Wireless Password HDU

Wireless Password

 HDU - 2825 

ac自动机　+ 状压DP

dp[i][j][k]表示　长度为ｉ在节点ｊ包含串状态为ｋ时的方案数

dp[i][j][k | s] += dp[i-1][u][k] , 其中ｕ是ｊ的父节点，　ｓ是走到ｊ节点时新加入的串的状态

#include <bits/stdc++.h>
using namespace std;
const int sigma = 26;
const int maxnode = 110;
const int mod = 20090717;
int dp[26][110][1050];
int n, m, k;
int sumbit[1050];

struct AC{
    int ch[maxnode][sigma];
    int f[maxnode], mc[maxnode];
    int sz;

    void init(){
        memset(mc, 0, sizeof(mc));
        memset(ch[0], 0, sizeof(ch[0]));
        sz = 1;
    }

    int idx(char c) {
        return c - 'a';
    }

    void insert(char *s, int id) {
        int n = strlen(s), u = 0;
        for(int i = 0; i < n; i++){
            int c = idx(s[i]);
            if(!ch[u][c]){
                memset(ch[sz], 0, sizeof(ch[sz]));
                ch[u][c] = sz++;
            }
            u = ch[u][c];
        }
        mc[u] |= (1<<id);
    }
    void getfail(){
        queue<int> q;
        f[0] = 0;
        for(int c = 0; c < sigma; c++){
            int u = ch[0][c];
            if(u){
                q.push(u);
                f[u] = 0;
            }
        }
        while(!q.empty()){
            int r = q.front();
            q.pop();
            for(int c = 0; c < sigma; c++){
                int u = ch[r][c];
                if(!u) {
                    ch[r][c] = ch[f[r]][c];
                    continue;
                }
                q.push(u);
                int v = f[r];
               // while(v && !ch[v][c]) v = f[v];
                f[u] = ch[v][c];
                mc[u] |= mc[f[u]];
            }
        }
    }
    int solve(){
        //memset(dp, 0, sizeof(dp));
        for(int i = 0; i <= n; i++){
            for(int j = 0; j < sz; j++){
                for(int sta = 0; sta < (1<<m); sta++){
                    dp[i][j][sta] = 0;
                }
            }
        }
        dp[0][0][0] = 1;
        for(int i = 1; i <= n; i++){
            for(int j = 0; j < sz; j++){
                for(int sta = 0; sta < (1<<m); sta++){
                    if(dp[i-1][j][sta] == 0) continue;
                    for(int c = 0; c < sigma; c++){
                        int v = ch[j][c];
                        dp[i][v][sta | mc[v]] += dp[i-1][j][sta];
                        dp[i][v][sta | mc[v]] %= mod; 
                    }
                }
            }
        }
        int ans = 0;
        for(int i = 0; i < (1<<m); i++){
            if(sumbit[i] < k) continue;
            for(int j = 0; j < sz; j++){
                ans = (ans + dp[n][j][i]) % mod;
            }
        }
        return ans;
    }
};
AC ac;
int main(){
    //freopen("in.txt", "r", stdin);
    for(int i = 0; i < 1050; i++){
        sumbit[i] = 0;
        for(int j = 0; j < 11; j++){
            if(i & (1<<j)) sumbit[i]++; 
        }
    }
    while(scanf("%d %d %d", &n, &m, &k) && (n+m+k)){
        ac.init();
        char s[15];
        for(int i = 0; i < m; i++){
            scanf("%s", s);
            ac.insert(s, i);
        }
        ac.getfail();
        int ans = ac.solve();
        printf("%d
", ans);
    }
    return 0;
}