HDU 2243 考研路茫茫——单词情结(AC自动机+DP+快速幂）

题目链接

错的上头了...

这题是DNA的加强版，26^1 +26^2... - A^1-A^2...

先去学了矩阵的等比数列求和，学的是第二种方法，扩大矩阵的方法。剩下就是各种模板，各种套。

#include <cstdio>

#include <cstring>

#include <iostream>

#include <map>

#include <algorithm>

#include <vector>

#include <string>

using namespace std;

#define LL unsigned __int64

#define N 100001

int t,trie[N][26],que[N],o[N],fail[N];

LL p[201][201],mat[201][201];

void CL()

{

    memset(trie,-1,sizeof(trie));

    memset(p,0,sizeof(p));

    memset(o,0,sizeof(o));

    t = 1;

}

void insert(char *s)

{

    int i,root,len;

    len = strlen(s);

    root = 0;

    for(i = 0; i < len; i ++)

    {

        if(trie[root][s[i]-'a'] == -1)

            trie[root][s[i]-'a'] = t ++;

        root = trie[root][s[i]-'a'];

    }

    o[root] = 1;

}

void build_ac()

{

    int head,tail,front,i;

    head = tail = 0;

    for(i = 0; i < 26; i ++)

    {

        if(trie[0][i] != -1)

        {

            fail[trie[0][i]] = 0;

            que[tail++] = trie[0][i];

        }

        else

        {

            trie[0][i] = 0;

        }

    }

    while(head != tail)

    {

        front = que[head++];

        if(o[fail[front]])

            o[front] = 1;

        for(i = 0; i < 26; i ++)

        {

            if(trie[front][i] != -1)

            {

                que[tail++] = trie[front][i];

                fail[trie[front][i]] = trie[fail[front]][i];

            }

            else

            {

                trie[front][i] = trie[fail[front]][i];

            }

        }

    }

}

void work()

{

    int i,j;

    for(i = 0; i < t; i ++)

    {

        if(o[i]) continue;

        for(j = 0; j < 26; j ++)

        {

            if(o[trie[i][j]]) continue;

            p[i][trie[i][j]] ++;

        }

    }

    for(i = 0; i < t; i ++)

    {

        p[i][t+i] = p[t+i][t+i] = 1;

    }

    t <<= 1;

}

LL qmod(LL n)

{

    int i,j,k;

    LL c[201][201],ans = 0;

    while(n)

    {

        if(n&1)

        {

            memset(c,0,sizeof(c));

            for(i = 0; i < t; i ++)

            {

                for(j = 0; j < t; j ++)

                {

                    for(k = 0; k < t; k ++)

                    {

                        c[i][j] += mat[i][k]*p[k][j];

                    }

                }

            }

            memcpy(mat,c,sizeof(mat));

        }

        memset(c,0,sizeof(c));

        for(i = 0; i < t; i ++)

        {

            for(j = 0; j < t; j ++)

            {

                for(k = 0; k < t; k ++)

                {

                    c[i][j] += p[i][k]*p[k][j];

                }

            }

        }

        memcpy(p,c,sizeof(p));

        n >>= 1;

    }

    for(i = t/2; i < t; i ++)

    {

        ans += mat[0][i];

    }

    return ans-1;

}



LL fastmod(LL a,LL k)

{

    LL b = 1;

    while(k)

    {

        if(k&1)

            b = a*b;

        a = a*a;

        k = k/2;

    }

    return b;

}

LL getnum(LL p,LL k)

{

    if (k <= 0) return 1;

    if (k%2 == 0)

        return ((getnum(p,k/2 - 1))*((1 + fastmod(p,k/2 + 1))) + fastmod(p,k/2));

    else

        return ((getnum(p,(k + 1)/2 - 1))*((1 + fastmod(p,(k + 1)/2))));

}

int main()

{

    int i,j,m;

    LL n;

    char ch[101];

    while(scanf("%d%I64u",&m,&n)!=EOF)

    {

        CL();

        for(i = 1; i <= m; i ++)

        {

            scanf("%s",ch);

            insert(ch);

        }

        build_ac();

        work();

        for(i = 0; i < t; i ++)

        {

            for(j = 0; j < t; j ++)

            {

                mat[i][j] = (i == j);

            }

        }

        printf("%I64u\n",getnum(26,n)-1-qmod(n+1));

    }

}

View Code