HDU2825 Wireless Password 【AC自动机】【状压DP】
HDU2825 Wireless Password
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
我觉得挺好的一道题。。一开始只知道是DP，然后写了一个循环，后面发现不能重复计算，然后改成了状压DP，改成了队列实现DP，上线是步数小于等于长度，然后每一次DP转移将步数加1就好了，然后或一下当前存在的状态。
处理fail指针的时候做两个小优化，一个是在处理的时候就把fail链上的id全部统计起来，二个是把失配后跳到的节点直接储存成儿子节点
然后注意开一下取模的操作，防止ＴＬＥ
#include<bits/stdc++.h>
using namespace std;
#define N 110
#define Mod 20090717
struct Node{
int fail,id,ch[26];
void clean(){
fail=id=0;
memset(ch,0,sizeof(ch));
}
}t[N];
struct Node1{int id,tmp,now;};
int n,m,p,tot=0;
int dp[N][26][1025],c[1025];
bool vis[N][26][1025];
char s[20][N];
void init(){
tot=1;
t[0].clean();t[1].clean();
for(int i=0;i<26;i++)t[0].ch[i]=1;
memset(dp,0,sizeof(dp));
}
int index(char c){return c-'a';}
void insert(char *s,int id){
int len=strlen(s),u=1;
for(int i=0;i<len;i++){
int tmp=index(s[i]);
if(!t[u].ch[tmp])
t[t[u].ch[tmp]=++tot].clean();
u=t[u].ch[tmp];
}
t[u].id=1<<(id-1);
}
void buildFail(){
queue<int> q;q.push(1);
while(!q.empty()){
int u=q.front();q.pop();
for(int i=0;i<26;i++){
int w=t[u].ch[i],v=t[u].fail;
while(!t[v].ch[i])v=t[v].fail;
v=t[v].ch[i];
if(w){
t[w].fail=v;
t[w].id|=t[v].id;//***
q.push(w);
}else t[u].ch[i]=v;//***
}
}
}
void solve(){
queue<Node1> qn;
qn.push((Node1){1,0,0});
dp[1][0][0]=1;
while(!qn.empty()){
Node1 u=qn.front();qn.pop();
vis[u.id][u.tmp][u.now]=0;
if(u.tmp>n)continue;
for(int i=0;i<26;i++){
int id=t[u.id].ch[i],tmp=u.tmp+1,now=u.now|t[id].id;
dp[id][tmp][now]+=dp[u.id][u.tmp][u.now];
if(dp[id][tmp][now]>Mod)dp[id][tmp][now]-=Mod;
if(!vis[id][tmp][now]){
vis[id][tmp][now]=1;
qn.push((Node1){id,tmp,now});
}
}
}
}
int count(int t){
int ans=0;
while(t){
if(t&1)ans++;
t>>=1;
}
return ans;
}
int main(){
for(int i=0;i<1024;i++)c[i]=count(i);
while(1){
scanf("%d%d%d",&n,&m,&p);
if(!n)break;
init();//初始化
for(int i=1;i<=m;i++){
scanf("%s",s[i]);
insert(s[i],i);
}
buildFail();
solve();
int ans=0,up=1<<m;
for(int i=0;i<up;i++)if(c[i]>=p)
for(int j=1;j<=tot;j++){
ans+=dp[j][n][i];
if(ans>Mod)ans-=Mod;
}
printf("%d\n",ans);
}
return 0;
}
