• BZOJ5297 CQOI2018社交网络(矩阵树定理)


      板子题。

    #include<iostream> 
    #include<cstdio>
    #include<cmath>
    #include<cstdlib>
    #include<cstring>
    #include<algorithm>
    using namespace std;
    #define ll long long
    #define N 256
    #define P 10007
    char getc(){char c=getchar();while ((c<'A'||c>'Z')&&(c<'a'||c>'z')&&(c<'0'||c>'9')) c=getchar();return c;}
    int gcd(int n,int m){return m==0?n:gcd(m,n%m);}
    int read()
    {
        int x=0,f=1;char c=getchar();
        while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();}
        while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar();
        return x*f;
    }
    int n,m,degree[N],inv[P];
    struct matrix
    {
        int a[N][N];
        int value()
        {
            int s=1;
            for (int i=2;i<=n;i++)
                for (int j=2;j<=n;j++)
                a[i][j]=(a[i][j]%P+P)%P;
            for (int i=2;i<=n;i++)
            {
                if (!a[i][i])
                for (int j=i+1;j<=n;j++)
                if (a[j][i]) {swap(a[i],a[j]);s=-s;break;}
                for (int j=i+1;j<=n;j++)
                {
                    int t=1ll*a[j][i]*inv[a[i][i]]%P;
                    for (int k=i;k<=n;k++)
                    a[j][k]=((a[j][k]-a[i][k]*t)%P+P)%P;
                }
            }
            for (int i=2;i<=n;i++) s=1ll*s*a[i][i]%P;
            return (s+P)%P;
        }
    }a;
    int main()
    {
    #ifndef ONLINE_JUDGE
        freopen("bzoj5297.in","r",stdin);
        freopen("bzoj5297.out","w",stdout);
        const char LL[]="%I64d
    ";
    #else
        const char LL[]="%lld
    ";
    #endif    
        n=read(),m=read();
        for (int i=1;i<=m;i++)
        {
            int x=read(),y=read();
            a.a[y][x]--;degree[x]++;
        }
        for (int i=1;i<=n;i++) a.a[i][i]+=degree[i];
        inv[1]=1;for (int i=2;i<P;i++) inv[i]=P-(P/i)*inv[P%i]%P;
        cout<<a.value();
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/Gloid/p/10107716.html
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