HDU2825 Wireless Password —— AC自动机 + 状压DP

题目链接：https://vjudge.net/problem/HDU-2825

Wireless Password

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7733    Accepted Submission(s): 2509

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：

给出m个单词，问长度为n且至少含有k个单词的字符串有多少个?

题解：

1.把这m个单词插入到AC自动机中。

2.设dp[i][j][status]为：长度为i，到达j状态（AC自动机中的状态），且含有单词的信息为status（状态压缩）的字符串有多少个。

3.模拟在AC自动机上的跳动，求出dp数组。

代码如下：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>
#include <queue>
#include <stack>
#include <map>
#include <string>
#include <set>
using namespace std;
typedef long long LL;
const double EPS = 1e-6;
const int INF = 2e9;
const LL LNF = 9e18;
const int MOD = 20090717;
const int MAXN = 100+10;

int num[1<<10], dp[30][110][1<<10];

struct Trie
{
    const static int sz = 26, base = 'a';
    int next[MAXN][sz], fail[MAXN], end[MAXN];
    int root, L;
    int newnode()
    {
        for(int i = 0; i<sz; i++)
            next[L][i] = -1;
        end[L++] = 0;
        return L-1;
    }
    void init()
    {
        L = 0;
        root = newnode();
    }
    void insert(char buf[], int id)
    {
        int len = strlen(buf);
        int now = root;
        for(int i = 0; i<len; i++)
        {
            if(next[now][buf[i]-base] == -1) next[now][buf[i]-base] = newnode();
            now = next[now][buf[i]-base];
        }
        end[now] |= (1<<id);
    }
    void build()
    {
        queue<int>Q;
        fail[root] = root;
        for(int i = 0; i<sz; i++)
        {
            if(next[root][i] == -1) next[root][i] = root;
            else fail[next[root][i]] = root, Q.push(next[root][i]);
        }
        while(!Q.empty())
        {
            int now = Q.front();
            Q.pop();
            end[now] |= end[fail[now]]; //当前串的后缀是否也包含单词
            for(int i = 0; i<sz; i++)
            {
                if(next[now][i] == -1) next[now][i] = next[fail[now]][i];
                else fail[next[now][i]] = next[fail[now]][i], Q.push(next[now][i]);
            }
        }
    }

    int query(int n, int m, int k)
    {
        for(int i = 0; i<=n; i++)
            for(int j = 0; j<L; j++)
                for(int s = 0; s<(1<<m); s++)
                    dp[i][j][s] = 0;
        dp[0][0][0] = 1;
        for(int i = 0; i<n; i++)    //模拟在AC自动机上的跳动
            for(int j = 0; j<L; j++)
                for(int s = 0; s<(1<<m); s++)
                {
                    if(dp[i][j][s]==0) continue;
                    for(int p = 0; p<sz; p++)
                    {
                        int newi = i+1;
                        int newj = next[j][p];
                        int news = (s|end[newj]);
                        dp[newi][newj][news] += dp[i][j][s];
                        dp[newi][newj][news] %= MOD;
                    }
                }
        int ret = 0;
        for(int s = 0; s<(1<<m); s++)
        {
            if(num[s]<k) continue;
            for(int i = 0; i<L; i++)
                ret = (ret+dp[n][i][s])%MOD;
        }
        return ret;
    }
};

Trie ac;
char buf[20];
int main()
{
    for(int s = 0; s<(1<<10); s++)
    {
        num[s] = 0;
        for(int i = 0; i<10; i++)
            if(s&(1<<i)) num[s]++;
    }

    int n, m, k;
    while(scanf("%d%d%d", &n,&m,&k)&&(n||m||k))
    {
        ac.init();
        for(int i = 0; i<m; i++)
        {
            scanf("%s", buf);
            ac.insert(buf, i);
        }
        ac.build();
        int ans = ac.query(n,m,k);
        printf("%d
", ans);
    }
    return 0;
}