无向图生成树计数 基尔霍夫矩阵 SPOJ Highways

Posted 2020-12-31 stranger-

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

 1 #include <bits/stdc++.h>
 2 #define pb push_back
 3 #define mp make_pair
 4 #define fi first
 5 #define se second
 6 #define all(a) (a).begin(), (a).end()
 7 #define fillchar(a, x) memset(a, x, sizeof(a))
 8 #define huan printf("
");
 9 #define debug(a,b) cout<<a<<" "<<b<<" ";
10 using namespace std;
11 typedef long long ll;
12 const int maxn=1e3+100,maxm=100,inf=0x3f3f3f3f;
13 const ll mod=998244353;
14 ll a[maxn][maxn],du[maxn];
15 ll det(int n)
16 {
17     ll ans = 1;
18     for (int i = 2; i <= n; i++)
19     {
20         for (int j = i + 1; j <= n; j++)
21         {
22             while (a[j][i] != 0)
23             {
24                 ll u = a[i][i] / a[j][i];
25                 for (int k = i; k <= n; k++)
26                 {
27                     ll t = (a[i][k] - (ll)a[j][k] * u);// % mod + mod) % mod;
28                     a[i][k] = a[j][k];
29                     a[j][k] = t;
30                 }
31                 ans = -ans;
32             }
33         }
34         ans = ans * a[i][i] ;//% mod;
35     }
36     if (ans < 0)
37     {
38         ans=-ans;
39         //ans += mod;
40     }
41     return ans;
42 }
43 int main()
44 {
45     int t,n,m;
46     scanf("%d",&t);
47     while(t--)
48     {
49         fillchar(a,0);
50         fillchar(du,0);
51         scanf("%d%d",&n,&m);
52         for(int i=0;i<m;i++)
53         {
54             int u,v;
55             scanf("%d%d",&u,&v);
56             du[u]++,du[v]++;
57             a[v][u]=-1,a[u][v]=-1;
58         }
59         for(int i=1;i<=n;i++)
60             a[i][i]=du[i];
61         printf("%lld
",det(n));
62     }
63 }

 行列式求值模板

int a[maxn][maxn];
ll det(int n)
{
    ll ans = 1;
    for (int i = 1; i <= n; i++)
    {
        for (int j = i + 1; j <= n; j++)
        {
            while (a[j][i] != 0)
            {
                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 < 0)
    {
        ans += mod;
    }
    return ans;
}