hdu2243考研路茫茫——单词情结（ac+二分矩阵）

链接

跟2778差不多，解决了那道题这道也不成问题如果做过基本的矩阵问题。

数比较大，需要用unsigned longlong 就不需要mod了 溢出就相当于取余

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 30
 #define LL  unsigned __int64
 #define INF 0xfffffff
 #pragma comment(linker, "/STACK:1024000000,1024000000")
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 const int mn = ;
 const int child_num = ;
 struct Mat
 {
     LL mat[N][N];
 };
 Mat operator *(Mat a,Mat b)
 {
     Mat c;
     memset(c.mat,,sizeof(c.mat));
     int i,j,k;
     for(k = ; k < mn ; k++)
     {
         for(i =  ; i < mn ;i++)
         {
             if(a.mat[i][k]==) continue;//优化
             for(j =  ;j < mn ;j++)
             {
                 if(b.mat[k][j]==) continue;//优化
                 c.mat[i][j] = c.mat[i][j]+(a.mat[i][k]*b.mat[k][j]);
             }
         }
     }
     return c;
 }
 Mat operator + (Mat a,Mat b)
 {
     Mat c;
     memset(c.mat,,sizeof(c.mat));
     int i,j;
     for(i =  ; i < mn ;i++)
         for(j =  ;j < mn ;j++)
             c.mat[i][j] = a.mat[i][j]+b.mat[i][j];
     return c;
 }
 Mat operator ^(Mat a,int k)
 {
     Mat c;
     int i,j;
     for(i = ; i < mn ;i++)
         for(j = ; j < mn ;j++)
         c.mat[i][j] = (i==j);
     for(; k ;k >>= )
     {
         if(k&) c = c*a;
         a = a*a;
     }
     return c;
 }
 Mat cal(Mat x,int k)
 {
     if(k==) return x;
     Mat c,cc;
     c = cal(x,k/);
     cc = x^(k/);
     c = c+cc*c;
     if(k&)
     c = c+(x^k);
     return c;
 }
 LL ex_mod(int a,int k)
 {
     if(k==) return a;
     LL t = ex_mod(a,k/);
     t = t*t;
     if(k&)
     t = t*a;
     return t;
 }
 LL solve(int a,int k)
 {
     if(k==) return a;
     LL t = solve(a,k/);
     t = t + ex_mod(a,k/)*t;
     if(k&) t = t+ex_mod(a,k);
     return t;
 }
 class AC
 {
     private:
     int ch[N][child_num];
     int Q[N];
     int fail[N];
     int val[N];
     int sz;
     int id[];
     public :
     void init()
     {
         fail[] = ;
         for(int i =  ; i < child_num ; i++)
         id[i+'a'] = i;
     }
     void reset()
     {
         memset(val,,sizeof(val));
         memset(ch[],,sizeof(ch[]));
         memset(fail,,sizeof(fail));
         sz = ;
     }
     void insert(char *a,int key)
     {
         int p = ;
         for(; *a ;a++)
         {
             int d = id[*a];
             if(ch[p][d]==)
             {
                 memset(ch[sz],,sizeof(ch[sz]));
                 ch[p][d] = sz++;
             }
             p = ch[p][d];
         }
         val[p] = key;
     }
     void construct()
     {
         int i;
         int head =, tail=;
         for(i = ;i < child_num ;i++)
         {
             if(ch[][i])
             {
                 fail[ch[][i]] = ;
                 Q[tail++] = ch[][i];
             }
         }
         while(head!=tail)
         {
             int u = Q[head++];
             val[u]|=val[fail[u]];
             for(i= ; i < child_num ; i++)
             {
                 if(ch[u][i])
                 {
                     fail[ch[u][i]] = ch[fail[u]][i];
                     Q[tail++] = ch[u][i];
                 }
                 else ch[u][i] = ch[fail[u]][i];
             }
         }
     }
     void work(int n)
     {
         int i;
         Mat x;
         memset(x.mat,,sizeof(x.mat));
         for(i =  ; i < sz; i++)
         {
             for(int j = ; j < child_num ; j++)
             if(val[ch[i][j]]==)
             x.mat[i][ch[i][j]]++;
         }
         x = cal(x,n);
         LL ans = ;
         for(i =  ; i < sz ; i++)
         ans = ans+x.mat[][i];
         ans = solve(,n)-ans;
         cout<<ans<<endl;
     }
 }ac;
 char vir[];
 int main()
 {
     int n,l,i;
     ac.init();
     while(scanf("%d%d",&n,&l)!=EOF)
     {
         ac.reset();
         for(i = ; i <= n; i++)
         {
             scanf("%s",vir);
             ac.insert(vir,);
         }
         ac.construct();
         ac.work(l);
     }
     return ;
 }