HDU 2825 AC自动机+DP

Wireless Password

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7250    Accepted Submission(s): 2384

Problem Description

Liyuan lives in a old apartment. One day, he suddenly found that there was a wireless network in the building. Liyuan did not know the password of the network, but he got some important information from his neighbor. He knew the password consists only of lowercase letters 'a'-'z', and he knew the length of the password. Furthermore, he got a magic word set, and his neighbor told him that the password included at least k words of the magic word set (the k words in the password possibly overlapping).

For instance, say that you know that the password is 3 characters long, and the magic word set includes 'she' and 'he'. Then the possible password is only 'she'.

Liyuan wants to know whether the information is enough to reduce the number of possible passwords. To answer this, please help him write a program that determines the number of possible passwords.

 

Input

There will be several data sets. Each data set will begin with a line with three integers n m k. n is the length of the password (1<=n<=25), m is the number of the words in the magic word set(0<=m<=10), and the number k denotes that the password included at least k words of the magic set. This is followed by m lines, each containing a word of the magic set, each word consists of between 1 and 10 lowercase letters 'a'-'z'. End of input will be marked by a line with n=0 m=0 k=0, which should not be processed.

 

Output

For each test case, please output the number of possible passwords MOD 20090717.

 

Sample Input

10 2 2 hello world 4 1 1 icpc 10 0 0 0 0 0

 

Sample Output

2 1 14195065

 

Source

2009 Multi-University Training Contest 1 - Host by TJU

 

 题意：

构造长度是n的字符串其中至少包含k个给定的单词的方案数。字符集a~z

代码：

//m只有10显然可以压缩，dp[i][j][k]表示构造到到字符串的i位置在树上的第j节点时包含的单词状态是k时的方案数
#include<iostream>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
typedef long long ll;
const int MOD=20090717;
const int MAXN=100000;
int node[109][27],val[109],sz,f[109],dp[26][109][1030];
void init()
{
    sz=0;
    memset(node[0],0,sizeof(node[0]));
    val[0]=f[0]=0;
}
void insert(char *s,int nu)
{
    int len=strlen(s),rt=0;
    for(int i=0;i<len;i++){
        int id=s[i]-'a';
        if(!node[rt][id]){
            node[rt][id]=++sz;
            memset(node[sz],0,sizeof(node[sz]));
            val[sz]=0;
        }
        rt=node[rt][id];
    }
    val[rt]|=(1<<nu);
}
void get_fail()
{
    queue<int>q;
    for(int i=0;i<26;i++){
        int u=node[0][i];
        if(u) { q.push(u);f[u]=0; }
    }
    while(!q.empty()){
        int rt=q.front();q.pop();
        for(int i=0;i<26;i++){
            int u=node[rt][i];
            if(!u){
                node[rt][i]=node[f[rt]][i];
                continue;
            }
            q.push(u);
            f[u]=node[f[rt]][i];
        }
        val[rt]|=val[f[rt]];
    }
}
void get_dp(int max_sta,int n)
{
    memset(dp,0,sizeof(dp));
    dp[0][0][0]=1;
    for(int i=0;i<n;i++){
        for(int j=0;j<=sz;j++){
            for(int k=0;k<max_sta;k++){
                if(dp[i][j][k]==0) continue;
                for(int h=0;h<26;h++){
                    dp[i+1][node[j][h]][k|val[node[j][h]]]+=dp[i][j][k];
                    dp[i+1][node[j][h]][k|val[node[j][h]]]%=MOD;
                }
            }
        }
    }
}
int one(int x)
{
    int s=0;
    while(x){
        s+=x&1;
        x>>=1;
    }
    return s;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int n,m,k;
    char ch[20];
    while(scanf("%d%d%d",&n,&m,&k)&&(n+m+k)){
        init();
        for(int i=0;i<m;i++){
            scanf("%s",ch);
            insert(ch,i);
        }
        int max_sta=(1<<m);
        get_fail();
        get_dp(max_sta,n);
        int ans=0;
        for(int j=0;j<max_sta;j++){
            int tmp=one(j);
            if(tmp<k) continue;
            for(int i=0;i<=sz;i++){
                ans+=dp[n][i][j];
                ans%=MOD;
            }
        }
        printf("%d
",ans);
    }
    return 0;
}