线性代数（矩阵乘法）：NOI 2007 生成树计数

　　这道题就是深搜矩阵，再快速幂。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <map>
using namespace std;
const int maxn=200;
const int mod=65521;
struct Matrix{
    long long mat[maxn][maxn];
    int r,c;
    Matrix(int r_=0,int c_=0,int on=0){
        memset(mat,0,sizeof(mat));
        r=r_;c=c_;
        if(on)for(int i=1;i<=r;i++)mat[i][i]=1;
    }
    Matrix operator *(Matrix a){
        Matrix ret(r,a.c);
        long long l;
        for(int i=1;i<=r;i++)
            for(int k=1;k<=c;k++){
                l=mat[i][k];
                for(int j=1;j<=a.c;j++)
                    (ret.mat[i][j]+=(l*a.mat[k][j])%mod)%=mod;
            }
        return ret;            
    }
    Matrix operator ^(long long k){
        Matrix ret(r,c,1),x(r,c);
        for(int i=1;i<=r;i++)
            for(int j=1;j<=c;j++)
                x.mat[i][j]=mat[i][j];
        while(k){
            if(k&1)
                ret=ret*x;
            k>>=1;
            x=x*x;
        }
        return ret;
    }
}A,B;
 
long long n;
map<int,int>ID;
map<int,bool>used;
int k,e,e1,E[maxn][2];
int cnt,st[1<<17],mem[1<<17];
int fa[maxn],sz[maxn],vis[maxn];
int Find(int x){
    return x==fa[x]?x:fa[x]=Find(fa[x]);
}
bool Check(int s){
    for(int i=1;i<maxn;i++)
        fa[i]=i,sz[i]=1;
    for(int i=0;i<e+e1;i++)
        if(s&(1<<i)){
            int u=Find(E[i][0]),v=Find(E[i][1]);
            if(u!=v){fa[u]=v;sz[v]+=sz[u];}
            else return false;
        }    
    return true;        
}
 
void Solve(int x){
    for(int i=1;i<=x;i++)
        for(int j=i+1;j<=x;j++)
            E[e][0]=i,E[e][1]=j,e++;
    for(int i=1;i<=x;i++)
        E[e+e1][0]=i,E[e+e1][1]=x+1,e1++;
    
    for(int s=(1<<e)-1,num;s>=0;s--)
        if(Check(s)){
            memset(vis,0,sizeof(vis));num=0;
            for(int i=1;i<=x;i++){
                if(vis[Find(i)])continue;
                vis[Find(i)]=++num;
            }
            num=0;
            for(int i=1;i<=x;i++)
                num=num*10+vis[Find(i)];
            if(ID[num])B.mat[ID[num]][1]+=1;
            else{
                A.r+=1;A.c+=1;B.r+=1;
                B.mat[ID[num]=B.r][1]=1;
                st[++cnt]=s;mem[cnt]=num;
            }
        }
    
    for(int t=1,s;t<=cnt;t++){    
        for(int p=(1<<e1)-1,num;p>=0;p--){
            s=st[t]^(p<<e);
            if(Check(s)&&sz[Find(1)]!=1){
                memset(vis,0,sizeof(vis));num=0;
                for(int i=2;i<=x+1;i++){
                    if(vis[Find(i)])continue;
                    vis[Find(i)]=++num;
                }
                num=0;
                for(int i=2;i<=x+1;i++)
                    num=num*10+vis[Find(i)];
                A.mat[ID[num]][ID[mem[t]]]+=1;
            }
        }
    }
    return;
}
  
int main(){
#ifndef ONLINE_JUDGE
    freopen("count.in","r",stdin);
    freopen("count.out","w",stdout);
#endif
    scanf("%d%lld",&k,&n);
    k=min(1ll*k,n);Solve(k);B.c=1;
    A=A^((n-k)%(1ll*(mod+1)*(mod-1)));B=A*B; 
    printf("%lld
",B.mat[1][1]);
    return 0;
}