http://acm.hdu.edu.cn/showproblem.php?pid=2825
Wireless Password Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 4560 Accepted Submission(s): 1381
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
Source
2009 Multi-University Training Contest 1 - Host
by TJU
题意：
给出m个模式串，要求构造一长度为n的文本串，至少包括k种模式串，求有多少种可能的模式串。
分析：
m个模式串构建AC自动机，然后要在这AC自动机中走n步，至少经过k个单词结点。因为m<=10,显然可以用状压表示已经有哪几个单词结点。用dp[i][j][k]表示走了i步到AC自动机中的第j个结点，单词状态为k，由计数原理可推出状态转移方程：dp[i][j][k]=sum(dp[i-1][last_j][last_k]),last_j表示可以抵达第j个结点的上一个结点，last_k表示上一步的状态;因为第i步只和第i-1步有关，所以可以用滚动数组优化空间。
/*
*
* Author : fcbruce <[email protected]>
*
* Time : Thu 20 Nov 2014 10:01:45 AM CST
*
*/
#include <cstdio>
#include <iostream>
#include <sstream>
#include <cstdlib>
#include <algorithm>
#include <ctime>
#include <cctype>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <set>
#define sqr(x) ((x)*(x))
#define LL long long
#define itn int
#define INF 0x3f3f3f3f
#define PI 3.1415926535897932384626
#define eps 1e-10
#ifdef _WIN32
#define lld "%I64d"
#else
#define lld "%lld"
#endif
#define maxm
#define maxn 255
using namespace std;
const int mod = 20090717;
int dp[2][maxn][1024];
inline int add(int a,int b)
{
return (a+b)%mod;
}
int q[maxn];
const int maxsize = 26;
struct ACauto
{
int ch[maxn][maxsize];
int val[maxn],nex[maxn],last[maxn];
int sz;
ACauto()
{
memset(ch[0],0,sizeof ch[0]);
val[0]=0;
sz=1;
}
void clear()
{
memset(ch[0],0,sizeof ch[0]);
val[0]=0;
sz=1;
}
int idx(char c)
{
return c-'a';
}
void insert(const char *s,int v)
{
int u=0;
for (int i=0;s[i]!='\0';i++)
{
int c=idx(s[i]);
if (ch[u][c]==0)
{
memset(ch[sz],0,sizeof ch[sz]);
val[sz]=0;
ch[u][c]=sz++;
}
u=ch[u][c];
}
val[u]=v;
}
void get_fail()
{
int f=0,r=-1;
nex[0]=0;
for (int c=0;c<maxsize;c++)
{
int u=ch[0][c];
if (u!=0)
{
nex[u]=0;
q[++r]=u;
last[u]=0;
}
}
while (f<=r)
{
int x=q[f++];
for (int c=0;c<maxsize;c++)
{
int u=ch[x][c];
if (u==0)
{
ch[x][c]=ch[nex[x]][c];
continue;
}
q[++r]=u;
int v=nex[x];
nex[u]=ch[v][c];
val[u]|=val[nex[u]];
}
}
}
int calc(int x)
{
int cnt=0;
for (int i=0;i<32;i++)
if (x&(1<<i)) cnt++;
return cnt;
}
int DP(int l,int m,int k)
{
memset(dp,0,sizeof dp);
dp[0][0][0]=1;
int x=1;
for (int i=0;i<l;i++,x^=1)
{
memset(dp[x],0,sizeof dp[x]);
for (int j=0;j<sz;j++)
{
for (int s=0;s<m;s++)
{
if (dp[x^1][j][s]==0) continue;
for (int c=0;c<maxsize;c++)
{
int &cur=dp[x][ch[j][c]][s|val[ch[j][c]]];
cur=add(cur,dp[x^1][j][s]);
}
}
}
}
int total=0;
for (int i=0;i<m;i++)
{
if (calc(i)<k) continue;
for (int j=0;j<sz;j++)
total=add(total,dp[x^1][j][i]);
}
return total;
}
}acauto;
char str[16];
int main()
{
#ifdef FCBRUCE
freopen("/home/fcbruce/code/t","r",stdin);
#endif // FCBRUCE
int n,m,k;
while (scanf("%d%d%d",&n,&m,&k),n||m||k)
{
acauto.clear();
for (int i=0;i<m;i++)
{
scanf("%s",str);
acauto.insert(str,1<<i);
}
acauto.get_fail();
printf("%d\n",acauto.DP(n,1<<m,k));
}
return 0;
}