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Kirchhoff.cpp
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151 lines (135 loc) · 3 KB
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#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
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 (998244353)
#define eps (1e-3)
#define MAXN (16+10)
#define MAXM (16*16+10)
typedef __int64 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;}
struct M
{
int n,m;
ll a[MAXN][MAXN];
M(int _n=0){n=m=_n;MEM(a);}
M(int _n,int _m){n=_n,m=_m;MEM(a);}
void mem (int _n=0){n=m=_n;MEM(a);}
void mem (int _n,int _m){n=_n,m=_m;MEM(a);}
friend M operator*(M a,M b)
{
M c;
For(k,a.m)
For(i,a.n)
For(j,b.m)
c.a[i][j]=(c.a[i][j]+a.a[i][k]*b.a[k][j])%F;
return c;
}
void make_I(int _n)
{
n=m=_n; MEM(a)
For(i,n) a[i][i]=1;
}
// 求行列式
long double mat[MAXN][MAXN],tmp[MAXN];
long double det()
{
For(i,n) For(j,m) mat[i][j]=a[i][j];
For(i,n)
{
int pos=i;
while (fabs(mat[pos][i])<eps&&pos<n) ++pos;
if (fabs(mat[pos][i])<eps) continue;
if (pos^i)
{
copy(mat[pos]+1,mat[pos]+1+m+1,tmp+1);
copy(mat[i]+1,mat[i]+1+m+1,mat[pos]+1);
copy(tmp+1,tmp+1+m+1,mat[i]+1);
}
For(j,n)
if (i^j)
{
long double p = mat[j][i]/mat[i][i];
For(k,m) mat[j][k]-=mat[i][k]*p;
}
}
long double ans=1;
For(i,n) ans*=mat[i][i];
return ans;
}
}A,C,D;
M pow2(M a,ll b)
{
M c;c.make_I(a.n);
static bool a2[1000000];
int n=0;while (b) a2[++n]=b&1,b>>=1;
For(i,n)
{
if (a2[i]) c=c*a;
a=a*a;
}
return c;
}
const ll p2[]={1,2,4,8,16,32,64,128,256,512,1024,2048,4096,8192,16384,32768,65536};
M a;
int n,m,t[MAXN];
void work()
{
ll ans=0,cnt;
//t[i] 表示t缩点的标号
//将C[G]的第a行,第b列同时去掉后得到的新矩阵 a,b为任意(1≤a,b≤n)
// 处理t 最大值为n-cnt+1
a.mem(n-cnt);
For(j,n)
For(l,n)
if (t[j]!=t[l]&&A.a[j][l])
{
a.a[t[j]][t[j]]++;
a.a[t[j]][t[l]]--;
}
ll t2=(ll)(fabs(a.det())+eps)%F;
cout<<ans<<endl;
}
int u[MAXN],v[MAXN];
void Kirchhoff()
{
while (cin>>n>>m) {
A.mem(n),D.mem(n),C.mem(n);
For(i,m)
{
scanf("%d%d",&u[i],&v[i]);
D.a[u[i]][u[i]]++;
D.a[v[i]][v[i]]++;
A.a[u[i]][v[i]]++;
A.a[v[i]][u[i]]++;
}
work();
}
}
int main()
{
// freopen(".in","r",stdin);
return 0;
}