[JSOI2009]密码 [AC自动机]

题面

bzoj
luogu

首先看到这题就知道随便暴枚
只要是多项式算法都能过
先常规建AC自动机
注意被别的单词包含的单词没有存在的价值
剩余单词状压

大力dp f[长度][节点编号][状态]
\(ans = \sum f[m][i][S]\)
这里把题面的l换成m了 表示密码长度

如果方案数小于等于42的话
说明这个密码是给定词拼成的 不会有自由字母
那么就逆向找到转移到它的状态 记录密码就好啦

注意比较那里原来写的是
if(x.s[i] > y.s[i]) return 1;
显然这样是不行的啊qvq

#include <cmath>
#include <cstring>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <complex>
#include <ctime>
#include <vector>
#include <queue>
#include <bitset>
#define mp(x, y) make_pair(x, y)
using namespace std;
const int N = 12;
const int M = 102;
const int Sig = 26;
int m, n, acsize, vsize, S;
int en[N], c[M], val[M];
long long ans, f[26][M][(1 << 10) + 5];
 
struct STR{
    char s[30];
    void print(){s[m + 1] = '\0'; printf("%s\n", s + 1);}
    friend bool operator <(STR x, STR y){
        for(int i = 1; i <= m; ++i)
            if(x.s[i] != y.s[i]) return x.s[i] > y.s[i];
        return 0;
    }//大于 
}stk;
vector<STR> str;
 
struct AC{
    int ch[M][Sig], fail[M];
    queue<int> que;
    void ins(char* ss, int id){
        int now = 0, len = strlen(ss + 1);
        for(int i = 1, cc; i <= len; ++i){
            cc = ss[i] - 'a';
            if(!ch[now][cc]) ch[now][cc] = ++acsize;
            now = ch[now][cc]; c[now] = cc;
        }
        en[id] = now, val[now] = 1;
    }
    void build(){
        int now = 0;
        for(int i = 0; i < Sig; ++i) if(ch[0][i]) que.push(ch[0][i]);
        while(!que.empty()){
            int fro = que.front(); que.pop();
            for(int i = 0; i < Sig; ++i){
                if(ch[fro][i]) fail[ch[fro][i]] = ch[fail[fro]][i], que.push(ch[fro][i]);//!!
                else ch[fro][i] = ch[fail[fro]][i];
            }
        }
        for(int i = 0; i <= acsize; ++i) val[fail[i]] = 0;
        for(int i = 0; i <= acsize; ++i) if(val[i])
            val[i] = (1 << vsize), ++vsize;
        S = (1 << vsize) - 1;
    }
    void DP(){ 
        f[0][0][0] = 1;
        for(int i = 0; i < m; ++i)
            for(int j = 0; j <= acsize; ++j)
                for(int k = 0; k <= S; ++k) if(f[i][j][k]){  
        for(int t = 0; t < Sig; ++t){
            f[i + 1][ch[j][t]][k | val[ch[j][t]]] += f[i][j][k];
              //  printf("%d %d %d %lld\n", i, j, k, f[i + 1][ch[j][t]][k | val[ch[j][t]]]);
        }
                }
    }
}ac;
 
void dfs(int x, int cur, int y, int z){
    stk.s[x] = z + 'a';
    if(x == 1){str.push_back(stk); return ;}
    for(int i = 0; i <= acsize; ++i)
        if(f[x - 1][i][y] > 0 && ac.ch[i][z] == cur)
            dfs(x - 1, i, y, c[i]);
    if(val[cur])
        for(int i = 0; i <= acsize; ++i)
            if(f[x - 1][i][y ^ val[cur]] > 0 && ac.ch[i][z] == cur)
                dfs(x - 1, i, y ^ val[cur], c[i]);
}
 
int main(){
    scanf("%d%d", &m, &n);
    char ss[N];
    for(int i = 1; i <= n; ++i){
        scanf("%s", ss + 1);
        ac.ins(ss, i);
    }
    ac.build();
    ac.DP();
     
    for(int i = 0; i <= acsize; ++i)
        ans += f[m][i][S];
    printf("%lld\n", ans);
    if(ans > 42) return 0;
     
    for(int i = 0; i <= acsize; ++i)
        if(f[m][i][S]) dfs(m, i, S, c[i]);
    sort(str.begin(), str.end());
    while(!str.empty()){
        (str.back()).print(); str.pop_back();
    }
    return 0;   
}