POJ 1625 AC自动机+DP

http://poj.org/problem?id=1625

Description

The alphabet of Freeland consists of exactly N letters. Each sentence of Freeland language (also known as Freish) consists of exactly M letters without word breaks. So, there exist exactly N^M different Freish sentences. 

But after recent election of Mr. Grass Jr. as Freeland president some words offending him were declared unprintable and all sentences containing at least one of them were forbidden. The sentence S contains a word W if W is a substring of S i.e. exists such k >= 1 that S[k] = W[1], S[k+1] = W[2], ...,S[k+len(W)-1] = W[len(W)], where k+len(W)-1 <= M and len(W) denotes length of W. Everyone who uses a forbidden sentence is to be put to jail for 10 years. 

Find out how many different sentences can be used now by freelanders without risk to be put to jail for using it. 

Input

The first line of the input file contains three integer numbers: N -- the number of letters in Freish alphabet, M -- the length of all Freish sentences and P -- the number of forbidden words (1 <= N <= 50, 1 <= M <= 50, 0 <= P <= 10). 

The second line contains exactly N different characters -- the letters of the Freish alphabet (all with ASCII code greater than 32). 

The following P lines contain forbidden words, each not longer than min(M, 10) characters, all containing only letters of Freish alphabet. 

Output

Output the only integer number -- the number of different sentences freelanders can safely use.

Sample Input

2 3 1abbb

Sample Output

5

/**POJ 1625 AC自动机+DP题目大意：给定n个模式串，求可以构造出多少长度为m的字符串使其不包括任一模式串作为其子串解题思路：这个题没有取模，大数用矩阵连乘会爆栈，可以用dp来转移，不过要是大数的处理*/#include <stdio.h>#include <string.h>#include <algorithm>#include <stdio.h>#include <map>#include <queue>using namespace std;map<char,int>mp;int N,M,P;struct Matrix{    int mat[110][110];    int n;    Matrix() {}    Matrix(int _n)    {        n=_n;        for(int i=0; i<n; i++)        {            for(int j=0; j<n; j++)            {                mat[i][j]=0;            }        }    }};struct Trie{    int next[110][256],fail[110];    bool end[110];    int L,root;    int newnode()    {        for(int i=0; i<256; i++)        {            next[L][i]=-1;        }        end[L++]=false;        return L-1;    }    void init()    {        L=0;        root=newnode();    }    void insert(char *buf)    {        int len=strlen(buf);        int now=root;        for(int i=0; i<len; i++)        {            if(next[now][mp[buf[i]]]==-1)                next[now][mp[buf[i]]]=newnode();            now=next[now][mp[buf[i]]];        }        end[now]=true;    }    void build()    {        queue<int>Q;        fail[root]=root;        for(int i=0; i<256; i++)        {            if(next[root][i]==-1)                next[root][i]=root;            else            {                fail[next[root][i]]=root;                Q.push(next[root][i]);            }        }        while(!Q.empty())        {            int now=Q.front();            Q.pop();            if(end[fail[now]]==true)                end[now]=true;            for(int i=0; i<256; i++)            {                if(next[now][i]==-1)                    next[now][i]=next[fail[now]][i];                else                {                    fail[next[now][i]]=next[fail[now]][i];                    Q.push(next[now][i]);                }            }        }    }    Matrix getMatrix()    {        Matrix res=Matrix(L);        for(int i=0; i<L; i++)        {            for(int j=0; j<N; j++)            {                if(end[next[i][j]]==false)                    res.mat[i][next[i][j]]++;            }        }        return res;    }} ac;/**高精度类（实现乘法和加法）*/struct BigInt{    const static int mod=10000;    const static int DLEN=4;    int a[600],len;    BigInt()    {        memset(a,0,sizeof(a));        len=1;    }    BigInt(int v)    {        memset(a,0,sizeof(a));        len=0;        do        {            a[len++]=v%mod;            v/=mod;        }        while(v);    }    BigInt(const char *s)    {        memset(a,0,sizeof(a));        int L=strlen(s);        len=L/DLEN;        if(L%DLEN)len++;        int index=0;        for(int i=L-1; i>=0; i-=DLEN)        {            int t=0;            int k=i-DLEN+1;            if(k<0)k=0;            for(int j=k; j<=i; j++)            {                t=t*10+s[j]-'0';            }            a[index++]=t;        }    }    BigInt operator +(const BigInt &b)const    {        BigInt res;        res.len=max(len,b.len);        for(int i=0; i<=res.len; i++)        {            res.a[i]=0;        }        for(int i=0; i<res.len; i++)        {            res.a[i]+=((i<len)?a[i]:0)+((i<b.len)?b.a[i]:0);            res.a[i+1]+=res.a[i]/mod;            res.a[i]%=mod;        }        if(res.a[res.len]>0)res.len++;        return res;    }    BigInt operator *(const BigInt &b)const    {        BigInt res;        for(int i=0; i<len; i++)        {            int up=0;            for(int j=0; j<b.len; j++)            {                int temp=a[i]*b.a[j]+res.a[i+j]+up;                res.a[i+j]=temp%mod;                up=temp/mod;            }            if(up!=0)                res.a[i+b.len]=up;        }        res.len=len+b.len;        while(res.a[res.len-1]==0&&res.len>1)res.len--;        return res;    }    void output()    {        printf("%d",a[len-1]);        for(int i=len-2; i>=0; i--)            printf("%04d",a[i]);        printf("\n");    }};char buf[1010];BigInt dp[2][110];int main(){    while(~scanf("%d%d%d%*c",&N,&M,&P))    {        gets(buf);        mp.clear();        int len=strlen(buf);        for(int i=0; i<len; i++)        {            mp[buf[i]]=i;        }        ac.init();        for(int i=0; i<P; i++)        {            gets(buf);            ac.insert(buf);        }        ac.build();        Matrix a=ac.getMatrix();        int now=0;        dp[now][0]=1;        for(int i=1; i<a.n; i++)        {            dp[now][i]=0;        }        for(int i=0; i<M; i++)        {            now^=1;            for(int j=0; j<a.n; j++)            {                dp[now][j]=0;            }            for(int j=0; j<a.n; j++)            {                for(int k=0; k<a.n; k++)                {                    if(a.mat[j][k]>0)                        dp[now][k]=dp[now][k]+dp[now^1][j]*a.mat[j][k];                }            }        }        BigInt ans=0;        for(int i=0; i<a.n; i++)        {            ans=ans+dp[now][i];        }        ans.output();    }    return 0;}