hdu 2825

题解：

ac自动机+dp的题目

差不多都一个套路

记录枚举了i位，匹配到自动机上的x位，然后对于匹配了哪些单词状态压缩一下就可以了

代码：

#include <bits/stdc++.h>
using namespace std;
#define mo 20090717 
int dp[27][250][1030],w[1000],c[500][27];
int cnt,fail[1000],pp[1030];
char s[1000009];
void insert(char s[1000009],int x)
{
    int len=strlen(s),now=0;
    for (int i=0;i<len;i++)
    {
       int v=s[i]-'a';
       if (!c[now][v]) c[now][v]=++cnt;
       now=c[now][v];  
    }  
    w[now]=1<<(x-1);
} 
queue<int> q;
void build()
{
  for (int i=0;i<26;i++)
    if (c[0][i]) fail[c[0][i]]=0,q.push(c[0][i]);
  while (!q.empty())
  {
    int u=q.front();q.pop();
    for (int i=0;i<26;i++)
    {
      if (c[u][i])
      {
          fail[c[u][i]]=c[fail[u]][i];
          q.push(c[u][i]);
      }  else c[u][i]=c[fail[u]][i];
      w[c[u][i]]|=w[c[fail[u]][i]];
    }
  }
}
int main()
{
  freopen("noip.in","r",stdin);
  freopen("noip.out","w",stdout);
  for (int i=1;i<=1028;i++)
  {
    int l=0;
    for (int j=0;j<=10;j++)
      if ((i>>j)%2==1) l++;
    pp[i]=l;
  }
  int n,m,k;
  while (cin>>n>>m>>k&&!(n==0&&m==0&&k==0))
  {
    memset(fail,0,sizeof(fail));
    memset(c,0,sizeof(c));
    memset(w,0,sizeof(w)); cnt=0;
    for (int i=1;i<=m;i++)
    {
      cin>>s; insert(s,i);
    }
    build();
    memset(dp,0,sizeof(dp));
    dp[0][0][0]=1;
    for (int i=0;i<=n;i++)
      for (int j=0;j<=cnt;j++)
        for (int k1=0;k1<=(1<<m)-1;k1++)
          if (dp[i][j][k1])
            for (int k2=0;k2<26;k2++)
            {
              int u=c[j][k2];
              dp[i+1][u][k1|w[u]]+=dp[i][j][k1];
              dp[i+1][u][k1|w[u]]%=mo;
              
            }
   /* for (int i=0;i<=n;i++)
      for (int j=0;j<=cnt;j++)
        for (int k1=1;k1<=(1<<m)-1;k1++)
          if (dp[i][j][k1])
            cout<<i<<" "<<j<<" "<<k1<<endl; */
    int ans=0;
    for (int i=0;i<=(1<<m)-1;i++)
      if (pp[i]>=k)
      for (int j=0;j<=cnt;j++) 
      {
        ans+=dp[n][j][i];
        ans%=mo;
      }
    cout<<ans<<" "<<cnt<<endl;
  }
  return 0;
}