sro_ptx_orz
qwq算是一个套路的记录
对于一个有向图来说
如果你要求一个外向生成树的话，那么如果存在一个(u
ightarrow v)的边
那么(a[u][v]--,a[v][v]++)
对应的去掉第(i)行和第(i)列的余子式，就是以(i)为根的生成树个数。
内向生成树也是同理。所有的反过来即可
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<cmath>
#include<queue>
#include<map>
#include<set>
#define mk make_pair
#define ll long long
//#define int long long
#define rint register int
using namespace std;
inline int read()
{
int x=0,f=1;char ch=getchar();
while (!isdigit(ch)) {if (ch=='-') f=-1;ch=getchar();}
while (isdigit(ch)) {x=(x<<1)+(x<<3)+ch-'0';ch=getchar();}
return x*f;
}
const int maxn = 310;
const int mod = 10007;
int qsm(int i,int j)
{
int ans=1;
while (j)
{
if (j&1) ans=ans*i%mod;
i=i*i%mod;
j>>=1;
}
return ans;
}
int a[maxn][maxn];
int n,m;
int d[maxn][maxn];
int b[maxn][maxn];
void gauss()
{
int k=1;
int ff=1;
int ans=1;
for (rint i=1;i<=n;++i)
{
int now = k;
while (now<=n && !(a[now][i])) ++now;
if (now==n+1) continue;
if (now!=k) ff*=-1;
for (rint j=1;j<=n;++j) swap(a[now][j],a[k][j]);
int inv = qsm(a[k][i],mod-2);
for (rint j=i+1;j<=n;++j)
{
int t = a[j][i]*inv%mod;
for (rint p=1;p<=n;++p) a[j][p]=(a[j][p]-a[k][p]*t%mod+mod)%mod;
}
++k;
}
for (rint i=1;i<=n;++i) ans=ans*a[i][i]%mod;
if(ff==-1) cout<<mod-ans;
else cout<<ans;
}
signed main()
{
n=read(),m=read();
for (rint i=1;i<=m;++i)
{
int x=read(),y=read();
a[x][x]++;
a[y][x]--;
}
for (rint i=1;i<=n;++i)
for (rint j=1;j<=n;++j) b[i][j]=a[i][j];
for (rint i=1;i<n;++i)
for (rint j=1;j<n;++j)
{
a[i][j]=b[i+1][j+1];
}
n--;
gauss();
return 0;
}