题目大意
【60%】n<=50,m<=200
【100%】
代码
#include<cstdio>
#include<cstring>
#include<algorithm>
#define fo(i,a,b) for(int i=a;i<=b;i++)
using namespace std;
typedef long long LL;
const int maxn=305, maxm=1e5+5;
const LL mo=1e9+7;
int n,m,u[maxm],v[maxm],w[maxm];
LL mi(LL x,LL y)
{
LL re=1;
for(; y; y>>=1, x=x*x%mo) if (y&1) re=re*x%mo;
return re;
}
LL D,J[maxn][maxn];
void Det()
{
D=1;
fo(i,1,n-1)
{
fo(j,i,n-1) if (J[j][i]!=0)
{
swap(J[i],J[j]);
if (i!=j) D*=-1;
break;
}
fo(j,i+1,n)
{
LL c=J[j][i]*mi(J[i][i],mo-2)%mo;
fo(k,i,n) (J[j][k]-=c*J[i][k])%=mo;
}
}
fo(i,1,n-1) (D*=J[i][i])%=mo;
D=(D+mo)%mo;
}
LL G[maxn][maxn],Gc[maxn][maxn],c[maxn];
void Gauss()
{
fo(i,1,n)
{
fo(j,i,n) if (G[j][i]!=0)
{
swap(G[i],G[j]), swap(Gc[i],Gc[j]);
break;
}
LL c=mi(G[i][i],mo-2);
fo(j,1,n) (G[i][j]*=c)%=mo, (Gc[i][j]*=c)%=mo;
fo(j,1,n) if (j!=i)
{
LL c=G[j][i];
fo(k,1,n) (G[j][k]-=c*G[i][k])%=mo, (Gc[j][k]-=c*Gc[i][k])%=mo;
}
}
}
void Pre()
{
fo(i,1,n)
{
Gc[i][i]=1;
fo(j,1,n) G[i][j]=J[j][i];
}
Gauss();
}
int main()
{
scanf("%d %d",&n,&m);
fo(i,1,m)
{
scanf("%d %d %d",&u[i],&v[i],&w[i]);
J[u[i]][u[i]]++;
J[u[i]][v[i]]--;
}
Pre();
Det();
LL ans=0;
fo(i,1,m) if (u[i]<n)
{
LL x=(Gc[u[i]][u[i]]-Gc[u[i]][v[i]])%mo;
(ans+=D*x%mo*w[i])%=mo;
}
printf("%lld\n",(ans+mo)%mo);
}