\(>Codeforces \space 1037\ H. Security<\)
题目大意 : 有一个串 \(S\) ，\(q\) 组询问，每一次给出一个询问串 \(T\) 和一个区间 \([l,r]\) ，要求找出 \(S\) 在 \([l,r]\) 之间的子串中字典序大于 \(T\) 且最小的
\(1 \leq |S|\leq 10^5, 1\leq q \leq 2 \times 10^5\)
解题思路 :
其实这个题一点意思都没有，就是一个 \(sam\) + 线段树合并裸题..
但是某位不得了的指导大人近期 \(AC\) 了此题，于是我就去顺手做了一下
考虑每次把询问串在 \(sam\) 上匹配，对于每一个在 \(sam\) 上出现的 \(T\) 的合法前缀，大力算出能否在后面加一个字符满足比 \(T\) 大且在 \(L, R\) 区间
后者线段树合并维护即可
/*program by mangoyang*/
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
template <class T>
inline void read(T &x){
int ch = 0, f = 0; x = 0;
for(; !isdigit(ch); ch = getchar()) if(ch == '-') f = 1;
for(; isdigit(ch); ch = getchar()) x = x * 10 + ch - 48;
if(f) x = -x;
}
const int N = 1000005;
char s[N]; int n;
struct SegmentTree{
int sz[N*25], lc[N*25], rc[N*25], size;
inline void ins(int &u, int l, int r, int pos){
if(!u) u = ++size;
if(l == r) return (void) (sz[u]++);
int mid = l + r >> 1;
if(pos <= mid) ins(lc[u], l, mid, pos);
else ins(rc[u], mid + 1, r, pos); sz[u] = sz[lc[u]] + sz[rc[u]];
}
inline int merge(int x, int y, int l, int r){
if(!x || !y) return x + y;
int o = ++size, mid = l + r >> 1;
if(l == r) sz[o] = sz[x] + sz[y];
else{
lc[o] = merge(lc[x], lc[y], l, mid);
rc[o] = merge(rc[x], rc[y], mid + 1, r);
sz[o] = sz[lc[o]] + sz[rc[o]];
}
return o;
}
inline int query(int u, int l, int r, int L, int R){
if(!u || l > r) return 0;
if(l >= L && r <= R) return sz[u];
int mid = l + r >> 1, res = 0;
if(L <= mid) res += query(lc[u], l, mid, L, R);
if(mid < R) res += query(rc[u], mid + 1, r, L, R);
return res;
}
}Seg;
struct SaffixAutomaton{
vector<int> g[N];
int ch[N][26], dep[N], rt[N], fa[N], size, tail;
inline SaffixAutomaton(){ size = tail = 1; }
inline int newnode(int x){ return dep[++size] = x, size; }
inline void ins(int c, int pos){
int p = tail, np = newnode(dep[p] + 1);
Seg.ins(rt[np], 1, n, pos);
for(; !ch[p][c] && p; p = fa[p]) ch[p][c] = np;
if(!p) return (void) (fa[np] = 1, tail = np);
int q = ch[p][c];
if(dep[q] == dep[p] + 1) fa[np] = q;
else{
int nq = newnode(dep[p] + 1);
fa[nq] = fa[q], fa[q] = fa[np] = nq;
for(int i = 0; i < 26; i++) ch[nq][i] = ch[q][i];
for(; ch[p][c] == q && p; p = fa[p]) ch[p][c] = nq;
}tail = np;
}
inline void dfs(int u){
for(int i = 0; i < g[u].size(); i++){
int v = g[u][i];
dfs(v), rt[u] = Seg.merge(rt[u], rt[v], 1, n);
}
}
inline void Prework(){
for(int i = 1; i <= size; i++) g[fa[i]].push_back(i); dfs(1);
}
inline void solve(char *s, int l, int r){
int len = strlen(s + 1), p = 1, pos = 0, res = 0;
for(int i = 1; i <= len + 1; i++){
int c = i > len ? (-1) : (s[i] - 'a'), flg = 0;
for(int j = c + 1; j < 26; j++) if(ch[p][j]){
int u = ch[p][j];
if(Seg.query(rt[u], 1, n, l + i - 1, r)){
res = j + 'a', pos = i - 1; break;
}
}
if(ch[p][c]) p = ch[p][c]; else break;
}
if(!res) return (void) puts("-1");
for(int i = 1; i <= pos; i++) putchar(s[i]);
putchar(res), putchar('\n');
}
}van;
int main(){
scanf("%s", s + 1); n = strlen(s + 1);
for(int i = 1; i <= n; i++) van.ins(s[i] - 'a', i);
van.Prework(); int Q; read(Q);
for(int i = 1, l, r; i <= Q; i++)
read(l), read(r), scanf("%s", s + 1), van.solve(s, l, r);
return 0;
}