AC自动机+DP+大数poj1625

Language: Default Censored! Time Limit: 5000MS   Memory Limit: 10000K Total Submissions: 8102   Accepted: 2191 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

这个题跟2778一样，但不知道问世么不能用矩阵，网上说会超时，但好像数据不会超啊？？？

dp[i][j]表示第i步在第j个节点的方法数。flag[k]=0表示不是非法节点,son是k的儿子节点的集合。

　　dp[i][j]=sum(dp[i-1][k]*mat[k][j]&&val[ch[k][j]]==0);

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=110;
int SIGMA_SIZE;
int n,m,p;
char P[60],s[60];
map<char,int> mp;

struct Matrix
{
    int mat[maxn][maxn];
    Matrix(){memset(mat,0,sizeof(mat));}

};
struct AC
{
    int ch[maxn][256],val[maxn];
    int fail[maxn],last[maxn];
    int sz;
    void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;}
    int idx(char x){return mp[x];}
    void insert(char *s)
    {
        int n=strlen(s);
        int u=0;
        for(int i=0;i<n;i++)
        {
            int c=idx(s[i]);
            if(!ch[u][c])
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
                ch[u][c]=sz++;
            }
            u=ch[u][c];
        }
        val[u]=1;
    }
    void getfail()
    {
        queue<int> q;
        fail[0]=0;
        int u=0;
        for(int i=0;i<SIGMA_SIZE;i++)
        {
            u=ch[0][i];
            if(u){fail[u]=last[u]=0;q.push(u);}
        }
        while(!q.empty())
        {
            int r=q.front();q.pop();
            if(val[fail[r]])val[r]=1;
            for(int c=0;c<SIGMA_SIZE;c++)
            {
                u=ch[r][c];
                if(!u){ch[r][c]=ch[fail[r]][c];continue;}
                q.push(u);
                int v=fail[r];
                while(v&&!ch[v][c])v=fail[v];
                fail[u]=ch[v][c];
                last[u]=val[fail[u]]?fail[u]:last[fail[u]];
            }
        }
    }
    Matrix getMatrix()
    {
        Matrix res;
        for(int i=0;i<sz;i++)
            for(int j=0;j<SIGMA_SIZE;j++)
                if(!val[ch[i][j]])res.mat[i][ch[i][j]]++;
        return res;
    }
}ac;
/*
 * 高精度，支持乘法和加法
 */
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");
    }
};
BigInt dp[2][maxn];
int main()
{
    while(scanf("%d%d%d",&n,&m,&p)!=EOF)
    {
        scanf("%s",P);
        ac.clear();
        mp.clear();
        for(int i=0;i<strlen(P);i++)
            mp[P[i]]=i;
        SIGMA_SIZE=n;
        for(int i=0;i<p;i++)
        {
            scanf("%s",s);
            ac.insert(s);
        }
        ac.getfail();
        Matrix A=ac.getMatrix();
        int now=0;
        dp[now][0]=1;
        for(int i=1;i<ac.sz;i++)dp[now][i]=0;
        for(int i=1;i<=m;i++)
        {
            now^=1;
            for(int j=0;j<ac.sz;j++)dp[now][j]=0;
            for(int j=0;j<ac.sz;j++)
                for(int k=0;k<ac.sz;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<ac.sz;i++)
            ans=ans+dp[now][i];
        ans.output();
    }
    return 0;
}