【HDU2825】Wireless Password (AC自动机+状压DP)

Wireless Password
Time Limit: 1000MS   Memory Limit: 32768KB   64bit IO Format: %I64d & %I64u

Description

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. 
 

Input

There 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.
 

Output

For 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
 

 

【题意】

　　有个人想破解他邻居的密码，他邻居告诉了一些关于这个密码的信息，并且给他一个单词集合，他用这些信息判断一下最少有多少种密码。
　　1->, 所有的密码都是有小写字母组成。
　　2->，密码的长度是 n （1<= n <=25）。
　　3->，密码至少包含 k 种字符集里面的单词。
 
　　比如，给集合｛"she", "he"｝，单词长度是3，最少包含两个单词的密码,很明显只能是“she”（题目表述的不清楚）。
 
【分析】

　　先建AC自动机。
　　DP部分　令 dp[i][j][l] 表示长度为 i ，跑到自动机节点 j ，已经包含单词为集合l的方案数。（因为k比较小）
 
代码如下：
  1 #include<cstdio>
  2 #include<cstdlib>
  3 #include<cstring>
  4 #include<iostream>
  5 #include<algorithm>
  6 #include<queue>
  7 using namespace std;
  8 #define Maxn 1010
  9 #define Maxl 1010
 10 #define INF 0xfffffff
 11 #define Mod 20090717
 12 
 13 int n,m,ak;
 14 char s[210];
 15 
 16 struct node
 17 {
 18     int cnt,fail,mark;
 19     int son[30];
 20 }t[Maxn];int tot;//=0;
 21 int num;
 22 int p[210];
 23 
 24 void upd(int x)
 25 {
 26     t[x].cnt=0;t[x].mark=0;
 27     memset(t[x].son,0,sizeof(t[x].son));
 28 }
 29 
 30 int mymin(int x,int y) {return x<y?x:y;}
 31 
 32 void read_trie()
 33 {
 34     scanf("%s",s+1);
 35     int len=strlen(s+1);
 36     int now=0;
 37     for(int i=1;i<=len;i++)
 38     {
 39         int ind=s[i]-'a'+1;
 40         if(!t[now].son[ind]) 
 41         {
 42             t[now].son[ind]=++tot;
 43             upd(tot);
 44         }
 45         now=t[now].son[ind];
 46         if(i==len) 
 47         {
 48             t[now].mark=1<<(++num)-1;
 49         }
 50     }
 51 }
 52 
 53 queue<int > q;
 54 void build_AC()
 55 {
 56     while(!q.empty()) q.pop();
 57     q.push(0);//inq[0]=1;
 58     while(!q.empty())
 59     {
 60         int x=q.front();q.pop();
 61         for(int i=1;i<=26;i++) 
 62         {
 63             if(t[x].son[i])
 64             {
 65                 t[t[x].son[i]].fail=x?t[t[x].fail].son[i]:0;
 66                 q.push(t[x].son[i]);
 67             }
 68             else t[x].son[i]=t[t[x].fail].son[i];
 69             t[x].mark|=t[t[x].fail].mark;
 70         }
 71     }
 72 }
 73 
 74 bool check(int x)
 75 {
 76     int h=0;
 77     for(int i=1;i<=num;i++) if(x&(1<<i-1)) h++;
 78     if(h>=ak) return 1;
 79     return 0;
 80 }
 81 
 82 int f[30][260][2500];
 83 void dp()
 84 {
 85     for(int i=0;i<=n;i++)
 86      for(int j=0;j<=tot;j++)
 87       for(int l=0;l<=(1<<num)-1;l++) f[i][j][l]=0;
 88     f[0][0][0]=1;
 89     for(int i=1;i<=n;i++)
 90      for(int j=0;j<=tot;j++) 
 91       for(int l=0;l<=(1<<num)-1;l++) if(f[i-1][j][l]!=0)
 92       {
 93          for(int k=1;k<=26;k++) 
 94          {
 95                f[i][t[j].son[k]][l|t[t[j].son[k]].mark]=(f[i][t[j].son[k]][l|t[t[j].son[k]].mark]+f[i-1][j][l])%Mod;
 96          }
 97      }
 98     int ans=0;
 99     for(int i=0;i<=(1<<num)-1;i++) if(check(i))
100     {
101         for(int j=0;j<=tot;j++) ans=(ans+f[n][j][i])%Mod;
102     }
103     printf("%d\n",ans);
104 }
105 
106 void init()
107 {
108     tot=0;num=0;
109     upd(0);
110     for(int i=1;i<=m;i++)
111     {
112         read_trie();
113     }
114     build_AC();
115 }
116 
117 int main()
118 {
119     int kase=0;
120     while(1)
121     {
122         scanf("%d%d%d",&n,&m,&ak);
123         if(n==0&&m==0&&ak==0) break;
124         init();
125         dp();
126     }
127     return 0;
128 }
[HDU2825]

 

2016-07-11 14:04:35

posted @ 2016-07-11 14:01  konjak魔芋  阅读(445)  评论(0)  编辑  收藏  举报