有一个串\(S\)，每次询问给你一个串\(T\)，两个数\(L\)和\(R\)，问你\(T\)有多少个本质不同的子串不是\(S[L,R]\)的子串
SOLUTION
如果你做过生成魔咒和CF1037H，就会做这道题了
有两个坑点：
1.线段树合并时必须每次都新建结点，因为两颗树都得保留
2.每次失配时必须先尝试减小已经匹配的长度，无法继续减少时再跳\(suflink\)
我的大常数代码
#include <algorithm> #include <iostream> #include <cstdlib> #include <cstring> #include <cstdio> #include <random> #include <string> #include <vector> #include <cmath> #include <ctime> #include <queue> #include <map> #include <set> #define IINF 0x3f3f3f3f3f3f3f3fLL #define u64 unsigned long long #define pii pair<int, int> #define mii map<int, int> #define u32 unsigned int #define lbd lower_bound #define ubd upper_bound #define INF 0x3f3f3f3f #define vi vector<int> #define ll long long #define mp make_pair #define pb push_back #define is insert #define se second #define fi first #define ps push #define $SHOW(x) cout << #x" = " << x << endl #define $DEBUG() printf("%d %s\n", __LINE__, __FUNCTION__) using namespace std; #define S 500000 #define Q 100000 #define T 1000000 #define mid ((l+r)>>1) int q, n, m, L, R; char s[S+5], t[T+5]; int nid1 = 1, lst1 = 1, ch1[26][2*S+5], link1[2*S+5], len1[2*S+5], nodecnt, root[2*S+5], ch[2][200*S+5]; int nid, lst, nxt[26][2*T+5], link[2*T+5], len[2*T+5], rest[2*T+5], val[T+5]; ll ans = 0; int a[2*T+5], buc[2*T+5]; void add(int &o, int l, int r, int x) { if (!o) o = ++nodecnt; if (l == r) return ; if (x <= mid) add(ch[0][o], l, mid, x); else add(ch[1][o], mid+1, r, x); } int merge(int x, int y, int l, int r) { int u = ++nodecnt; if (x * y == 0) u = x ? x : y; else { ch[0][u] = merge(ch[0][x], ch[0][y], l, mid); ch[1][u] = merge(ch[1][x], ch[1][y], mid + 1, r); } return u; } int query(int u, int l, int r, int L, int R) { if (!u) return 0; if (L <= l && r <= R) return 1; int ret = 0; if (L <= mid) ret |= query(ch[0][u], l, mid, L, R); if (R > mid) ret |= query(ch[1][u], mid+1, r, L, R); return ret; } void build() { for (int i = 1, c; i <= n; ++i) { c = s[i] - 'a'; int cur = ++nid1; len1[cur] = len1[lst1] + 1; add(root[cur], 1, n, i); while (lst1 && !ch1[c][lst1]) ch1[c][lst1] = cur, lst1 = link1[lst1]; if (!lst1) link1[cur] = 1; else { int p = lst1, q = ch1[c][lst1]; if (len1[q] == len1[p] + 1) link1[cur] = q; else { int clone = ++nid1; len1[clone] = len1[p] + 1; for (int j = 0; j < 26; ++j) ch1[j][clone] = ch1[j][q]; link1[clone] = link1[q], link1[cur] = link1[q] = clone; while (p && ch1[c][p] == q) ch1[c][p] = clone, p = link1[p]; } } lst1 = cur; } for (int i = 1; i <= nid1; ++i) buc[len1[i]]++; for (int i = 1; i <= n; ++i) buc[i] += buc[i-1]; for (int i = 1; i <= nid1; ++i) a[buc[len1[i]]--] = i; for (int i = nid1; i >= 2; --i) root[link1[a[i]]] = merge(root[link1[a[i]]], root[a[i]], 1, n); } void extend(int c) { int cur = ++nid; len[cur] = len[lst] + 1; for(int i = 0; i < 26; ++i) nxt[i][cur] = 0; while (lst && !nxt[c][lst]) nxt[c][lst] = cur, lst = link[lst]; if (!lst) link[cur] = 1; else { int p = lst, q = nxt[c][lst]; if (len[q] == len[p] + 1) link[cur] = q; else { int clone = ++nid; len[clone] = len[p] + 1; for (int i = 0; i < 26; ++i) nxt[i][clone] = nxt[i][q]; link[clone] = link[q], link[q] = link[cur] = clone; while (p && nxt[c][p] == q) nxt[c][p] = clone, p = link[p]; } } lst = cur; } void calc() { m = strlen(t+1); lst = nid = 1; int i, j, c; for (i = 0; i < 26; ++i) nxt[i][1] = 0; for (j = 1; j <= m; ++j) extend(t[j] - 'a'); for (i = 2; i <= nid; ++i) rest[i] = 0; int u = 1, match = 0; for (i = 1, c; i <= m; ++i) { c = t[i] - 'a'; while (u && !query(root[ch1[c][u]], 1, n, L + match, R)) { if(match > len1[link1[u]]) match--; else u = link1[u], match = len1[u]; } if (!u) u = 1, match = 0; else u = ch1[c][u], match++; val[i] = match; } u = 1; for (i = 1; i <= m; ++i) u = nxt[t[i] - 'a'][u], rest[u] = val[i]; for (i = 0; i <= m; ++i) buc[i] = 0; for (i = 1; i <= nid; ++i) buc[len[i]]++; for (i = 1; i <= m; ++i) buc[i] += buc[i-1]; for (i = 1; i <= nid; ++i) a[buc[len[i]]--] = i; for (i = nid; i >= 2; --i) rest[link[a[i]]] = max(rest[link[a[i]]], rest[a[i]]); ll ans = 0; for (i = 2; i <= nid; ++i) ans += max(0, min(len[i] - rest[i], len[i] - len[link[i]])); printf("%lld\n", ans); } int main() { scanf("%s", s+1); n = strlen(s+1); build(); scanf("%d", &q); for (int i = 1; i <= q; ++i) { scanf("%s%d%d", t+1, &L, &R); calc(); } return 0; }