无向图生成树计数 基尔霍夫矩阵 SPOJ Highways

基尔霍夫矩阵

https://blog.csdn.net/w4149/article/details/77387045

https://blog.csdn.net/qq_29963431/article/details/51236064

题目链接  https://vjudge.net/problem/SPOJ-HIGH

AC代码

 #include <bits/stdc++.h>
 #define pb push_back
 #define mp make_pair
 #define fi first
 #define se second
 #define all(a) (a).begin(), (a).end()
 #define fillchar(a, x) memset(a, x, sizeof(a))
 #define huan printf("\n");
 #define debug(a,b) cout<<a<<" "<<b<<" ";
 using namespace std;
 typedef long long ll;
 const int maxn=1e3+,maxm=,inf=0x3f3f3f3f;
 const ll mod=;
 ll a[maxn][maxn],du[maxn];
 ll det(int n)
 {
     ll ans = ;
     for (int i = ; i <= n; i++)
     {
         for (int j = i + ; j <= n; j++)
         {
             while (a[j][i] != )
             {
                 ll u = a[i][i] / a[j][i];
                 for (int k = i; k <= n; k++)
                 {
                     ll t = (a[i][k] - (ll)a[j][k] * u);// % mod + mod) % mod;
                     a[i][k] = a[j][k];
                     a[j][k] = t;
                 }
                 ans = -ans;
             }
         }
         ans = ans * a[i][i] ;//% mod;
     }
     if (ans < )
     {
         ans=-ans;
         //ans += mod;
     }
     return ans;
 }
 int main()
 {
     int t,n,m;
     scanf("%d",&t);
     while(t--)
     {
         fillchar(a,);
         fillchar(du,);
         scanf("%d%d",&n,&m);
         for(int i=;i<m;i++)
         {
             int u,v;
             scanf("%d%d",&u,&v);
             du[u]++,du[v]++;
             a[v][u]=-,a[u][v]=-;
         }
         for(int i=;i<=n;i++)
             a[i][i]=du[i];
         printf("%lld\n",det(n));
     }
 }

 行列式求值模板

int a[maxn][maxn];
ll det(int n)
{
    ll ans = ;
    for (int i = ; i <= n; i++)
    {
        for (int j = i + ; j <= n; j++)
        {
            while (a[j][i] != )
            {
                ll u = a[i][i] / a[j][i];
                for (int k = i; k <= n; k++)
                {
                    int t = (a[i][k] - (ll)a[j][k] * u % mod + mod) % mod;
                    a[i][k] = a[j][k];
                    a[j][k] = t;
                }
                ans = -ans;
            }
        }
        ans = ans * a[i][i] % mod;
    }
    if (ans < )
    {
        ans += mod;
    }
    return ans;
}