Barty's Computer 字典树

https://nanti.jisuanke.com/t/17122

Barty have a computer, it can do these two things.

1. Add a new string to its memory, the length of this string is even.

2. For given 44 strings a,b,c,da,b,c,d, find out how many strings that can be product by a+s1+b+c+s2+da+s1+b+c+s2+d, and |a| + |s1| + |b| = |c| + |s2| + |d|∣a∣+∣s1∣+∣b∣=∣c∣+∣s2∣+∣d∣. |s|∣s∣ means the length of string ss, s1s1 and s2s2 can be any string, including "".

Please help your computer to do these things.

Input Format

Test cases begins with T(T \le 5)T(T≤5).

Then TT test cases follows.

Each test case begins with an integer Q(Q \le 30000)Q(Q≤30000).

Then QQ lines,

1 s: add a new string ss to its memory.

2 a b c d: find how many strings satisfying the requirement above.

\sum |s| + |a| + |b| + |c| + |d| \le 2000000∑∣s∣+∣a∣+∣b∣+∣c∣+∣d∣≤2000000.

Output Format

For type 22 query. Output the answer in one line.

样例输入

1
10
1 abcqaq
1 abcabcqaqqaq
2 ab bc qa aq
2 a c q q
1 abcabcqaqqwq
2 ab bc qa aq
2 a c q q
1 abcq
2 a c q q
2 a b c q

样例输出

1
2
1
3
3
1

题目来源

2017 ACM-ICPC 亚洲区（西安赛区）网络赛

首先对于每一个主串，都把它对半砍，前缀加入字典树0，前缀逆序加入字典树1，后缀加入字典树2，后缀逆序加入字典树3

所以每一个节点都开一个vector存着有哪些主串能经过这个节点。内存复杂度玄学

然后每次询问，用a去字典树0找，就能知道有哪些主串能匹配a

同理b、c、d，然后求一个交集即可。

这样会有些不合法

比如就是会使得

ab  bc这样，结合成abc

所以要用长度来判断是否合法，

lenstr(a) + lenstr(b)不能大于lenstr (s) / 2

#include <bits/stdc++.h>
#define IOS ios::sync_with_stdio(false)
using namespace std;
#define inf (0x3f3f3f3f)
typedef long long int LL;
const int N = , maxn = 2e6 + ;
struct Node {
    vector<int> vc;
    struct Node * pNext[N];
} tree[][ + ];
int t[];
int len[maxn];
char str[maxn];
char a[maxn], b[maxn], c[maxn], d[maxn];
struct Node * create(int id) {
    struct Node * p = &tree[id][t[id]++];
    p->vc.clear();
    for (int i = ; i < N; ++i) p->pNext[i] = NULL;
    return p;
}
void toInset(struct Node **T, char str[], int be, int en, int flag, int id, int which) {
    struct Node *p = *T;
    if (p == NULL) {
        p = *T = create(id);
    }
    if (flag == -) {
        for (int i = en; i >= be; --i) {
            int tid = str[i] - 'a';
            if (!p->pNext[tid]) p->pNext[tid] = create(id);
            p = p->pNext[tid];
            p->vc.push_back(which);
        }
    } else {
        for (int i = be; i <= en; ++i) {
            int tid = str[i] - 'a';
            if (!p->pNext[tid]) {
                p->pNext[tid] = create(id);
            }
            p = p->pNext[tid];
            p->vc.push_back(which);
        }
    }
}
int vis[][maxn], DFN;
bool flag;
void ask(struct Node *T, char str[], int be, int en, int flag, int which) {
    struct Node *p = T;
    if (p == NULL) {
        flag = true;
        return;
    }
    if (flag == -) {
        for (int i = en; i >= be; --i) {
            int id = str[i] - 'a';
            if (!p->pNext[id]) {
                flag = true;
                return ;
            }
            p = p->pNext[id];
        }
        for (int i = ; i < p->vc.size(); ++i) {
            vis[which][p->vc[i]] = DFN;
        }
    } else {
        for (int i = be; i <= en; ++i) {
            int id = str[i] - 'a';
            if (!p->pNext[id]) {
                flag = true;
                return ;
            }
            p = p->pNext[id];
        }
//        printf("fff");
//        printf("%d * * ** \n", p->vc.size());
        for (int i = ; i < p->vc.size(); ++i) {
            vis[which][p->vc[i]] = DFN;
        }
    }
}
void work() {
    t[] = t[] = t[] = t[] = ;
    struct Node *T[];
    for (int i = ; i < ; ++i) T[i] = NULL;
    int q;
    scanf("%d", &q);
    int now = ;
    while (q--) {
        int op;
        scanf("%d", &op);
        if (op == ) {
            scanf("%s", str + );
            len[++now] = strlen(str + );
            toInset(&T[], str, , len[now] / , , , now);
            toInset(&T[], str, , len[now] / , -, , now);
            toInset(&T[], str, len[now] /  + , len[now], , , now);
            toInset(&T[], str, len[now] /  + , len[now], -, , now);
        } else {
            scanf("%s%s%s%s", a + , b + , c + , d + );
            ++DFN, flag = false;
            int lena = strlen(a + ), lenb = strlen(b + ), lenc = strlen(c + ), lend = strlen(d + );
            ask(T[], a, , lena, , );
            if (flag) {
                printf("0\n");
                continue;
            }
            ask(T[], b, , lenb, -, );
            if (flag) {
                printf("0\n");
                continue;
            }
            ask(T[], c, , lenc, , );
            if (flag) {
                printf("0\n");
                continue;
            }
            ask(T[], d, , lend, -, );
            if (flag) {
                printf("0\n");
                continue;
            }
            int ans = ;
//            for (int i = 1; i <= now; ++i) {
//                printf("%d %d %d %d\n", vis[0][i], vis[1][i], vis[2][i], vis[3][i]);
//            }
//            printf("*************\n");
            for (int i = ; i <= now; ++i) {
                if (vis[][i] == DFN && vis[][i] == DFN && vis[][i] == DFN && vis[][i] == DFN && lena + lenb <= len[i] /  && lenc + lend <= len[i] / )
                    ans++;
            }
            printf("%d\n", ans);
        }
    }
}

int main() {
#ifdef local
    freopen("data.txt", "r", stdin);
//    freopen("data.txt", "w", stdout);
#endif
    int t;
    scanf("%d", &t);
    while (t--) work();
    return ;
}