其实很水的一道题吧....
题意是：每次给定一个串\(T\)以及\(l, r\)，询问有多少个字符串\(s\)满足，\(s\)是\(T\)的子串，但不是\(S[l .. r]\)的子串
统计\(T\)本质不同的串，建个后缀自动机
然后自然的可以想到，对于每个\(T\)的子串，它对应了一个\(right\)集合
那么，它应该会被这个\(right\)集合所限制
考虑对于每个\(i\)，求出最小的\(l\)使得\(T[l .. i]\)存在于\(S[l..r]\)中
这个可以套个线段树转移
然后就没了.....
如果不需要统计\(T\)本质不同的串，又怎么做呢？
统计的时候乘上\(right\)集合大小就行
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;#define ri register int
#define ll long long
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)#define gc getchar
inline int read() {int p = 0, w = 1; char c = gc();while(c > '9' || c < '0') { if(c == '-') w = -1; c = gc(); }while(c >= '0' && c <= '9') p = p * 10 + c - '0', c = gc();return p * w;
}const int sid = 1005000;
const int eid = 30000000 + 5;struct SAM {int id, fa[sid], mx[sid];int go[sid][26], mc[sid];inline int newnode() {++ id;fa[id] = mx[id] = mc[id] = 0;memset(go[id], 0, sizeof(go[id]));return id;}inline void init() {id = 0;newnode();}inline int extend(int lst, int c, int pos) {int np = newnode(), p = lst;mx[np] = mx[p] + 1; mc[np] = pos;for( ; p && !go[p][c]; p = fa[p]) go[p][c] = np;if(!p) fa[np] = 1;else {int q = go[p][c];if(mx[p] + 1 == mx[q]) fa[np] = q;else {int nq = newnode(); mx[nq] = mx[p] + 1;fa[nq] = fa[q]; fa[np] = fa[q] = nq;memcpy(go[nq], go[q], sizeof(go[q]));for( ; p && go[p][c] == q; p = fa[p]) go[p][c] = nq;}}return np;}} S, T;int q, n, m, seg;
char s[sid], t[sid];
int nc[sid], ip[sid], w[sid], val[sid];
int rt[sid], ls[eid], rs[eid];inline int merge(int x, int y) {if(!x || !y) return x + y;int o = ++ seg;ls[o] = merge(ls[x], ls[y]);rs[o] = merge(rs[x], rs[y]);return o;
}inline void ins(int &o, int l, int r, int p) {o = ++ seg;if(l == r) return;int mid = (l + r) >> 1;if(p <= mid) ins(ls[o], l, mid, p);else ins(rs[o], mid + 1, r, p);
}inline bool qry(int o, int l, int r, int ml, int mr) {if(ml > r || mr < l || ml > mr || !o) return 0;if(ml <= l && mr >= r) return 1;int mid = (l + r) >> 1;if(qry(ls[o], l, mid, ml, mr)) return 1;else return qry(rs[o], mid + 1, r, ml, mr);
}inline void init() {S.init();int lst = 1;rep(i, 1, n) lst = S.extend(lst, s[i] - 'a', i);int id = S.id;rep(i, 1, id) nc[S.mx[i]] ++;rep(i, 1, n) nc[i] += nc[i - 1];rep(i, 1, id) ip[nc[S.mx[i]] --] = i;rep(i, 1, id) if(S.mc[i]) ins(rt[i], 1, n, S.mc[i]);drep(i, id, 1) {int o = ip[i], f = S.fa[o];rt[f] = merge(rt[f], rt[o]);}
}void Match(int l, int r) {int o = 1, nl = 0;rep(i, 1, m) {int c = t[i] - 'a';while(1) {int nxt = S.go[o][c], f = S.fa[o];if(nxt && qry(rt[nxt], 1, n, l + nl, r)) {nl ++; o = nxt;break;}if(!nl) break; nl --;if(nl == S.mx[f]) o = f;}w[i] = nl;}
}int main() {scanf("%s", s + 1);n = strlen(s + 1);init(); q = read();rep(i, 1, q) {scanf("%s", t + 1);m = strlen(t + 1);T.init();int lst = 1;rep(j, 1, m) lst = T.extend(lst, t[j] - 'a', j);int l = read(), r = read();Match(l, r);int id = T.id;rep(i, 1, id) nc[i] = val[i] = 0;rep(i, 1, id) nc[T.mx[i]] ++;rep(i, 1, id) nc[i] += nc[i - 1];rep(i, 1, id) ip[nc[T.mx[i]] --] = i;drep(i, id, 1) {int o = ip[i], f = T.fa[o];if(T.mc[o]) val[o] = w[T.mc[o]];val[f] = max(val[f], val[o]);}ll ans = 0;rep(i, 1, id) ans += max(T.mx[i] - max(T.mx[T.fa[i]], val[i]), 0);printf("%lld\n", ans);}return 0;
}
