POJ 1625 Censored!

题意：给一个字符集，构成长N的字符串，不含M个危险串的任一个，求合法字符串总数。

AC自动机+DP。。

dp[i][j]=在字符串长i，在节点j的总数。

dp[i][j]=sigma(dp[i-1][k]*move[k][j])，move[k][j]表示节点k到节点j路径数。

Censored!

Time Limit: 5000MS		Memory Limit: 10000K
Total Submissions: 7862		Accepted: 2132

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 1
ab
bb

Sample Output

5

Source

Northeastern Europe 2001, Northern Subregion

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
using namespace std;

const int N=1022;
int ch[N][128],fail[N];
bool end[N];
int root,L;
int Char;
char charset[N];
int newnode() { memset(ch[L],-1,sizeof ch[L]);end[L++]=0; return L-1; }
void init(char s[])
{
    strcpy(charset,s); Char=strlen(s); L=0; root=newnode();
}
int idx(char a)
{
    for(int i=0;i<Char;i++) if(charset[i]==a)return i;
}
void insert(char s[])
{
    int n=strlen(s),u=root;
    for(int i=0;i<n;i++)
    {
        int& tmp=ch[u][idx(s[i])];
        if(tmp==-1) tmp=newnode();
        u=tmp;
    }
    end[u]=1;
}
void BUILD()
{
    queue<int> q;
    for(int i=0;i<Char;i++)
    {
        int& tmp=ch[root][i];
        if(tmp==-1) tmp=root;
        else { fail[tmp]=root; q.push(tmp); }
    }
    while(!q.empty())
    {
        int u=q.front(); q.pop();
        end[u]|=end[fail[u]];
        for(int i=0;i<Char;i++)
        {
            int& tmp=ch[u][i];
            if(tmp==-1) tmp=ch[fail[u]][i];
            else { fail[tmp]=ch[fail[u]][i]; q.push(tmp); }
        }
    }
}
int move[N][N];
void getmove()
{
    memset(move,0,sizeof move);
    for(int i=0;i<L;i++)
        for(int j=0;j<Char;j++)
        move[i][ch[i][j]]+=!end[ch[i][j]];
}

struct Big
{
    int a[121]; int len;
    Big() { len=1;memset(a,0,sizeof a); }
    void add(Big b)
    {
        len=max(len,b.len);
        for(int i=0;i<len;i++)
        {
            a[i+1]+=(a[i]+b.a[i])/10;
            a[i]=(a[i]+b.a[i])%10;
        }
        while(a[len]) len++;
    }
    void print()
    {
        for(int i=len-1;i>=0;i--) printf("%d",a[i]);
        putchar(10);
    }
};
Big dp[2][N];
char s[N];
int main()
{
    int tt,n,m;
    while(cin>>tt>>n>>m)
    {
        scanf("%s",s);
        init(s);
        for(int i=0;i<m;i++) { scanf("%s",s);insert(s); }
        BUILD();
        getmove();
        int pos=0;
        for(int i=0;i<L;i++) dp[0][i]=Big();
        dp[0][0].a[0]=1; dp[0][0].len=1;
        for(int i=0;i<n;i++,pos=!pos)
        {
            for(int j=0;j<L;j++)
            {
                dp[!pos][j]=Big();
                for(int k=0;k<L;k++)
                    for(int t=0;t<move[k][j];t++)
                    dp[!pos][j].add(dp[pos][k]);
            }
        }
        Big res=Big();
        for(int i=0;i<L;i++) res.add(dp[pos][i]);
        res.print();
    }
}