HDU - 2825 Wireless Password AC自动机+状压dp

Wireless Password

  HDU - 2825

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. 

InputThere 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. OutputFor 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 HDU - 2825
My Solution
题意：给出一个字符串集合，集合里包含m(m <= 10)个长度不大于10的字符串，要求构造长度为n(1<=n<=25)的字符串且子串中至少出现k个集合里的字符串，求方案数 MOD 20090717。AC自动机+状压dp
AC自动机上统计类的dp问题，dpijs表示在处理主串第i个字符遍历到了AC自动机上的j号节点s且当前构造成的字符串中包含的集合信息s时的方案数。这里的s是一个长度为10的二进制数(表现形式为32位的带符号)，第i位为1则表示构造出的字符串中包含了给定集合中的第i个字符串。若dpijs != 0, dp[i+1][j][s | danger[ch[j][k]] += dp[i][j][s];然后危险节点记录当前以此节点为字符串结束节点的字符串集合(同样是一个长度为10的二进制数)。然后对s里包含至少k个集合中的字符串的dp[n][j][s]求和。复杂度 O(T*n*ac.sz*(1<<10)*CHAR_SIZE)

#include 
#include 
#include 
#include 
#include 
using namespace std;
typedef long long LL;
const int MOD = 20090717;
const int CHAR_SIZE = 26;
const int MAX_SIZE = 1e2 + 8;

inline int mp(char ch){
    return ch - 'a';
}
inline int mod(int x)
{
    return x - x / MOD * MOD;
}

struct AC_Machine{
    int ch[MAX_SIZE][CHAR_SIZE], danger[MAX_SIZE], fail[MAX_SIZE];
    int sz;
    void init(){
        sz = 1;
        memset(ch[0], 0, sizeof ch[0]);
        memset(danger, 0, sizeof danger);
    }
    void _insert(const string &s, int m){
        int n = s.size();
        int u = 0, c;
        for(int i = 0; i < n; i++){
            c = mp(s[i]);
            if(!ch[u][c]){
                memset(ch[sz], 0, sizeof ch[sz]);
                danger[sz] = 0;
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        danger[u] |= 1 << m;
    }
    void _build(){
        queue Q;
        fail[0] = 0;
        for(int c = 0, u; c < CHAR_SIZE; c++){
            u = ch[0][c];
            if(u){Q.push(u); fail[u] = 0;}
        }
        int r;
        while(!Q.empty()){
            r = Q.front();
            Q.pop();
            danger[r] |= danger[fail[r]];
            for(int c = 0, u; c < CHAR_SIZE; c++){
                u = ch[r][c];
                if(!u){ch[r][c] = ch[fail[r]][c]; continue; }
                fail[u] = ch[fail[r]][c];
                Q.push(u);
            }
        }
    }
}ac;

string ss;
int dp[2][MAX_SIZE][1<<10];
inline bool check(int x, int k)
{
    int t, cnt = 0;
    while(x){
        if(x&1) cnt++;
        x >>= 1;
    }
    if(cnt >= k) return true;
    else return false;
}

int main()
{
    #ifdef LOCAL
    freopen("26.in", "r", stdin);
    //freopen("26.out", "w", stdout);
    #endif // LOCAL
    ios::sync_with_stdio(false); cin.tie(0);

    int n, m, kk, len, i, j, k, s, x, u;
    int ans;
    while(true){
        cin >> n >> m >> kk;

        if(n == 0 && m == 0 && kk == 0) break;
        ac.init();
        for(i = 0; i < m; i++){
            cin >> ss;
            //cout << ss << endl;
            ac._insert(ss, i);
        }
        ac._build();

        memset(dp, 0, sizeof dp);
        dp[0][0][0] = 1;
        for(i = 0, x = 1; i < n; i++, x ^= 1){
            memset(dp[x], 0, sizeof dp[x]);
            for(j = 0; j < ac.sz; j++){
                for(s = 0; s < (1<

Thank you!                                                                                                                                              ------from  ProLights