126 lines
3.6 KiB
C++
126 lines
3.6 KiB
C++
#include <algorithm>
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#include <iostream>
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#include <cstring>
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#include <cstdio>
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#include <cmath>
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#define QWQ cout << "QwQ" << endl;
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#define ll long long
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#include <vector>
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#include <queue>
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#include <stack>
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#include <map>
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#define ls now << 1
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#define rs now << 1 | 1
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using namespace std;
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const int N = 401010;
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const int qwq = 4003030;
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const int inf = 0x3f3f3f3f;
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int n, m;
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int cnt;
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char s[N];
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struct T {
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int to[28], faith, from;
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} f[N]; // trie树
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vector<int> d[N]; // fail指针树
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vector<int> at[N]; // tire树上点是哪些串的结束。
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vector<int> wen[N]; //这个点（即y串）有哪些问题。。.
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vector<int> e[N]; //原来的树，因为我BFS时将原来的trie树改成了trie图，所以我需要记录最开始时的树。
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queue<int> q;
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int wei[N]; //此字符串结束位置。
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int askman[N], ans[N]; //此问题：谁发起问题（即x串），问题答案；
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int tree[qwq], id[N], siz[N];
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void BFS() {
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for (int i = 1; i <= 26; i++)
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if (f[0].to[i]) q.push(f[0].to[i]);
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while (!q.empty()) {
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int u = q.front();
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q.pop();
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for (int i = 1; i <= 26; i++) {
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int v = f[u].to[i];
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if (v) {
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f[v].faith = f[f[u].faith].to[i];
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q.push(v);
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} else
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f[u].to[i] = f[f[u].faith].to[i];
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}
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}
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for (int i = 1; i <= cnt; i++) d[f[i].faith].push_back(i); //建fail树。
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}
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void DFS1(int u) { //遍历fail树，算出每个点子树大小，并给每个点一个编号。
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siz[u] = 1;
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id[u] = ++cnt;
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for (int i = 0; i < d[u].size(); i++) {
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int v = d[u][i];
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DFS1(v);
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siz[u] += siz[v];
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}
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}
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void insert(int now, int l, int r, int we, int g) { //单点修改
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if (l == r) {
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tree[now] += g;
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return;
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}
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int mid = (l + r) >> 1;
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if (we <= mid)
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insert(ls, l, mid, we, g);
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else
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insert(rs, mid + 1, r, we, g);
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tree[now] = tree[ls] + tree[rs];
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}
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int query(int now, int l, int r, int x, int y) { //区间查询
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if (x <= l && r <= y) { return tree[now]; }
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int mid = l + r >> 1, res = 0;
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if (x <= mid) res += query(ls, l, mid, x, y);
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if (y > mid) res += query(rs, mid + 1, r, x, y);
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return res;
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}
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void TREE(int now) { //遍历原树
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insert(1, 1, cnt, id[now], 1); //每到一个点就加上这个点的贡献
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for (int i = 0; i < at[now].size(); i++) {
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int u = at[now][i];
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for (int j = 0; j < wen[u].size(); j++) {
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int qu = wen[u][j], v = wei[askman[qu]]; // qu为问题编号，v为串x的结尾点
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ans[qu] = query(1, 1, cnt, id[v], id[v] + siz[v] - 1);
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}
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}
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for (int i = 0; i < e[now].size(); i++) TREE(e[now][i]);
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insert(1, 1, cnt, id[now], -1);
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}
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int main() {
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int x, y;
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scanf("%s", s);
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int len = strlen(s), now = 0;
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for (int i = 0; i < len; i++) {
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int v = s[i] - 'a' + 1;
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if (s[i] == 'P') {
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at[now].push_back(++n);
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wei[n] = now;
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continue;
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}
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if (s[i] == 'B') {
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now = f[now].from;
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continue;
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}
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if (!f[now].to[v]) f[now].to[v] = ++cnt, e[now].push_back(f[now].to[v]);
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f[f[now].to[v]].from = now;
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now = f[now].to[v]; //涉及到退格，需要记录from。
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}
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BFS();
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cnt = 0;
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DFS1(0);
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scanf("%d", &m);
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for (int i = 1; i <= m; i++) {
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scanf("%d%d", &askman[i], &x);
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wen[x].push_back(i);
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}
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TREE(0);
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for (int i = 1; i <= m; i++)
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printf("%d\n", ans[i]);
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return 0;
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} |