建立完后缀自动机过后用f[i]表示节点i以及i的儿子缩代表的串的个数
排序过后由深的点往浅的点更新
询问的时候进去跑一跑就好
SPOJ时间限制真是紧
#include<cstdio>
#include<algorithm>
#include<cstring>
#include<iostream>
#include<queue>
#define SF scanf
#define PF printf
#define idx(c) (c-'a')
using namespace std;
typedef long long LL;
const int MAXN = 180000;
int n;
char s[MAXN+10];
struct SAM {
int fa[MAXN+10], ch[MAXN+10][26], step[MAXN+10];
int f[MAXN+10], last, ncnt, v[MAXN+10], sa[MAXN+10];
void extend(int c) {
int p = last, np = last = ++ncnt;
n = step[np] = step[p]+1;
while(!ch[p][c] && p) ch[p][c] = np, p = fa[p];
if(!p) fa[np] = 1;
else {
int q = ch[p][c], nq;
if(step[q] == step[p]+1) fa[np] = q;
else {
nq = ++ncnt;
step[nq] = step[p]+1;
memcpy(ch[nq], ch[q], sizeof(ch[nq]));
fa[nq] = fa[q]; fa[np] = fa[q] = nq;
while(ch[p][c] == q && p) ch[p][c] = nq, p = fa[p];
}
}
}
inline void build() {
SF("%s", s);
for(char *c = s; *c; c++) extend(*c-'a');
}
inline void solve() {
for(int i = 1; i <= ncnt; i++) v[step[i]]++;
for(int i = 1; i <= n; i++) v[i] += v[i-1];
for(int i = 1; i <= ncnt; i++) sa[v[step[i]]--] = i;
for(int i = ncnt; i >= 1; i--) {
f[sa[i]] = 1;
for(int j = 0; j < 26; j++)
f[sa[i]] += f[ch[sa[i]][j]];
}
}
inline void query() {
int k, u = 1;
SF("%d", &k);
while(k) {
for(int c = 0; c < 26; c++)
if(ch[u][c]) {
if(f[ch[u][c]] < k) k -= f[ch[u][c]];
else {
k--;
putchar('a'+c);
u = ch[u][c];
break;
}
}
}
puts("");
}
} ac;
int main() {
ac.last = ac.ncnt = 1;
ac.build();
ac.solve();
int Q; SF("%d", &Q);
for(int i = 1; i <= Q; i++)
ac.query();
}