Wireless Password
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)Total Submission(s): 3737 Accepted Submission(s): 1133
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.
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
dp[i][j][k]表示长度为i的串在状态j下各串出现情况。
代码：
/* ***********************************************
Author :xianxingwuguan
Created Time :2014-2-3 13:21:05
File Name :1.cpp
************************************************ */
#pragma comment(linker, "/STACK:102400000,102400000")
#include <stdio.h>
#include <iostream>
#include <algorithm>
#include <sstream>
#include <stdlib.h>
#include <string.h>
#include <limits.h>
#include <string>
#include <time.h>
#include <math.h>
#include <queue>
#include <stack>
#include <set>
#include <map>
using namespace std;
#define INF 0x3f3f3f3f
#define eps 1e-8
#define pi acos(-1.0)
typedef long long ll;
int dp[30][110][1<<10],num[1<<10];
const int mod=20090717;
struct Trie{
int next[110][26],fail[110],end[110];
int root,L;
int newnode(){
for(int i=0;i<26;i++)
next[L][i]=-1;
end[L++]=0;
return L-1;
}
void init(){
L=0;
root=newnode();
}
void insert(char *str,int id){
int len=strlen(str);
int now=root;
for(int i=0;i<len;i++){
int p=str[i]-‘a‘;
if(next[now][p]==-1)
next[now][p]=newnode();
now=next[now][p];
}
end[now]|=(1<<id);
}
void build(){
queue<int> q;
fail[root]=root;
for(int i=0;i<26;i++)
if(next[root][i]==-1)
next[root][i]=root;
else {
fail[next[root][i]]=root;
q.push(next[root][i]);
}
while(!q.empty()){
int now=q.front();
q.pop();
end[now]|=end[fail[now]];
for(int i=0;i<26;i++)
if(next[now][i]==-1)
next[now][i]=next[fail[now]][i];
else {
fail[next[now][i]]=next[fail[now]][i];
q.push(next[now][i]);
}
}
}
int solve(int n,int m,int k){
for(int i=0;i<n;i++)
for(int j=0;j<L;j++)
for(int p=0;p<(1<<m);p++)
if(dp[i][j][p])
for(int q=0;q<26;q++){
int newi=i+1;
int newj=next[j][q];
int newp=p|end[newj];
dp[newi][newj][newp]+=dp[i][j][p];
dp[newi][newj][newp]%=mod;
}
int ans=0;
for(int i=0;i<(1<<m);i++){
if(num[i]<k)continue;
for(int j=0;j<L;j++){
ans=(ans+dp[n][j][i])%mod;
}
}
return ans;
}
}ac;
char str[1000];
int main()
{
//freopen("data.in","r",stdin);
//freopen("data.out","w",stdout);
int i,j,k,m,n;
for(i=0;i<1024;i++)
for(j=0;j<10;j++)
if(i&(1<<j))
num[i]++;
while(~scanf("%d%d%d",&n,&m,&k)){
if(n==0&&m==0&&k==0)break;
ac.init();
for(i=0;i<m;i++){
scanf("%s",str);
ac.insert(str,i);
}
ac.build();
for(i=0;i<=n;i++)
for(j=0;j<ac.L;j++)
for(int p=0;p<(1<<m);p++)
dp[i][j][p]=0;
dp[0][0][0]=1;
printf("%d\n",ac.solve(n,m,k));
}
return 0;
}