL num=0;
        flag=0;
        for(LL i=1; i<=m; i++)
        {
            scanf("%lld",&ma[i][1]);
            if(ma[i][1]==-1) num++;
            else
            {
                memset(vis,0,sizeof(vis));
                vis[ma[i][1]]=1;
                for(LL j=2; j<=n; j++)
                {
                    scanf("%lld",&ma[i][j]);
                    if(ma[i][j]<1||ma[i][j]>n||vis[ma[i][j]])
                        flag=1;
                    vis[ma[i][j]]=1;
                }
            }
        }
        if(flag){
            printf("0\n");
            continue;
        }
        if(num)
        {
            LL ans=get(n,num);
            printf("%lld\n",ans);
        }
        else
        {
            flag=0;
            for(LL i=1; i<=n; i++)
            {
                LL t=i;
                for(LL j=m; j>=1; j--)
                    t=ma[j][t];
                if(t!=i)
                {
                    flag=1;
                    break;
                }
            }
            if(flag) printf("0\n");
            else printf("1\n");
        }
    }
    return 0;
}