题意
Sol
(10^5)次询问每次询问(10^5)个区间。。这种题第一感觉就是根号/数据分治的模型。
(K)是个定值这个很关键。
考虑(K)比较小的情况，可以直接暴力建SAM，(n^2)枚举(w)的子串算出现次数。询问用个(n^2)的vector记录一下每次在vector里二分就好。
(K)比较大的情况我没想到什么好的做法，网上的做法复杂度也不是很好。。
然后写了个广义SAM + 暴力跳parent就过了。。
不过这题思想还是很好的
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP(x, y) make_pair(x, y)
#define fi first
#define se second
#define LL long long
#define ull unsigned long long
#define Fin(x) {freopen(#x".in","r",stdin);}
#define Fout(x) {freopen(#x".out","w",stdout);}
using namespace std;
const int MAXN = 4e5 + 10, INF = 1e9 + 1, mod = 1e9 + 7;
const double eps = 1e-9, pi = acos(-1);
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline LL add(A x, B y) {if(x + y < 0) return x + y + mod; return x + y >= mod ? x + y - mod : x + y;}
template <typename A, typename B> inline void add2(A &x, B y) {if(x + y < 0) x = x + y + mod; else x = (x + y >= mod ? x + y - mod : x + y);}
template <typename A, typename B> inline LL mul(A x, B y) {return 1ll * x * y % mod;}
template <typename A, typename B> inline void mul2(A &x, B y) {x = (1ll * x * y % mod + mod) % mod;}
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M, Q, K;
char s[MAXN], w[MAXN];
string q[MAXN];
int l[MAXN], r[MAXN], a[MAXN], b[MAXN], pos[MAXN];
vector<int> line[1001][1001], v[MAXN];
int ch[MAXN][26], siz[MAXN], len[MAXN], fa[MAXN], las = 1, root = 1, tot = 1;
void insert(int x, int opt) {
int now = ++tot, pre = las; las = now; len[now] = len[pre] + 1; siz[now] = opt;
for(; pre && !ch[pre][x]; pre = fa[pre]) ch[pre][x] = now;
if(!pre) fa[now] = root;
else {
int q = ch[pre][x];
if(len[pre] + 1 == len[q]) fa[now] = q;
else {
int nq = ++tot; len[nq] = len[pre] + 1; fa[nq] = fa[q];
memcpy(ch[nq], ch[q], sizeof(ch[q]));
fa[now] = fa[q] = nq;
for(; pre && ch[pre][x] == q; pre = fa[pre]) ch[pre][x] = nq;
}
}
}
void dfs(int x) {
for(auto &to : v[x]) {
dfs(to);
siz[x] += siz[to];
}
}
int Query(vector<int> &q, int a, int b) {
return (upper_bound(q.begin(), q.end(), b) - lower_bound(q.begin(), q.end(), a));
}
LL solve1(int a, int b) {
LL ret = 0;
for(int i = 1; i <= K; i++) {
int now = root;
for(int j = i; j <= K; j++) {
int x = w[j] - 'a'; now = ch[now][x];
if(!now) break;
else ret += 1ll * siz[now] * Query(line[i][j], a, b);
}
}
return ret;
}
void Build() {
for(int i = 1; i <= tot; i++) v[fa[i]].push_back(i);
dfs(root);
}
signed main() {
// freopen("string9.in", "r", stdin); freopen("b.out", "w", stdout);
N = read(); M = read(); Q = read(); K = read();
scanf("%s", s + 1);
for(int i = 1; i <= N; i++) insert(s[i] - 'a', 1);
if(K <= 1000) {
Build();
for(int i = 1; i <= M; i++) l[i] = read() + 1, r[i] = read() + 1, line[l[i]][r[i]].push_back(i);
for(int i = 1; i <= Q; i++) {
scanf("%s", w + 1); int a = read() + 1, b = read() + 1;
cout << solve1(a, b) << '
';
}
}
else {
for(int i = 1; i <= M; i++) l[i] = read() + 1, r[i] = read() + 1;
for(int i = 1; i <= Q; i++) {
cin >> q[i]; a[i] = read() + 1, b[i] = read() + 1;
las = 1;
for(auto &x : q[i]) insert(x - 'a', 0);
}
Build();
for(int i = 1; i <= Q; i++) {
memset(pos, 0, sizeof(pos));
int now = root;
for(int j = 0; j < q[i].length(); j++) {
int x = q[i][j] - 'a'; now = ch[now][x];
if(!now) break;
else pos[j + 1] = now;
}
LL ans = 0;
for(int j = a[i]; j <= b[i]; j++) {
int cur = pos[r[j]];
while(len[fa[cur]] >= r[j] - l[j] + 1) cur = fa[cur];//这里可以卡掉
ans += siz[cur];
}
cout << ans << '
';
}
}
return 0;
}