【MZ】ZOJ 3494 BCD Code AC自动机+数位DP

problem：

给你 n 个由 01 串组成的病毒。(0<=n<=100)

一个数的 BCD 编码为每个数字变成四位的二进制(如127 would be:0001 0010 0111)。

问从x到y有多少个数其BCD编码中没有病毒。（0<x<=y<10^(200)）

think:

CODE：

const int kind = 2;
const LL mod = 1000000009;

LL f[222][2222];//用于数位DP
int bit[222];
char ch[55], str[222];

int fail[2222];//每个状态的fail指针
int child[2222][2];//01两种
int end[2222];//每个状态是否已经包含一整个病毒
int dp[2222][12];//i状态后面加数字j变成dp[i][j]状态
int cnt;//用于记录状态

int newNode(){
    ++cnt;
    child[cnt][0] = child[cnt][1] = -1;
    end[cnt] = 0;
    return cnt;
}

void insert(char *str, int root){
    int p = root;
    for(int i=0; str[i]; ++i){
        int k = str[i] - '0';
        if(child[p][k] == -1) child[p][k] = newNode();
        p = child[p][k];
    }
    end[p] = 1;
}

//这个建立fail指针不同于hdu2222
//这里的fail不能有空指针，必须要指向一个状态
queue<int>q;
void build_fail(int root){
    while(!q.empty()) q.pop();
    fail[root] = root;
    q.push(root);
    while(!q.empty()){
        int tmp = q.front();
        q.pop();
        if(end[fail[tmp]]==1) end[tmp] = 1;//tmp也已经包括病毒
        for(int k = 0; k < kind; ++k){
            if(child[tmp][k]!=-1){
                if(tmp==root) fail[child[tmp][k]] = root;
                else fail[child[tmp][k]] = child[fail[tmp]][k];
                q.push(child[tmp][k]);
            } else {
                if(tmp==root) child[root][k] = root;
                else child[tmp][k] = child[fail[tmp]][k];
            }
        }
    }
}

int change(int pre, int num){
    if(end[pre]) return -1;
    int cur = pre;
    for(int i = 3; i >= 0; --i){
        cur = child[cur][(num>>i)&1];
        if(end[cur]) return -1;
    }
    return cur;
}

void init_dp(){
    for(int i = 0; i <= cnt; ++i){
        for(int j = 0; j < 10; ++j){
            dp[i][j] = change(i, j);
        }
    }
}

LL dfs(int pos, int sta, bool e, bool z){
    if(pos == 0) return 1LL;
    if(!e && ~f[pos][sta]) return f[pos][sta];
    LL res = 0;
    if(z) res += dfs(pos-1, sta, e&&bit[pos]==0, true);
    else if(dp[sta][0]!=-1) res += dfs(pos-1, dp[sta][0], e&&bit[pos]==0, false);
    res %= mod;
    int u = e ? bit[pos] : 9;
    for(int d = 1; d <= u; ++d){
        if(dp[sta][d]==-1) continue;
        res += dfs(pos-1, dp[sta][d], e && d==u, false);
        res %= mod;
    }
    if(!e && !z) f[pos][sta] = res;
    return res;
}

LL solve(char *str){
    int len = strlen(str);
    for(int i = 0; i < len; ++i){
        bit[len-i] = str[i] - '0';
    }
    return dfs(len, 0, 1, 1);
}

int main(){
    int t;
    scanf("%d", &t);
    while(t--){
        cnt = -1;
        int root = newNode();
        int n;
        scanf("%d", &n);
        for(int i = 0; i < n; ++i){
            scanf("%s", ch);
            insert(ch, root);
        }
        build_fail(root);
        init_dp();
        memset(f, -1, sizeof(f));
        scanf("%s", str);
        int len = strlen(str);
        for(int i = len-1; i>=0; --i){
            if(str[i]>'0'){
                --str[i];
                break;
            }
            else str[i] = '9';
        }
        LL res = -solve(str);
        scanf("%s", str);
        res += solve(str);
        res = (res + mod) % mod;
        printf("%lld\n", res);
    }
    return 0;
}