poj 1625 Censored!

Posted 2020-09-22 日拱一卒 功不唐捐

Censored!

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

Time Limit: 5000MS		Memory Limit: 10000K

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

 

题意：

给出n个字符，构造长为m的字符串s，有p个模板串

字符串s中不能包含模板串，问能构造的字符串的个数

 

AC自动机+DP+压位高精

 

dp[i][j]表示 当前在AC自动机的i节点，还要走j步，构造出的字符串个数

dp[i][j]=Σ dp[k][j-1] k不是单词节点

记忆化搜索即可

加上高精度

 

#include<queue>
#include<cstdio>
#include<cstring>

using namespace std;

const int mod=1000000000;

int n,m,p,id,len,root,tot=1;
char s[101],ss[501];
bool mark[1001],v[1001][501];
int f[1001],trie[1001][501],mp[1001];
queue<int>q;

struct bigint
{
    int num[20];
    bigint()
    {
        memset(num,0,sizeof(num));
        num[0]=1;
        num[1]=0;
    }
    bigint(int i)
    {
        memset(num,0,sizeof(num));
        num[0]=1;
        num[1]=i;
    }
    //void operator += (bigint b) 
    //{
        /*for(int i=num[0];i;i--) printf("%d",num[i]);
        putchar(‘+‘);
        for(int i=b.num[0];i;i--) printf("%d",b.num[i]);
        putchar(‘=‘);*/
//        num[0]=max(b.num[0],num[0]);
//        for(int i=1;i<=num[0];i++)
//        {
//            num[i]+=b.num[i];
//            num[i+1]+=num[i]/10;
//            num[i]%=10;
//        }
//        if(num[num[0]+1])     num[0]++;
        /*for(int i=num[0];i;i--) printf("%d",num[i]);
        puts("");*/
//    }
    void operator += (bigint b)
    {
        /*for(int i=num[0];i;i--) printf("%d",num[i]);
        putchar(‘+‘);
        for(int i=b.num[0];i;i--) printf("%d",b.num[i]);
        putchar(‘=‘);*/
        num[0]=max(num[0],b.num[0]);
        for(int i=1;i<=num[0];i++) num[i]+=b.num[i];
        for(int i=1;i<=num[0];i++) 
        {
            num[i+1]+=num[i]/mod;
            num[i]%=mod;
        }    
        if(num[num[0]+1]) num[0]++;
        /*for(int i=num[0];i;i--) printf("%d",num[i]);
        puts("");*/
    }
    void operator = (bigint b)
    {
        int lb=b.num[0];
        for(int i=0;i<=lb;i++) num[i]=b.num[i];
    }
    void output()
    {
        printf("%d",num[num[0]]);
        for(int i=num[0]-1;i;i--) printf("%09d",num[i]); 
    }
};
bigint dp[101][51];

bigint dfs(int now,int l)
{
    if(!l) return bigint(1);
    if(v[now][l]) return dp[now][l];
    v[now][l]=1;
    bigint x;
    for(int i=0;i<n;i++)
       if(!mark[trie[now][i]]) x+=dfs(trie[now][i],l-1);
/*    printf("%d %d: ",now,l);
    dp[now][l].output();
    puts("");*/
    dp[now][l]=x;
    return x;
}

struct ACautomata
{
    int get(char c)
    {
        return mp[c];
    }
    void insert()
    {
        len=strlen(s);
        root=1;
        for(int i=0;i<len;i++)
        {
            id=get(s[i]);
            if(!trie[root][id]) 
            {
                trie[root][id]=++tot;
                memset(trie[tot],0,sizeof(trie[tot]));
                mark[tot]=0;
            }
            root=trie[root][id];
        }
        mark[root]=true;
    }
    void getfail()
    {
        memset(f,0,sizeof(f));
        for(int i=0;i<n;i++) trie[0][i]=1;
        q.push(1);
        int now,j;
        while(!q.empty())
        {
            now=q.front(); q.pop();
            for(int i=0;i<n;i++)
            {
                if(!trie[now][i]) 
                {
                    trie[now][i]=trie[f[now]][i];
                    //if(mark[trie[f[now]][i]]) mark[trie[now][i]]=true;
                    continue;
                }
                q.push(trie[now][i]);
                j=f[now];
                f[trie[now][i]]=trie[j][i];
                if(mark[trie[j][i]]) mark[trie[now][i]]=true;
            }
        }
    }
};
ACautomata AC;

int main()
{
    while(scanf("%d%d%d",&n,&m,&p)!=EOF)
    {
        memset(v,0,sizeof(v));
        memset(trie[1],0,sizeof(trie[1]));
        tot=1;
        scanf("%s",ss);
        for(int i=0;i<n;i++)  mp[ss[i]]=i;
        while(p--)
        {
            scanf("%s",s);
            AC.insert();
        }
        AC.getfail();
        bigint out=dfs(1,m);
        out.output();
        puts("");
    }
}