Description

题库链接

给你 \(n\) 个单词， \(m\) 组询问，每组询问形同 \((x,y)\) ，询问 \(x\) 串在 \(y\) 串中出现多少次。

\(1\leq n,m\leq10^5\)

Solution

比较暴力的做法就是建好 \(AC\) 自动机后，对于每个 \(y\) 串暴力跑一遍。看看查询的时候有多少次落在了 \(x\) 串的末尾。

我们可以构建 \(fail\) 树，那么其实题目可以转变为对于 \(x\) 串末尾节点，其子树中有多少个节点位于 \(y\) 串上。

由于题目的特殊性，我们可以离线询问按照 \(y\) 来排序。并且预处理出 \(AC\) 自动机的 \(dfn\) 。

我们按照构建 \(Trie\) 树的操作再按原字符串走一遍。入栈时对应的 \(dfn\) 处 \(+1\) ，出时对应的 \(dfn\) 处 \(-1\) 。那么走到一个单词节点，所有打上标记的 \(dfn\) 都是该单词上的。

注意到一个子树内的 \(dfn\) 都是连续的，显然就可以回答所有 \(y\) 等于该单词的询问了。树状数组维护 \(dfn\) 的标记的前缀和即可。

Code

//It is made by Awson on 2018.3.18

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 100000;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(int x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(int x) {if (x < 0) putchar('-'); print(Abs(x)); }

char ch[N+5];

int n, m, idx, dfn[N+5], size[N+5], ans[N+5], mp[N+5];

struct tt {int to, next; }edge[(N<<1)+5];

struct qu {

    int x, y, id;

    bool operator < (const qu &b) const {return y < b.y; }

}que[N+5];

int path[N+5], top;

void add(int u, int v) {edge[++top].to = v, edge[top].next = path[u], path[u] = top; }

struct bittree {

    int c[N+5];

    void add(int o, int val) {for (; o <= idx; o += lowbit(o)) c[o] += val; }

    int count(int o) {int ans = 0; for (; o; o -= lowbit(o)) ans += c[o]; return ans; }

}BT;

struct Trie {

    int ch[N+5][26], pre[N+5], f[N+5], val[N+5], pos;

    void build(char *S) {

    int u = 0;

    for (int i = 0, len = strlen(S); i < len; i++) {

        if (S[i] == 'P') {val[u] = ++n, mp[n] = u; continue; }

        if (S[i] == 'B') {u = pre[u]; continue; }

        if (ch[u][S[i]-'a'] == 0) ++pos, pre[pos] = u, ch[u][S[i]-'a'] = pos;

        u = ch[u][S[i]-'a'];

    }

    }

    void get_fail() {

    queue<int>Q;

    for (int i = 0; i < 26; i++) if (ch[0][i]) f[ch[0][i]] = 0, Q.push(ch[0][i]);

    while (!Q.empty()) {

        int u = Q.front(); Q.pop();

        for (int i = 0; i < 26; i++) {

        if (ch[u][i]) f[ch[u][i]] = ch[f[u]][i], Q.push(ch[u][i]);

        else ch[u][i] = ch[f[u]][i];

        }

    }

    for (int i = 1; i <= pos; i++) add(f[i], i);

    }

    void query(char *S) {

    int loc = 1, u = 0;

    for (int i = 0, len = strlen(S); i < len; i++) {

        if (S[i] == 'P') {

        while (loc <= m && que[loc].y == val[u])

            ans[que[loc].id] = BT.count(dfn[mp[que[loc].x]]+size[mp[que[loc].x]]-1)-BT.count(dfn[mp[que[loc].x]]-1), ++loc;

        }else if (S[i] == 'B') BT.add(dfn[u], -1), u = pre[u];

        else u = ch[u][S[i]-'a'], BT.add(dfn[u], 1);

    }

    }

}T;

void dfs(int o) {

    size[o] = 1, dfn[o] = ++idx;

    for (int i = path[o]; i; i = edge[i].next) {

    dfs(edge[i].to); size[o] += size[edge[i].to];

    }

}

void work() {

    scanf("%s", ch); T.build(ch); T.get_fail();

    dfs(0); read(m);

    for (int i = 1; i <= m; i++) read(que[i].x), read(que[i].y), que[i].id = i;

    sort(que+1, que+1+m); T.query(ch);

    for (int i = 1; i <= m; i++) writeln(ans[i]);

}

int main() {

    work(); return 0;

}