bzoj 2434 阿狸的打字机

题目传送门

　　传送站I

　　传送站II

题目大意

　　阿狸有一个打字机，它有3种键：

1. 向缓冲区追加小写字母
2. P：打印当前缓冲区（缓冲区不变）
3. B：删除缓冲区中最后一个字符

　　然后多次询问第$x$个被打印出来的串在第$y$个被打印出来的串中出现多少次。

　　每次查询相当于询问串$x$的结束节点在fail树中的子树包含多少串$y$的点。

　　又因为这些串的构造比较另类。所以考虑把询问离线，然后用树状数组维护子树中关键点的个数。

Code

/**
 * bzoj
 * Problem#2434
 * Accepted
 * Time: 620ms
 * Memory: 17792k
 */
#include <algorithm>
#include <iostream>
#include <cstdlib>
#include <cstdio>
#include <vector>
#include <queue>
#include <stack>
using namespace std;
typedef bool boolean;

const int N = 1e5 + 5;

typedef class IndexedTree {
    public:
        int s;
        int ar[N];

        void add(int idx, int val) {
            for ( ; idx <= s; idx += (idx & (-idx)))
                ar[idx] += val;
        }

        int getSum(int idx) {
            int rt = 0;
            for ( ; idx; idx -= (idx & (-idx)))
                rt += ar[idx];
            return rt;
        }
}IndexedTree;

typedef class TrieNode {
    public:
        int in, out;
        TrieNode* ch[26];
        TrieNode* fail;
}TrieNode;

TrieNode pool[N];
TrieNode *top = pool;

TrieNode* newnode() {
    return ++top;
}

typedef class AhoCorasick {
    public:
        TrieNode *rt;

        AhoCorasick():rt(newnode()) {    }

        void travel(char* str, TrieNode** ts) {
            stack<TrieNode*> s;
            TrieNode* p;
            s.push(rt);
            for (int i = 0, c, cp = 0; str[i]; i++) {
                if (str[i] == 'B')
                    s.pop();
                else if (str[i] == 'P')
                    ts[++cp] = s.top();
                else {
                    p = s.top(), c = str[i] - 'a';
                    if (!p->ch[c])
                        p->ch[c] = newnode();
                    s.push(p->ch[c]);
                }
            }
        }

        void build(vector<int> *g) {
            queue<TrieNode*> que;
            que.push(rt);
            while (!que.empty()) {
                TrieNode* p = que.front();
                que.pop();
                for (int i = 0; i < 26; i++) {
                    TrieNode *np = p->ch[i], *f = p->fail;
                    if (!np)    continue;
                    while (f && !f->ch[i])    f = f->fail;
                    if (!f)
                        np->fail = rt;
                    else
                        np->fail = f->ch[i];
                    que.push(np);
                }
            }
            for (TrieNode* p = top; p != pool + 1; p--)
                g[p->fail - pool].push_back(p - pool);
        }
}AhoCorasick;

typedef class Query {
    public:
        int x, y, id;

        boolean operator < (Query b) const {
            return y < b.y;
        }
}Query;

int m, cnt = 0;
char str[N];
int *res;
Query *qs;
TrieNode** ts;
vector<int> g[N];
IndexedTree it;
AhoCorasick ac;

inline void init() {
    gets(str);
    scanf("%d", &m);
    qs = new Query[(m + 1)];
    res = new int[(m + 1)];
    ts = new TrieNode*[(m + 1)];
    for (int i = 1; i <= m; i++) {
        scanf("%d%d", &qs[i].x, &qs[i].y);
        qs[i].id = i;
    }
}

void dfs(int p) {
    pool[p].in = ++cnt;
    for (int i = 0; i < (signed) g[p].size(); i++)
        dfs(g[p][i]);
    pool[p].out = cnt;
}

inline void solve() {
    ac.travel(str, ts);
    ac.build(g);
    it.s = (top - pool + 1);
    dfs(1);
    sort(qs + 1, qs + m + 1);
    stack<TrieNode*> s;
    s.push(ac.rt);
    for (int i = 0, cur = 0, cq = 1; cq <= m && str[i]; i++) {
        TrieNode *p = s.top();
        if (str[i] == 'B') {
            it.add(p->in, -1);
            s.pop();
        } else if (str[i] == 'P') {
            cur++;
            while (cq <= m && qs[cq].y == cur)
                res[qs[cq].id] = it.getSum(ts[qs[cq].x]->out) - it.getSum(ts[qs[cq].x]->in - 1), cq++;
        } else {
            p = p->ch[str[i] - 'a'];
            it.add(p->in, 1);
            s.push(p);
        }
    }
    for (int i = 1; i <= m; i++)
        printf("%d
", res[i]);
}

int main() {
    init();
    solve();
    return 0;
}