zoj3494BCD Code(ac自动机+数位dp)

l链接

这题想了好一会呢。。刚开始想错了，以为用自动机预处理出k长度可以包含的合法的数的个数，然后再数位dp一下就行了，写到一半发现不对，还要处理当前走的时候是不是为合法的，这一点无法移到trie树上去判断。

之后想到应该在trie树上进行数位dp，走到第i个节点且长度为j的状态是确定的，所以可以根据trie树上的节点来进行确定状态。

dp[i][j]表示当前节点为i，数第j位时可以包含多少个合法的数。

#include <iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<stdlib.h>
#include<vector>
#include<cmath>
#include<queue>
#include<set>
using namespace std;
#define N 2010
#define LL long long
#define INF 0xfffffff
const double eps = 1e-8;
const double pi = acos(-1.0);
const double inf = ~0u>>2;
const int child_num = 2;
const int mod = 1000000009;
int dp[210][N];
char s1[210],s2[210];
class AC
{
    private:
    int ch[N][child_num];
    int Q[N];
    int fail[N];
    int val[N];
    int id[127];
    int sz;
    int dd[810][N];
    public:
    void init()
    {
        fail[0] = 0;
        id['0'] = 0;id['1'] = 1;
    }
    void reset()
    {
        memset(val,0,sizeof(val));
        memset(ch[0],0,sizeof(ch[0]));
        sz=1;
    }
    void insert(char *a,int key)
    {
        int p =0 ;
        for( ; *a ; a++)
        {
            int d = id[*a];
            if(ch[p][d]==0){
                memset(ch[sz],0,sizeof(ch[sz]));
                ch[p][d] = sz++;
            }
            p = ch[p][d];
        }
        val[p] = key;
    }
    void construct()
    {
        int i,head=0,tail = 0;
        for(i = 0 ;i < child_num ; i++)
        {
            if(ch[0][i])
            {
                fail[ch[0][i]] = 0;
                Q[tail++] = ch[0][i];
            }
        }
        while(head!=tail)
        {
            int u = Q[head++];
            val[u]|=val[fail[u]];
            for(i =0 ;i < child_num ; i++)
            {
                if(ch[u][i])
                {
                    fail[ch[u][i]] = ch[fail[u]][i];
                    Q[tail++] = ch[u][i];
                }
                else ch[u][i] = ch[fail[u]][i];
            }
        }
    }
    int dfs(char *s,int i,int c,int e,int k)
    {
        if(i==-1)
        {
            return 1;
        }
        if(!e&&~dp[i][c])
        {
            return dp[i][c];
        }
        int mk = e?s[i]-'0':9;
        int ans = 0;
        for(int j = 0; j <= mk ; j++)
        {
            if(!k&&j==0&&i)
            {
                ans = (ans+dfs(s,i-1,c,e&&j==mk,k));
                continue;
            }
            int p = c,flag = 1;
            for(int g = 3; g >=0 ; g--)
            {
                int o = (j&(1<<g))?1:0;
                p = ch[p][o];
                int tmp = p;
                while(tmp!=0)
                {
                    if(val[tmp])
                    {
                        flag = 0;
                        break;
                    }
                    tmp = fail[tmp];
                }
                if(!flag) break;
            }
            if(flag)
            {
                ans = (ans+dfs(s,i-1,p,e&&j==mk,1))%mod;
            }
        }
        return e?ans:dp[i][c] = ans;
    }
    void work(char *s1,char *s2)
    {
        memset(dp,-1,sizeof(dp));
        printf("%d
",(dfs(s2,strlen(s2)-1,0,1,0)-dfs(s1,strlen(s1)-1,0,1,0)+mod)%mod);
    }
}ac;
char vir[22];
char ss1[210],ss2[210];
int main()
{
    int t,n,i;
    ac.init();
    scanf("%d",&t);
    while(t--)
    {
        ac.reset();
        scanf("%d",&n);
        while(n--)
        {
            scanf("%s",vir);
            ac.insert(vir,1);
        }
        ac.construct();
        scanf("%s%s",s1,s2);
        int k = strlen(s1),kk= strlen(s2);
        for(i = k-1 ; i >= 0; i--)
        {
            if(s1[i]>'0')
            {
                s1[i]-=1;
                break;
            }
            else
            s1[i] = '9';
        }
        for(i = 0; i < k ; i++)
        ss1[k-1-i] = s1[i];
        ss1[k] = '';
        for(i = 0; i < kk ; i++)
        ss2[kk-1-i] = s2[i];
        ss2[kk] = '';
        ac.work(ss1,ss2);
    }
    return 0;
}