Problem H
Organising the Organisation

Input: Standard Input

Output: Standard Output

Time Limit: 2 Second

I am the chief of the Personnel Division of a moderate-sized company that wishes to remain anonymous, and I am currently facing a small problem for which I need a skilled programmer's help.

One of the divisions is Central Management (division A in the figure above), and should of course be at the top of the hierarchy, but the relative importance of the remaining divisions is not determined, so in Figure 1 above, division C and D could equally well have switched places so that C was in charge over division D. One complication, however, is that it may be impossible to get some divisions to cooperate with each other, and in such a case, neither of these divisions can be directly in charge of the other. For instance, if in the example above A and D are unable to cooperate, Figure 1 is not a valid way to organise the company.

In general, there can of course be many different ways to organise the organisation, and thus it is desirable to find the best one (for instance, it is not a good idea to let the programming people be in charge of the marketing people). This job, however, is way too complicated for you, and your job is simply to help us find out how much to pay the consultant that we hire to find the best organisation for us. In order to determine the consultant's pay, we need to find out exactly how difficult the task is, which is why you have to count exactly how many

Oh, and I need the answer in five hours.

Input

The input consists of a series of test cases, at most 50, terminated by end-of-file. Each test cases begins with three integers n, m, k (1 ≤ n ≤ 50, 0 ≤ m ≤ 1500, 1 ≤ k ≤ n). n denotes the number of divisions in the company (for convenience, the divisions are numbered from 1

Output

For each test case, print the number of possible ways to organise the company on a line by itself. This number will be at least 1 and at most 10¹⁵.

Sample Input                               Output for Sample Input

5 5 2 

3 1 

3 4 

4 5 

1 4 

5 3 

4 1 1 

1 4 

3 0 2 

3 

8 

建圖簡單 但是這題不取模 可能會得到的值比較大 要使用long double 

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <string.h>
#include <string>
#include <vector>
#include <queue>

#define MEM(a,x) memset(a,x,sizeof a)
#define eps 1e-10
#define MOD 10009
#define MAXN 110
#define MAXM 100010
#define INF 99999999
#define ll __int64
#define bug cout<<"here"<<endl
#define fread freopen("ceshi.txt","r",stdin)
#define fwrite freopen("out.txt","w",stdout)

using namespace std;

int Read()
{
    char ch;
    int a = 0;
    while((ch = getchar()) == ' ' | ch == '\n');
    a += ch - '0';
    while((ch = getchar()) != ' ' && ch != '\n')
    {
        a *= 10;
        a += ch - '0';
    }
    return a;
}

void Print(int a)
{
     if(a>9)
         Print(a/10);
     putchar(a%10+'0');
}

int sgn(long double x)
{
    if(fabs(x)<eps) return 0;
    if(x<0) return -1;
    else return 1;
}
long double b[MAXN][MAXN];
long double det(long double a[][MAXN],int n)
{
    int i,j,k,sign=0;
    long double ret=1;
    for(i=0;i<n;i++)
        for(j=0;j<n;j++)
            b[i][j]=a[i][j];
    for(i=0;i<n;i++)
    {
        if(sgn(b[i][i])==0)
        {
            for(j=i+1;j<n;j++)
                if(sgn(b[j][i])!=0)
                    break;
            if(j==n) return 0;
            for(k=i;k<n;k++)
                swap(b[i][k],b[j][k]);
            sign++;
        }
        ret*=b[i][i];
        for(k=i+1;k<n;k++)
            b[i][k]/=b[i][i];
        for(j=i+1;j<n;j++)
            for(k=i+1;k<n;k++)
                b[j][k]-=b[j][i]*b[i][k];
    }
    if(sign&1) ret=-ret;
    return ret;
}
long double a[MAXN][MAXN];
int g[MAXN][MAXN];

int main()
{
//    fread;
    int n,m,k;
    while(scanf("%d%d%d",&n,&m,&k)!=EOF)
    {
        MEM(g,1);
        while(m--)
        {
            int u,v;
            scanf("%d%d",&u,&v);
            u--; v--;
            g[u][v]=g[v][u]=0;
        }
        MEM(a,0);
        for(int i=0;i<n;i++)
            for(int j=0;j<n;j++)
                if(i!=j&&g[i][j])
        {
            a[i][i]++;
            a[i][j]=-1;
        }
        long double ans=det(a,n-1);
        printf("%.0Lf\n",ans);
    }
    return 0;
}