BCD Code ZOJ

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("******
")
#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 = 13331;
const LL INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int maxn = 1e6 + 7;
const int maxm = 8e6 + 10;
const int INF = 0x3f3f3f3f;
const int mod = 1000000009;
int T, n;
char buf[maxn], num1[maxn], num2[maxn];
int bcd[2020][12], bit[maxn];
LL dp[205][2020];

struct Aho_Corasick {
    int next[2010][2], fail[2010], End[2010];
    int root, cnt;

    int newnode() {
        for (int i = 0; i < 2; i++) next[cnt][i] = -1;
        End[cnt++] = 0;
        return cnt - 1;
    }

    void init() {
        cnt = 0;
        root = newnode();
    }

    void insert(char buf[]) {
        int len = strlen(buf);
        int now = root;
        for (int i = 0; i < len; i++) {
            if (next[now][buf[i] - '0'] == -1) next[now][buf[i] - '0'] = newnode();
            now = next[now][buf[i] - '0'];
        }
        End[now] = 1;
    }

    void build() {
        queue<int> Q;
        fail[root] = root;
        for (int i = 0; i < 2; i++)
            if (next[root][i] == -1) 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] = 1;
            for (int i = 0; i < 2; i++)
                if (next[now][i] == -1) 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 -1;
        int now = cur;
        for (int i = 3; i >= 0; --i) {
            if (End[next[now][1 & (num >> i)]]) return -1;
            now = next[now][1 & (num >> i)];
        }
        return now;
    }

    void get_bcd() {
        for (int i = 0; i < cnt; ++i)
            for (int j = 0; j < 10; ++j)
                bcd[i][j] = get_num(i, j);
    }

    LL dfs(int pos, int idx, int flag, int limit) {
        if (pos == -1) return 1;
        if (!limit && dp[pos][idx] != -1) return dp[pos][idx];
        int num = limit ? bit[pos] : 9;
        LL ans = 0;
        for (int i = 0; i <= num; ++i) {
            if (flag && i == 0) ans = (ans + dfs(pos - 1, idx, 1, limit && i == num)) % mod;
            else if (bcd[idx][i] != -1) ans = (ans + dfs(pos - 1, bcd[idx][i], 0, 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 = len1-1; i>=0; --i) {
            if (num1[i] > '0') {
                num1[i]--;
                break;
            } else num1[i] = '9';
        }
        for (int i = 0; i < len1; ++i) bit[i] = num1[len1 - 1 - i] - '0';
        LL ans1 = dfs(len1 - 1, 0, 1, 1);
        for (int i = 0; i < len2; ++i) bit[i] = num2[len2 - 1 - i] - '0';
        LL ans2 = dfs(len2 - 1, 0, 1, 1);
        return (ans2 - ans1 + mod) % mod;
    }

    void debug() {
        for (int i = 0; i < cnt; i++) {
            printf("id = %3d,fail = %3d,end = %3d,chi = [", i, fail[i], End[i]);
            for (int j = 0; j < 26; j++) printf("%2d", next[i][j]);
            printf("]
");
        }
    }
} ac;

int main() {
   // FIN;
    sfi(T);
    while (T--) {
        sfi(n);
        ac.init();
        for (int i = 0; i < n; ++i) {
            sfs(buf);
            ac.insert(buf);
        }
        ac.build();
        sffs(num1, num2);
        mem(dp, -1);
        printf("%lld
", ac.solve());
    }
    return 0;
}