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-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
const int mn = 30;
const int child_num = 26;
struct Mat
{
    LL mat[N][N];
};
Mat operator *(Mat a,Mat b)
{
    Mat c;
    memset(c.mat,0,sizeof(c.mat));
    int i,j,k;
    for(k =0 ; k < mn ; k++)
    {
        for(i = 0 ; i < mn ;i++)
        {
            if(a.mat[i][k]==0) continue;//优化
            for(j = 0 ;j < mn ;j++)
            {
                if(b.mat[k][j]==0) 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,0,sizeof(c.mat));
    int i,j;
    for(i = 0 ; i < mn ;i++)
        for(j = 0 ;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 =0 ; i < mn ;i++)
        for(j = 0; j < mn ;j++)
        c.mat[i][j] = (i==j);
    for(; k ;k >>= 1)
    {
        if(k&1) c = c*a;
        a = a*a;
    }
    return c;
}
Mat cal(Mat x,int k)
{
    if(k==1) return x;
    Mat c,cc;
    c = cal(x,k/2);
    cc = x^(k/2);
    c = c+cc*c;
    if(k&1)
    c = c+(x^k);
    return c;
}
LL ex_mod(int a,int k)
{
    if(k==1) return a;
    LL t = ex_mod(a,k/2);
    t = t*t;
    if(k&1)
    t = t*a;
    return t;
}
LL solve(int a,int k)
{
    if(k==1) return a;
    LL t = solve(a,k/2);
    t = t + ex_mod(a,k/2)*t;
    if(k&1) 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[228];
    public :
    void init()
    {
        fail[0] = 0;
        for(int i = 0 ; i < child_num ; i++)
        id[i+'a'] = i;
    }
    void reset()
    {
        memset(val,0,sizeof(val));
        memset(ch[0],0,sizeof(ch[0]));
        memset(fail,0,sizeof(fail));
        sz = 1;
    }
    void insert(char *a,int key)
    {
        int p = 0;
        for(; *a ;a++)
        {
            int d = id[*a];
            if(ch[p][d]==0)
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                ch[p][d] = sz++;
            }
            p = ch[p][d];
        }
        val[p] = key;
    }
    void construct()
    {
        int i;
        int head =0, tail=0;
        for(i = 0;i < child_num ;i++)
        {
            if(ch[0][i])
            {
                fail[ch[0][i]] = 0;
                Q[tail++] = ch[0][i];
            }
        }
        while(head!=tail)
        {
            int u = Q[head++];
            val[u]|=val[fail[u]];
            for(i= 0; 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,0,sizeof(x.mat));
        for(i = 0 ; i < sz; i++)
        {
            for(int j = 0; j < child_num ; j++)
            if(val[ch[i][j]]==0)
            x.mat[i][ch[i][j]]++;
        }
        x = cal(x,n);
        LL ans = 0;
        for(i = 0 ; i < sz ; i++)
        ans = ans+x.mat[0][i];
        ans = solve(26,n)-ans;
        cout<<ans<<endl;
    }
}ac;
char vir[20];
int main()
{
    int n,l,i;
    ac.init();
    while(scanf("%d%d",&n,&l)!=EOF)
    {
        ac.reset();
        for(i = 1; i <= n; i++)
        {
            scanf("%s",vir);
            ac.insert(vir,1);
        }
        ac.construct();
        ac.work(l);
    }
    return 0;
}