AC自动机+状态压缩DP+hdu2825

最新推荐文章于 2019-10-14 20:23:54 发布

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寻人启事：2014级新生看过来！

Wireless Password

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4381 Accepted Submission(s): 1323

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

题意：构造一个长度为n的字符串，其中至少包含k个给出的串，文有多少种构造方式

思路：这个跟前面的几个题很像，前面的是构造的串不包含给出的字符串，可以矩阵或者dp，这里可以用状态压缩，记录下构造出来的串包含哪些字符串，就跟前面一样了

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
const int maxn=110;
const int MOD=20090717;
const int SIGMA_SIZE=26;
int N,M,K;
char word[20];
int dp[30][110][1<<10],num[5000];

struct AC
{
    int ch[maxn][SIGMA_SIZE];
    int fail[maxn];
    int val[maxn];
    int sz;
    void clear(){memset(ch[0],0,sizeof(ch[0]));sz=1;}
    int idx(char x){return x-'a';}
    void insert(char *s,int id)
    {
        int u=0,n=strlen(s);
        for(int i=0;i<n;i++)
        {
            int c=s[i]-'a';
            if(!ch[u][c])
            {
                memset(ch[sz],0,sizeof(ch[sz]));
                val[sz]=0;
                ch[u][c]=sz++;
            }
            u=ch[u][c];
        }
        val[u]|=(1<<id);
    }
    void getfail()
    {
        queue<int> q;
        int u=0;
        fail[0]=0;
        for(int i=0;i<SIGMA_SIZE;i++)
        {
            u=ch[0][i];
            if(u){fail[u]=0;q.push(u);}
        }
        while(!q.empty())
        {
            int r=q.front();q.pop();
            val[r]|=val[fail[r]];//不要忘记这里
            for(int c=0;c<SIGMA_SIZE;c++)
            {
                u=ch[r][c];
                if(!u){ch[r][c]=ch[fail[r]][c];continue;}
                q.push(u);
                int v=fail[r];
                while(v&&!ch[v][c])v=fail[v];
                fail[u]=ch[v][c];
            }
        }
    }
    void solve()
    {
        for(int i = 0;i <= N;i++)
            for(int j = 0;j < sz;j++)
                for(int p = 0;p < (1<<M);p++)
                    dp[i][j][p]=0;
        //memset(dp,0,sizeof(dp);用这个会超时
        dp[0][0][0]=1;
        for(int i=0;i<N;i++)
        {
            for(int j=0;j<sz;j++)
            {
                for(int k=0;k<(1<<M);k++)
                {
                    if(dp[i][j][k]<=0)continue;
                    for(int c=0;c<SIGMA_SIZE;c++)
                    {
                        int p=(k|val[ch[j][c]]);
                        dp[i+1][ch[j][c]][p]+=dp[i][j][k];
                        dp[i+1][ch[j][c]][p]%=MOD;
                    }
                }
            }
        }
        int ans=0;
        for(int p=0;p<(1<<M);p++)
        {
            if(num[p]<K)continue;
            for(int i=0;i<sz;i++)
                ans=(ans+dp[N][i][p])%MOD;
        }
        printf("%d\n",ans);
    }
};
AC ac;
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]++;
    }
    while(scanf("%d%d%d",&N,&M,&K)==3)
    {
        if(N==0&&M==0&&K==0)break;
        ac.clear();
        for(int i=0;i<M;i++)
        {
            scanf("%s",word);
            ac.insert(word,i);
        }
        ac.getfail();
        ac.solve();
    }
    return 0;
}