Wireless Password
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 5640 Accepted Submission(s): 1785
Problem 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
/*
hdu 2825 aC自动机+状压dp
给你m个子串,求长度为n的主串中至少出现k个子串的方案数
首先通过AC自动机构建关系图. 然后用dp解决状态转移,需要知道用过哪些子串
因为k比较小,我们直接转换成二进制来记录当前状态包含了哪些子串。用ed对各子串进行标记
dp[i][j][t]就表示长度为i,当前位置上是j时,所包含子串的情况t
hhh-2016-04-24 17:13:36
*/
#include <iostream>
#include <vector>
#include <cstring>
#include <string>
#include <cstdio>
#include <queue>
#include <algorithm>
#include <functional>
#include <map>
using namespace std;
#define lson (i<<1)
#define rson ((i<<1)|1)
typedef unsigned long long ll;
typedef unsigned int ul;
const int mod = 20090717;
const int INF = 0x3f3f3f3f;
int tot;
int dp[30][111][1<<10];
struct Matrix
{
int len;
int ma[111][111];
Matrix() {};
Matrix(int L)
{
len = L;
}
};
struct Tire
{
int nex[110][26],fail[110],ed[110];
int root,L;
int newnode()
{
for(int i = 0; i < 26; i++)
nex[L][i] = -1;
ed[L++] = 0;
return L-1;
}
void ini()
{
L = 0,root = newnode();
}
int cal(char ch)
{
if(ch == ‘A‘)
return 0;
else if(ch == ‘C‘)
return 1;
else if(ch == ‘G‘)
return 2;
else if(ch == ‘T‘)
return 3;
}
void inser(char buf[],int id)
{
int len = strlen(buf);
int now = root;
for(int i = 0; i < len; i++)
{
int ta = buf[i] - ‘a‘;
if(nex[now][ta] == -1)
nex[now][ta] = newnode();
now = nex[now][ta];
}
ed[now] |= (1<<id);
}
void build()
{
queue<int >q;
fail[root] = root;
for(int i = 0; i < 26; i++)
if(nex[root][i] == -1)
nex[root][i] = root;
else
{
fail[nex[root][i]] = root;
q.push(nex[root][i]);
}
while(!q.empty())
{
int now = q.front();
q.pop();
if(ed[fail[now]])
ed[now] |= ed[fail[now]];
for(int i = 0; i < 26; i++)
{
if(nex[now][i] == -1)
nex[now][i] = nex[fail[now]][i];
else
{
fail[nex[now][i]] = nex[fail[now]][i];
q.push(nex[now][i]);
}
}
}
}
Matrix to_mat()
{
Matrix mat(L);
memset(mat.ma,0,sizeof(mat.ma));
for(int i = 0; i < L; i++)
{
for(int j = 0; j < 4; j++)
{
if(!ed[nex[i][j]])
mat.ma[i][nex[i][j]] ++;
}
}
return mat;
}
};
//Matrix mat;
Tire ac;
char buf[22];
void debug()
{
Matrix t = ac.to_mat();
for(int i = 0; i < t.len; i++)
{
for(int j = 0; j < 26; j++)
{
printf("%d ",t.ma[i][ac.nex[i][j]]);
}
printf("\n");
}
}
int num[1<<10];
int main()
{
for(int i=0; i<(1<<10); i++)
{
num[i] = 0;
for(int j = 0; j < 10; j++)
if(i & (1<<j))
num[i]++;
}
int n,m,p;
while(scanf("%d%d%d",&n,&m,&p) != EOF)
{
if(!n && !m && !p)
break;
ac.ini();
for(int i = 0; i < m; i++)
{
scanf("%s",buf);
ac.inser(buf,i);
}
ac.build();
for(int i = 0; i <= n; i++)
{
for(int j = 0; j <ac.L; j++)
{
for(int k = 0; k < (1<<m); k++)
dp[i][j][k] = 0;
}
}
dp[0][0][0] = 1;
for(int i = 0; i < 2; i++)
{
for(int j = 0; j < ac.L; j++)
{
for(int t = 0; t < (1<<m); t++)
{
if(dp[i][j][t] > 0)
for(int k = 0; k < 26; k++)
{
int nexi = i+1;
int nexj = ac.nex[j][k];
int nexk = (t|ac.ed[nexj]);
dp[nexi][nexj][nexk] = (dp[nexi][nexj][nexk] + dp[i][j][t])%mod;
}
}
}
}
int ans = 0;
for(int j = 0; j < (1<<m); j++)
{
if(num[j] < p)
continue;
for(int i = 0; i < ac.L; i++)
ans = (ans+dp[n][i][j])%mod;
}
printf("%d\n",ans);
}
return 0;
}