【AC自动机】【矩阵乘法】【等比数列】hdu2243 考研路茫茫——单词情结

题解：http://blog.csdn.net/xingyeyongheng/article/details/10005923

这里采用了二分法求等比数列前n项和。

等比数列前n项和也可以用矩乘快速幂来求[a 1]    [Sn]    =    [Sn+1]

　　　　　　　　　　　　　　　　　　  [0 1]    [a  ]          [   a   ]

#include<cstdio>
#include<cstring>
#include<vector>
#include<queue>
#include<iostream>
using namespace std;
typedef unsigned long long ull;
typedef vector<ull> vec;
typedef vector<vec> mat;
typedef pair<mat,mat> Point2;
typedef pair<ull,ull> Point;
int N;
mat I;
mat operator * (const mat &a,const mat &b)
{
    mat c(N,vec(N));
    for(int i=0;i<N;++i)
      for(int j=0;j<N;++j)
        for(int k=0;k<N;++k)
          c[i][j]=c[i][j]+a[i][k]*b[k][j];
    return c;
}
mat operator - (const mat &a,const mat &b)
{
    mat c(N,vec(N));
    for(int i=0;i<N;++i)
      for(int j=0;j<N;++j)
        c[i][j]=a[i][j]-b[i][j];
    return c;
}
mat operator + (const mat &a,const mat &b)
{
    mat c(N,vec(N));
    for(int i=0;i<N;++i)
      for(int j=0;j<N;++j)
        c[i][j]=a[i][j]+b[i][j];
    return c;
}
Point sum_a_n(ull a,ull n)
{
	if(n==0)
	  return Point(1,1);
	Point t=sum_a_n(a,n>>1);
	if(n&1)
	  return Point(t.first*t.first*a,t.second*(t.first*a+1));
	else
	  return Point(t.first*t.first,(t.second-t.first)*(t.first*a+1)+t.first);
}
Point2 sum_A_n(mat a,ull n)
{
	if(n==0)
	  return Point2(I,I);
	Point2 t=sum_A_n(a,n>>1);
	if(n&1)
	  return Point2(t.first*t.first*a,t.second*(t.first*a+I));
	else
	  return Point2(t.first*t.first,(t.second-t.first)*(t.first*a+I)+t.first);
}
queue<int>q;
int child[40][26],fail[40],size=1;
bool word[40];
void Insert(char S[])
{
    int len=strlen(S);
    int now=0;
    for(int i=0;i<len;++i)
      {
        if(!child[now][S[i]-'a'])
          child[now][S[i]-'a']=size++;
        now=child[now][S[i]-'a'];
      }
    word[now]=1;
}
void build()
{
    fail[0]=-1;
    q.push(0);
    while(!q.empty())
      {
        int U=q.front(); q.pop();
        for(int i=0;i<26;++i)
          if(child[U][i])
            {
              int V=fail[U];
              while(V!=-1)
                {
                  if(child[V][i])
                    {
                      fail[child[U][i]]=child[V][i];
                      break;
                    }
                  V=fail[V];
                }
              if(V==-1)
                fail[child[U][i]]=0;
              if(word[fail[child[U][i]]])
                word[child[U][i]]=1;
              q.push(child[U][i]);
            }
          else if(U)
            child[U][i]=child[fail[U]][i];
      }
}
int n,ma2[40];
ull m;
void Init()
{
	memset(child,0,sizeof(child));
	memset(fail,0,sizeof(fail));
	memset(word,0,sizeof(word));
	N=0;
	size=1;
}
int main()
{
	//freopen("hdu2243.in","r",stdin);
    char s[8];
    while(cin>>n>>m){
    Init();
    for(int i=1;i<=n;++i)
      {
        scanf("%s",s);
        Insert(s);
      }
    build();
    for(int i=0;i<size;++i)
      if(!word[i])
        ma2[i]=N++;
    I.assign(N,vec(N));
    for(int i=0;i<N;++i)
      I[i][i]=1;
    mat A(N,vec(N));
    for(int i=0;i<size;++i)
      for(int j=0;j<26;++j) if((!word[i]) && (!word[child[i][j]]))
        ++A[ma2[i]][ma2[child[i][j]]];
    A=sum_A_n(A,m).second-I;
    ull ans=sum_a_n(26,m).second-1;
    for(int i=0;i<N;++i)
      ans=ans-A[0][i];
    cout<<ans<<endl;
	}
    return 0;
}