ZOJ 3494 BCD Code 數位DP+AC自動機

其實還是比較裸的數位DP，只不過需要用AC自動機來預先處理一下，寫了3個小時，各種調試，蛋碎了一地啊。不過幸好最後還是做出來了。

思路：先用AC自動機預處理字符串，數位DP時以dp[index][pos][fg]來表示狀態，分別爲層數，判斷到AC自動機的位置，以及是否前面出現過非0位。

代碼略搓：

#include <cstdio>
#include <cstring>
#include <string>
#include <iostream>
#include <map>
#include <set>
#include <vector>
#include <cmath>
#include <stack>
#include <queue>
#include <cstdlib>
#include <algorithm>
using namespace std;
//typedef __int64 int64;
typedef long long ll;
#define M 500005
#define N 1000005
#define max_inf 0x7f7f7f7f
#define min_inf 0x80808080
#define mod 1000000009

struct node//樹節點
{
	node *fail , *next[2];
	int flag , pos;
	node()
	{
		fail = NULL;
		memset(next , NULL , sizeof next);
		flag = 0;
	}
}*q[2005] , *rt;
int n , dig[1005] , dp[805][2005][2] , cnt , table[1005];
char A[205] , B[205];

void Insert(char *str)//構造樹
{
	int i = 0;
	node *p = rt;
	while (str[i])
	{
		int index = str[i]-'0';
		if (!p->next[index])
		{
			p->next[index] = new node;
			p->next[index]->pos = cnt++;//記錄每個點的位置
		}
		p = p->next[index];
		i++;
	}
	p->flag = 1;//標記爲一個字符串的末尾
}

void Build()//廣搜建立失敗指針
{
	int i , head , tail;
	rt->fail = NULL;
	head = tail = 0;
	q[tail++] = rt;
	while (head < tail)
	{
		node *p = q[head++];
		for (i = 0 ; i < 2 ; i++)
		{
			if (!p->next[i])continue;
			if (p == rt)p->next[i]->fail = rt;
			else
			{
				node *temp = p->fail;
				while (temp)
				{
					if (temp->next[i])
					{
						p->next[i]->fail = temp->next[i];
						break;
					}
					temp = temp->fail;
				}
				if (!temp)p->next[i]->fail = rt;
			}
			q[tail++] = p->next[i];
		}
	}
}

int Dfs(int index , node *p , int pos, int fg , int lim)
{
	if (!index)return fg;
	if (!lim && dp[index][pos][fg] != -1)return dp[index][pos][fg];
	int i , j , up = lim ? dig[index] : 9;
	int ret = 0;
	for (i = 0 ; i <= up ; i++)
	{ 
		if (i == 0 && !fg)
		{
			ret += Dfs(index-1 , p , pos , fg||i!=0 , lim&&i==up);
			ret %= mod;
			continue;
		}
		
		//判斷這個數字是否能與禁止字符匹配
		int t = i , bit[4];
		for (j = 3 ; j >= 0 ; j--)
		{
			bit[j] = t&1;
			t >>= 1;
		}
		int f = 0;
		node *temp = p;
		for (j = 0 ; j <= 3 ; j++)
		{
			while (temp != rt && !temp->next[bit[j]])temp = temp->fail;
			temp = temp->next[bit[j]];
			if (!temp)temp = rt;
			node *q = temp;
			while (q != rt)
			{
				if (q->flag)f = 1;
				q = q->fail;
			}
		}

		if (!f)//若不匹配則繼續往下搜
		{
			ret += Dfs(index-1 , temp , temp->pos , fg||i!=0 , lim&&i==up);
			ret %= mod;
		}
	}
	if (!lim)dp[index][pos][fg] = ret;
	return ret;
}

int Solve(char *num)
{
	int len = 0;
	int i;
	for (i = strlen(num)-1 ; i >= 0 ; i--)dig[++len] = num[i]-'0';
	return Dfs(len , rt , rt->pos , 0 , 1);
}

int Judge()//單獨判斷數字A是否滿足條件
{
	int i = 0 , j;
	node *p = rt;
	while (A[i])
	{
		int t = A[i]-'0' , bit[4];
		for (j = 3 ; j >= 0 ; j--)
		{
			bit[j] = t&1;
			t >>= 1;
		}
		for (j = 0 ; j <= 3 ; j++)
		{
			while (p != rt && !p->next[bit[j]])p = p->fail;
			p = p->next[bit[j]];
			if (!p)p = rt;
			node *q = p;
			while (q != rt)
			{
				if (q->flag)return 0;
				q = q->fail;
			}
		}
		i++;
	}
	return 1;
}

int main()
{
	int t;
	scanf("%d",&t);
	while (t--)
	{
		cnt = 0;
		rt = new node;
		rt->pos = cnt++;
		memset(dp , -1 , sizeof dp);
		scanf("%d",&n);
		while (n--)
		{
			char str[25];
			scanf("%s",str);
			Insert(str);
		}
		Build();
		scanf("%s%s",A,B);
		printf("%d\n",(Solve(B)-Solve(A)+Judge()+mod)%mod);
	}
	return 0;
}