HDU - 2825 Wireless Password(AC自己主动机+DP)

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 

题意：让你构造一个长度为n的字符串。使得至少包括共m个的集合里的k个，求个数。

思路：AC自己主动机构造。须要用到DP的思想，设dp[i][j][k]表示长度为i到达自己主动机状态为j的时候且状态k时的个数。

我们须要用一个来表示状态j包括的字符串的个数。也就是走到字符串末尾的个数，二进制表示。

然后还要注意一点的是：我们在构造自己主动机的时候有fail指针的转移。所以状态j也须要结合它失配时候的状态。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
using namespace std;
const int mod = 20090717;
int n, m, k;
int dp[30][250][1<<10];
int num[5000];
struct Trie {
    int nxt[110][26], fail[110], end[110];
    int root, sz;
    int newNode() {
        for (int i = 0; i < 26; i++)
            nxt[sz][i] = -1;
        end[sz++] = 0;
        return sz - 1;
    }
    void init() {
        sz = 0;
        root = newNode();
    }
    void insert(char buf[], int id) {
        int now = root;
        for (int i = 0; buf[i]; i++) {
            if (nxt[now][buf[i]-'a'] == -1)
                nxt[now][buf[i]-'a'] = newNode();
            now = nxt[now][buf[i]-'a'];
        }
        end[now] |= (1<<id);
    }
    void build() {
        queue<int> q;
        fail[root] = root;
        for (int i = 0; i < 26; i++) {
            if (nxt[root][i] == -1)
                nxt[root][i] = root;
            else {
                fail[nxt[root][i]] = root;
                q.push(nxt[root][i]);
            }
        }
        while (!q.empty()) {
            int now = q.front();
            q.pop();
            end[now] |= end[fail[now]];
            for (int i = 0; i < 26; i++) {
                if (nxt[now][i] == -1)
                    nxt[now][i] = nxt[fail[now]][i];
                else {
                    fail[nxt[now][i]] = nxt[fail[now]][i];
                    q.push(nxt[now][i]);
                }
            }
        }
    }
    int solve() {
        for (int i = 0; i <= n; i++)
            for (int j = 0; j < sz; j++)
                for (int k = 0; k < (1<<m); k++)
                    dp[i][j][k] = 0;
        dp[0][0][0] = 1;
        for (int i = 0; i < n; i++)
            for (int j = 0; j < sz; j++)
                for (int k = 0; k < (1<<m); k++)
                    if (dp[i][j][k] > 0) {
                        for (int x = 0; x < 26; x++) {
                            int ni = i+1;
                            int nj = nxt[j][x];
                            int nk = k | end[nj];
                            dp[ni][nj][nk] += dp[i][j][k];
                            dp[ni][nj][nk] %= mod;
                        }
                    }
        int ans = 0;
        for (int p = 0; p < (1<<m); p++) {
            if (num[p] < k) continue;
            for (int i = 0; i < sz; i++)
                ans = (ans + dp[n][i][p]) % mod;
        }
        return ans;
    }
} ac;
char buf[20];
int main() {
    for (int i = 0; i < (1<<10); i++) {
        num[i] = 0;
        for (int j = 0; j < 10; j++)
            if (i & (1<<j))
                num[i]++;
    }
    while (scanf("%d%d%d", &n, &m, &k) != EOF && n+m+k) {
        ac.init();
        for (int i = 0; i < m; i++) {
            scanf("%s", buf);
            ac.insert(buf, i);
        }
        ac.build();
        printf("%d\n", ac.solve());
    }
    return 0;
}

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