POJ 1625Censored!

给定p个模式串，求长度为m<=50的串中不包含任何模式串的串的种类数，字符仅由给出的n个字符构成，用mp数组标记下。然后和之前的几道类似，利用end和next数组得到转态转移数组，然后由于题目数据不像之前的那么大，可以用dp[i][j]来进行状态转移，i代表当前串长度，j为当前状态。但是由于答案很大需要用到高精度。代码如下

#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>#include<queue>using namespace std;int N;int mp[300];struct Matrix{    long long 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;    }    Matrix operator *(const Matrix &b)const    {        Matrix ret = Matrix(n);        for(int i=0;i<n;i++)            for(int j=0;j<n;j++)                for(int k=0;k<n;k++)                    ret.mat[i][j]+=mat[i][k]*b.mat[k][j];        return ret;    }};struct Tree{int next[110][256],end[110],fail[110];int L,root;int newnode(){for(int i=0;i<255;i++){next[L][i]=-1;}end[L++]=0;return L-1;}void init(){L=0;root=newnode();}void insert(char *s){int len=strlen(s);int p=root;for(int i=0;i<len;i++){if(next[p][mp[s[i]]]==-1){next[p][mp[s[i]]]=newnode();}p=next[p][mp[s[i]]];}end[p]++;}void build(){queue<int>q;int p=root;fail[root]=root;for(int i=0;i<255;i++){if(next[p][i]==-1){next[p][i]=root;}else{fail[next[p][i]]=root;q.push(next[p][i]);}}while(!q.empty()){p=q.front();q.pop();if(end[fail[p]]) end[p]=1;for(int i=0;i<255;i++){if(next[p][i]==-1){next[p][i]=next[fail[p]][i];}else{fail[next[p][i]]=next[fail[p]][i];q.push(next[p][i]);}}}}void query(char *s){int len=strlen(s);int p=root,ans=0;for(int i=0;i<len;i++){int id=s[i]-'a';p=next[p][id];int temp=p;while(temp!=root){if(end[temp]){ans+=end[temp];end[temp]=0;}temp=fail[temp];}}}Matrix getMatrix(){Matrix a(L);for(int i=0;i<a.n;i++){for(int j=0;j<N;j++){if(end[next[i][j]]==0)a.mat[i][next[i][j]]++;}}return a;}void debug()    {        for(int i = 0;i < L;i++)        {            printf("id = %3d,fail = %3d,end = %3d,chi = [",i,fail[i],end[i]);            for(int j = 0;j < 26;j++)                printf("%2d",next[i][j]);            printf("]\n");        }    }};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");    }};Tree ac;char s[55];BigInt dp[2][110];int main(){int m,p;while(~scanf("%d %d %d",&N,&m,&p)){scanf("%s",s);for(int i=0;i<N;i++)mp[s[i]]=i;ac.init();for(int i=0;i<p;i++){scanf("%s",s);ac.insert(s);}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();}}