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 ;

 }

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BCD Code ZOJ - 3494 AC自动机+数位DP

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