hdu2825Wireless Password(ac+dp)

链接

状压dp+ac

dp[i+1][next[j]][st|tt]表示第i+1长度结点为next[j]状态为st|tt的时候的ans;

dp[i+1][next[j]][st|tt]+=dp[i][j][tt]; st记录当前结点是否为给定单词的结束点

后一维用01状态表示截止到目前结点为止所包含的单词数量。

需要修改ac模板的一个地方，val[u]|=val[fail[u]];

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 115
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 const int child_num = ;
 const int mod = ;
 class AC
 {
     private:
     int ch[N][child_num];
     int fail[N];
     int Q[N];
     int val[N];
     int sz;
     int id[];
     int dp[][N][<<];
     public:
     void init()
     {
         fail[] =;
         for(int i =  ;i < child_num ; i++)
         id[i+'a'] = i;
     }
     void reset()
     {
         memset(val,,sizeof(val));
         memset(ch[],,sizeof(ch[]));
         //memset(fail,0,sizeof(fail));
         sz = ;
     }
     void insert(char *a,int key)
     {
         int p = ;
         for(; *a ; a++)
         {
             int d = id[*a];
             if(ch[p][d]==)
             {
                 memset(ch[sz],,sizeof(ch[sz]));
                 ch[p][d] = sz++;
             }
             p = ch[p][d];
         }
         val[p] = (<<key);
     }
     void construct()
     {
         int i,head=,tail = ;
         for(i =  ;i < child_num ; i++)
         {
             if(ch[][i])
             {
                 fail[ch[][i]] = ;
                 Q[tail++] = ch[][i];
             }
         }
         while(head!=tail)
         {
             int u  = Q[head++];
             val[u]|=val[fail[u]];
             for(i =  ;i < child_num ;i++)
             {
                 if(ch[u][i])
                 {
                     fail[ch[u][i]] = ch[fail[u]][i];
                     Q[tail++] = ch[u][i];
                 }
                 else ch[u][i] = ch[fail[u]][i];
             }
         }
     }
     void work(int n,int k,int m)
     {
         int i,j,g,o;
         for(i =  ; i <= n ;i++)
             for(j =  ; j <= sz ; j++)
                 for(g =  ;g  <=(<<m) ; g++)
                 dp[i][j][g] = ;
         dp[][][] = ;
         for(i = ; i < n ;i++)
             for(j = ; j < sz ; j++)
             {
                 for(int e = ; e < (<<m) ; e++)
                 {
                     if(!dp[i][j][e]) continue;
                     for(g =  ;g < child_num ; g++)
                     {
                         o = ;
                         if(val[ch[j][g]])
                         {
                             o = val[ch[j][g]];
                         }
                         dp[i+][ch[j][g]][e|o]=(dp[i+][ch[j][g]][e|o]+dp[i][j][e])%mod;
                     }
                 }
             }
         int ans = ;
         for(i =  ; i < sz;i ++)
         {
             for(j =  ;j < (<<m) ; j++)
             {
                 if(!dp[n][i][j]) continue;
                 int o = ;
                 for(g =  ;g < m ; g++)
                 if(j&(<<g)) o++;
                 if(o>=k)
                 ans = (ans+dp[n][i][j])%mod;
             }
         }
         printf("%d\n",ans);
     }
 }ac;
 char vir[];
 int main()
 {
     int n,m,k,i;
     ac.init();
     while(scanf("%d%d%d",&n,&m,&k)!=EOF)
     {
         if(!n&&!m&&!k) break;
         ac.reset();
         for(i = ; i <= m; i++)
         {
             scanf("%s",vir);
             ac.insert(vir,i-);
         }
         ac.construct();
         ac.work(n,k,m);
     }
     return ;
 }