ZOJ3494: BCD Code 题解

要使得数字的每一位二进制表示连起来的字符串中不出现某些给定的字符串，这是AC自动机+dp的模板题

所以我们考虑数位dp,dp[i][0/1]表示当前考虑到十进制数的第i位，在AC自动机上的第j号节点，当前数是否紧贴上界的合法数的个数

在AC自动机上判断一下是否合法，然后转移就好

注意卡常数(比如提前对AC自动机的每个节点预处理好是否危险，再例如提前预处理从AC自动机的j出发加十进制位i会走到哪里)

#include <cstdio>
#include <iostream>
#include <cstring>
#include <string>
#include <cstdlib>
#include <utility>
#include <cctype>
#include <algorithm>
#include <bitset>
#include <set>
#include <map>
#include <vector>
#include <queue>
#include <deque>
#include <stack>
#include <cmath>
#define LL long long
#define LB long double
#define x first
#define y second
#define Pair pair<int,int>
#define pb push_back
#define pf push_front
#define mp make_pair
#define LOWBIT(x) x & (-x)
using namespace std;

const int MOD=1e9+9;
const LL LINF=2e16;
const int INF=2e9;
const int magic=348;
const double eps=1e-10;
const double pi=3.14159265;

inline int getint()
{
    char ch;int res;bool f;
    while (!isdigit(ch=getchar()) && ch!='-') {}
    if (ch=='-') f=false,res=0; else f=true,res=ch-'0';
    while (isdigit(ch=getchar())) res=res*10+ch-'0';
    return f?res:-res;
}

struct node
{
	bool isdanger[5];
	int fail;bool danger;
	int next[5];bool exist[5];
	int father;
}trie[10048];
int tot=1;

LL DP[248][2048][2];
LL dp[248][2048][2];
LL tans[248];
char A[248],B[248];int lena,lenb;
char s[248];
int num[10][10];
int To[2048][12];
int n;

inline void TEN_TO_TWO(int x)
{
	int ind=x;
	for (int i=1;i<=4;i++) num[ind][i]=0;
	int pt=4;
	while (x)
	{
		num[ind][pt--]=x%2;
		x/=2;
	}
}

void Clear()
{
	tot=1;
	trie[tot].father=tot;
	trie[tot].fail=0;trie[tot].danger=false;
	for (int i=1;i<=2;i++) trie[tot].next[i]=0,trie[tot].exist[i]=false;
}

void Insert(int len)
{
	int i,j,cur=1;
	for (i=1;i<=len;i++)
	{
		if (!trie[cur].next[s[i]-'0'+1])
		{
			trie[cur].next[s[i]-'0'+1]=++tot;
			trie[cur].exist[s[i]-'0'+1]=true;
			trie[tot].father=cur;
			trie[tot].fail=0;trie[tot].danger=false;
			for (j=1;j<=2;j++) trie[tot].next[j]=0,trie[tot].exist[j]=false;
		}
		cur=trie[cur].next[s[i]-'0'+1];
		if (i==len) trie[cur].danger=true;
	}
}

queue<int> q;
void construct_fail()
{
	int i,cur,tmp;
	q.push(1);
	while (!q.empty())
	{
		cur=q.front();q.pop();
		for (i=1;i<=2;i++)
			if (trie[cur].next[i])
			{
				tmp=trie[cur].fail;
				while (tmp && !trie[tmp].next[i]) tmp=trie[tmp].fail;
				trie[trie[cur].next[i]].fail=(tmp?trie[tmp].next[i]:1);
				q.push(trie[cur].next[i]);
			}
			else
				trie[cur].next[i]=(cur==1?1:trie[trie[cur].fail].next[i]);
	}
}

void dfs(int cur)
{
	int i,tmp;
	for (i=1;i<=2;i++)
	{
		tmp=cur;
		trie[cur].isdanger[i]=false;
		while (tmp)
		{
			if (trie[trie[tmp].next[i]].danger)
			{
				trie[cur].isdanger[i]=true;
				break;
			}
			tmp=trie[tmp].fail;
		}
	}
	for (i=1;i<=2;i++)
		if (trie[cur].exist[i]) dfs(trie[cur].next[i]);
}

int Go(int cur,int dir)
{
	if (trie[cur].isdanger[dir+1]) return -1;
	return trie[cur].next[dir+1];
}

int check(int cur,int add)
{
	int i;
	for (i=1;i<=4;i++)
	{
		cur=Go(cur,num[add][i]);
		if (cur==-1) return cur;
	}
	return cur;
}

void init_To()
{
	int i,cur;
	for (i=1;i<=tot;i++)
		for (cur=0;cur<=9;cur++)
			To[i][cur]=check(i,cur);
}

void init_dp()
{
	int i,j,cur,to;
	memset(dp,0,sizeof(dp));
	dp[0][1][0]=1;
	for (i=0;i<=max(lena,lenb)-1;i++)
		for (j=1;j<=tot;j++)
			if(dp[i][j][0])
				for (cur=0;cur<=9;cur++)
				{
					if (cur==0 && i==0) continue;
					to=To[j][cur];
					if (to!=-1)
						dp[i+1][to][0]=(dp[i+1][to][0]+dp[i][j][0])%MOD;
				}
	for (i=1;i<=max(lena,lenb);i++)
	{
		tans[i]=0;
		for (j=1;j<=tot;j++)
			tans[i]=(tans[i]+dp[i][j][0])%MOD;
	}
}

void Minus(int &len)
{
	int i,j,cur=len;
	s[cur]--;
	while (s[cur]<'0')
	{
		s[cur]+=10;
		s[--cur]--;
	}
	for (i=1;s[i]=='0' && i<=len;i++) {}
	i--;
	for (j=i+1;j<=len;j++) s[j-i]=s[j];
	len-=i;
}

LL solve(int type)
{
	int i,j,to,cur,len;
	if (type)
	{
		for (i=1;i<=lenb;i++) s[i]=B[i];
		len=lenb;
	}
	else
	{
		for (i=1;i<=lena;i++) s[i]=A[i];
		len=lena;
		Minus(len);
	}
	if (len==0) return 0;
	LL res=0;
	for (i=1;i<=len-1;i++) res=(res+tans[i])%MOD;
	memset(DP,0,sizeof(DP));
	DP[0][1][1]=1;
	for (i=0;i<=len-1;i++)
		for (j=1;j<=tot;j++)
		{
			if (DP[i][j][0])
			{
				for (cur=0;cur<=9;cur++)
				{
					if (cur==0 && i==0) continue;
					to=To[j][cur];
					if (to!=-1)
						DP[i+1][to][0]=(DP[i+1][to][0]+DP[i][j][0])%MOD;
				}
			}
			if (DP[i][j][1])
			{
				for (cur=0;cur<=s[i+1]-'0'-1;cur++)
				{
					if (cur==0 && i==0) continue;
					to=check(j,cur);
					if (to!=-1)
						DP[i+1][to][0]=(DP[i+1][to][0]+DP[i][j][1])%MOD;
				}
				to=check(j,s[i+1]-'0');
				if (to!=-1)
					DP[i+1][to][1]=(DP[i+1][to][1]+DP[i][j][1])%MOD;
			}
		}
	for (i=1;i<=tot;i++)
		res=(res+DP[len][i][0]+DP[len][i][1])%MOD;
	return res;
}

int main ()
{
	int ca,i,len;
	ca=getint();
	for (i=0;i<=9;i++) TEN_TO_TWO(i);
	while (ca--)
	{
		n=getint();
		Clear();
		for (i=1;i<=n;i++)
		{
			scanf("%s",s+1);
			len=strlen(s+1);
			//cout<<len<<endl;
			Insert(len);
		}
		construct_fail();
		dfs(1);
		init_To();
		scanf("%s%s",A+1,B+1);
		lena=strlen(A+1);lenb=strlen(B+1);
		init_dp();
		cout<<((solve(1)-solve(0))%MOD+MOD)%MOD<<endl;
	}
	return 0;
}