BCD Code ZOJ - 3494 AC自动机+数位DP

题意:

问A到B之间的所有整数，转换成BCD Code后，

有多少个不包含属于给定病毒串集合的子串,A,B <=10^200,病毒串总长度<= 2000.

BCD码这个在数字电路课上讲了，题干也讲的很详细。

数位DP的实现是通过0~9 ,并不是通过BCD码

所有我们需要先把字串放入AC自动机，建立一个BCD数组

因为BCD码是一个4位二进制数，但是tire图上全是0，1，

所以对于一个数字，我们的要在转移4次，

如果中间出现了病毒串就return -1 表示不能转移，

BCD【i】【j】表示在AC自动机 i 这个节点转移到数字 j 对应的在AC自动机上的节点标号。

然后就是简单的数位DP了，然而我写搓了，由于没有前导0所以前导0要处理掉。

但是你不转移的时候，不能 bcd[idx][i] != -1 就直接continue ，

因为有0的情况，i==0 但是(bcd[idx][i] == -1) 但是这个0是前导0所以不影响。

 #include <set>
 #include <map>
 #include <stack>
 #include <queue>
 #include <cmath>
 #include <ctime>
 #include <cstdio>
 #include <string>
 #include <vector>
 #include <cstring>
 #include <iostream>
 #include <algorithm>

 #define  pi    acos(-1.0)
 #define  eps   1e-9
 #define  fi    first
 #define  se    second
 #define  rtl   rt<<1
 #define  rtr   rt<<1|1
 #define  bug                printf("******\n")
 #define  mem(a, b)          memset(a,b,sizeof(a))
 #define  name2str(x)        #x
 #define  fuck(x)            cout<<#x" = "<<x<<endl
 #define  sfi(a)             scanf("%d", &a)
 #define  sffi(a, b)         scanf("%d %d", &a, &b)
 #define  sfffi(a, b, c)     scanf("%d %d %d", &a, &b, &c)
 #define  sffffi(a, b, c, d) scanf("%d %d %d %d", &a, &b, &c, &d)
 #define  sfL(a)             scanf("%lld", &a)
 #define  sffL(a, b)         scanf("%lld %lld", &a, &b)
 #define  sfffL(a, b, c)     scanf("%lld %lld %lld", &a, &b, &c)
 #define  sffffL(a, b, c, d) scanf("%lld %lld %lld %lld", &a, &b, &c, &d)
 #define  sfs(a)             scanf("%s", a)
 #define  sffs(a, b)         scanf("%s %s", a, b)
 #define  sfffs(a, b, c)     scanf("%s %s %s", a, b, c)
 #define  sffffs(a, b, c, d) scanf("%s %s %s %s", a, b,c, d)
 #define  FIN                freopen("../in.txt","r",stdin)
 #define  gcd(a, b)          __gcd(a,b)
 #define  lowbit(x)          x&-x
 #define  IO                 iOS::sync_with_stdio(false)

 using namespace std;
 typedef long long LL;
 typedef unsigned long long ULL;
 const ULL seed = ;
 const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
 const int maxn = 1e6 + ;
 const int maxm = 8e6 + ;
 const int INF = 0x3f3f3f3f;
 const int mod = ;
 int T, n;
 char buf[maxn], num1[maxn], num2[maxn];
 int bcd[][], bit[maxn];
 LL dp[][];

 struct Aho_Corasick {
     int next[][], fail[], End[];
     int root, cnt;

     int newnode() {
         for (int i = ; i < ; i++) next[cnt][i] = -;
         End[cnt++] = ;
         return cnt - ;
     }

     void init() {
         cnt = ;
         root = newnode();
     }

     void insert(char buf[]) {
         int len = strlen(buf);
         int now = root;
         for (int i = ; i < len; i++) {
             if (next[now][buf[i] - ''] == -) next[now][buf[i] - ''] = newnode();
             now = next[now][buf[i] - ''];
         }
         End[now] = ;
     }

     void build() {
         queue<int> Q;
         fail[root] = root;
         for (int i = ; i < ; i++)
             if (next[root][i] == -) next[root][i] = root;
             else {
                 fail[next[root][i]] = root;
                 Q.push(next[root][i]);
             }
         while (!Q.empty()) {
             int now = Q.front();
             Q.pop();
             if (End[fail[now]]) End[now] = ;
             for (int i = ; i < ; i++)
                 if (next[now][i] == -) next[now][i] = next[fail[now]][i];
                 else {
                     fail[next[now][i]] = next[fail[now]][i];
                     Q.push(next[now][i]);
                 }
         }
     }

     int get_num(int cur, int num) {
         if (End[cur]) return -;
         int now = cur;
         for (int i = ; i >= ; --i) {
             if (End[next[now][ & (num >> i)]]) return -;
             now = next[now][ & (num >> i)];
         }
         return now;
     }

     void get_bcd() {
         for (int i = ; i < cnt; ++i)
             for (int j = ; j < ; ++j)
                 bcd[i][j] = get_num(i, j);
     }

     LL dfs(int pos, int idx, int flag, int limit) {
         if (pos == -) return ;
         if (!limit && dp[pos][idx] != -) return dp[pos][idx];
         int num = limit ? bit[pos] : ;
         LL ans = ;
         for (int i = ; i <= num; ++i) {
             if (flag && i == ) ans = (ans + dfs(pos - , idx, , limit && i == num)) % mod;
             else if (bcd[idx][i] != -) ans = (ans + dfs(pos - , bcd[idx][i], , limit && i == num)) % mod;
         }
         if (!limit && !flag) dp[pos][idx] = ans;
         return ans;
     }

     LL solve() {
         get_bcd();
         int len1 = strlen(num1), len2 = strlen(num2);
         for (int i = len1-; i>=; --i) {
             if (num1[i] > '') {
                 num1[i]--;
                 break;
             } else num1[i] = '';
         }
         for (int i = ; i < len1; ++i) bit[i] = num1[len1 -  - i] - '';
         LL ans1 = dfs(len1 - , , , );
         for (int i = ; i < len2; ++i) bit[i] = num2[len2 -  - i] - '';
         LL ans2 = dfs(len2 - , , , );
         return (ans2 - ans1 + mod) % mod;
     }

     void debug() {
         for (int i = ; i < cnt; i++) {
             printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
             for (int j = ; j < ; j++) printf("%2d", next[i][j]);
             printf("]\n");
         }
     }
 } ac;

 int main() {
    // FIN;
     sfi(T);
     while (T--) {
         sfi(n);
         ac.init();
         for (int i = ; i < n; ++i) {
             sfs(buf);
             ac.insert(buf);
         }
         ac.build();
         sffs(num1, num2);
         mem(dp, -);
         printf("%lld\n", ac.solve());
     }
     return ;
 }